## What is Algebra?

This past month has been very busy here for The Bearded Math Man.  I’ve learned a lot about things I have merely taken for granted and have shared most of them with you here on my site.  And while I have a goal and a mission, the methods of achieving that goal are still forming.  I’m learning what works best and what doesn’t work.  One such thing I’ve discovered is the purpose of this blog.

This blog is meant for two audiences.  Those interested in math and those teaching math.  Now that I have that defined, I’ll keep a more focused range of topics.  I just thought that was worth mentioning.

Now, for today’s topic, Algebra.  I do not intend to teach you Algebra, but would like to share something I did not know about the subject.  Algebra means to make complete, or to resolve.  I knew it was named after a Persian mathematician in the early 9th century, but that the branch of mathematics goes farther back in time than the name itself, even the Babylonians used Algebriac concepts.  But I thought the name was just that, a name.

It is stunningly powerful to recognize what Algebra means.  Everything operation we perform in Algebra is to meet this end, to complete or resolve an equation!  That’s what we do when we’re solving an Algebraic Equation.

One other thing you may not have known about Algebra is the equal sign.  The symbol itself never appeared until the 16th century and it traveled the entire width of the page.  It is hard to imagine how this would be a more efficient way of describing the equality present between two things, but it was.  Over time it was shortened to what we have today.  This is more than just an interesting factoid, too.  It goes to show that sometimes great ideas are so revolutionary that they seem obvious in hindsight.  First, we have a symbol that means equals, then we have, over time, an easier way of writing that symbol.

In many ways, isn’t that what makes mathematics so difficult, the jargon and abstraction?  That’s why one of my main points of focus is instilling mathematical literacy in students.  If they can read the math for what it says, not just as a funky collection of shapes and symbols, the mathematical ideas present themselves in a sensible and approachable fashion.

That’s what I’ve tried to do with my introduction to Algebra as a branch of mathematics, which is taught in Algebra 1, the class.  Here’s the link to the page.

As always, I thank you for reading and hope I’ve stirred some curiosity in you.

PS:  If you are interested in some of the history behind Algebra, the following book is highly recommended.  If you purchase it through this link you will help support the mission here of changing math from a hurdle in the way of young peoples’ dreams to a platform upon which success is built, and at no additional cost to you.

## Is Infinity Real?

### How Many Primes are There Is Infinity Real Part 1

Teachers: The following is a discussion that can be had with students to create interest in mathematics by discussing two very easy to understand, but perplexing problems in mathematics.  First, the nature of infinity.  The second is the lack of pattern and order in the prime numbers.

The number of primes is infinite.  Euclid proved it in a beautiful, easily understood proof by contradiction.  Paraphrasing, he said that there are either infinitely many primes, or a finite number of primes.  So let’s pick one and explore it.  Say there are a finite number of prime numbers.  If you were to list them all, then take their product you would have a very large number.  But if you just add one to that number, it would be prime because none of the other prime numbers would be a factor of it.  It would have exactly two factors, one and itself.

In case you don’t believe this works, let’s say we can list all of the primes, but there are only four.  Let’s say the entire list of primes was 2, 3, 5, and 7.  Their product, 2 × 3 × 5 × 7 = 210.  This number is composite because all of the primes are factors of it.  Add one to it, arriving and 211 and none of the prime numbers are a factor of it…making it have the factors of 211 and 1.  That means it is prime.

So it is false that there are a finite number of primes. Therefore, the are infinitely many prime numbers.

Beautiful, right?  Case closed. … or is it?

The case is closed, if you believe infinity exists.  To be clear, infinity is not a number, it’s a concept.  A set can only approach infinity, nothing ever equals infinity because it’s an idea.  The idea behind infinity is that the collection of things just keeps growing and growing.

We, as humans, have a very big problem with very big numbers, even large groups of things.  For example, there are some things that we only have a plural word for, we do not possess a singular word for these things.  A few examples are rice, sand, hair, shrimp and fish.  You can have a single hair, a grain of sand (or rice), and so on.  They are so vast in quantity they become indistinguishable.

And yet, they’re finite. You could conceivably collect all of the sand in the world and count every grain.  More sand does not magically appear once it is all collected.

What about stars in the sky?  What we call the observable universe is how far we can see.  We don’t know if it goes on forever, or if it is somehow contained.  Perhaps the word, universe, is misleading.  Perhaps there are multiples of it, maybe as many as there are grains of sand on the earth.

Before we chase that rabbit down its hole, let’s get back to earth.  Euclid’s proof that there are infinitely many prime numbers is beautiful.  But is he right?  Surely his proof is flawless, but what about infinity.  We have no examples of infinity, it might just be a human construction.  Now, if mathematics can discover things that are real and applicable from such a thing, that’s all the more powerful the tool it is, but what if we’re wrong about infinity?  There are two things I want you to consider as we explore prime numbers and their relationship with infinity.

The first thing is:  There’s an axiom (a statement we just accept as truth), called the Axiom of Infinity.  It basically says that there are infinite sets of things, like natural numbers.  We just say it’s true and roll with it until we discover a problem.  Then, we either adjust our axiom or start a new one.

The second thing is:  In the early 20th century a man named Kurt Gödel showed that we cannot actually prove any system of mathematics is true without assuming some supporting evidence is true.  We have to assume something is true in order to know if other things are true, roughly speaking.  In order to know if the thing we assumed to be true is actually true or not (like infinity), we have to assume that something else, more basic, is true.  So, and I’m taking some liberties here to make my point, but a conclusion, like the number of primes being infinite, is only as worthy as the presupposition (infinities exist).

Let’s look at a few strings of prime numbers and see if we can’t get our heads around this whole infinity thing.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

The gaps between these prime numbers are below.

1, 2, 2, 4, 2, 4, 2, 4, 6

Another string would be:

907, 911, 919, 929, 937, 941, 947

The gaps here are listed below.

4, 8, 10, 8, 4, 6

They are still relatively close.  Many mathematicians have tried to find a pattern in prime numbers.  After all, if you can find a pattern, then you can find the next one.  How cool would that be, right?

You might be thinking, uh, why would that be cool?

Well, there’s big money being paid if you can find the next prime number.  There is a project called GIMPS (Great Internet Mersenne Prime Search), where you can participate in the search.  And if your computer finds the next prime, you get some cash!

The last prime found with GIMPS was in 2013.  (At the time of this being written, it is 2017.) The number is massive.  The text file of the digits in the number is 7.7 MB.  That’s more data that a song and this is just a list of numbers.  The number is 257,885,161 – 1.  The number is huge that to verify that it is prime takes massive super computers days upon days to perform the calculation.  Finding the next prime number is a huge undertaking, very complicated and difficult, requiring computers all over the world working together before one is discovered.

Why all the fuss? What good are they?

Well, they keep you from being robbed, for one.  Internet security uses prime numbers to encrypt (code) your banking information.  The merchant will have a huge number that they multiply your card number by (kind of).  The huge number is the product of two of these gigantic prime numbers.  It’s so big that even though everybody (would be thieves) know it’s the product of two primes, they can’t figure out which two numbers.  The encrypted number is sent to your financial institution, who knows which two primes were used, which is basically like a key.

It’s also weird, and cool, that some bugs have a life cycle that only occurs in prime numbers!  Cicadas only come out and breed, and then die, in prime number years.  Incredible.

Back on track, forgive me.  It feels there are infinitely many tangents I can follow with math!  We have not been able to find a pattern in the prime numbers yet and let’s take a look at why.  You see, as these primes get huge, the gaps get larger and larger…approaching infinity!

Let’s take a look at one more string of prime numbers.

10009, 10037, 10039, 10061, 10067, 10069, 10079, 10091, 10093, 10099

The differences here are as follows.

28, 2, 22, 6, 10, 12, 2, 6

No discernable pattern, right?  If you can find one, you stand to make significant history, no one has found one yet.  We have some approximations that work within certain constraints but they all break down eventually.

But, to be clear, if you could find a pattern in the gaps between the primes a formula could be created that would generate prime numbers.  We can generate natural numbers by just adding one to the largest we have come up with so far.  But primes, as you’ve seen with the GIMPS project, aren’t so easily discovered.

And here’s one of the issues.  The gaps between prime numbers can get huge, perhaps infinitely huge.  Consider this.

Fact 1:  5! = 5×4×3×2 = 120

Fact 2:  120 is not prime because it is divisible by 5 and 4 and 3 and 2.

Fact 3:  5! + 5 is not prime because it is divisible by 5.  (When we add another 5, it’s like skip counting when you first learned multiplication.)

The same is true for 5! + 4 being divisible by 4, because 120/4 = 30.  5! + 4 is 4 × 31, there’s one more four.

The same holds true for 5! + 3 being divisible by 3 and 5! + 2 being divisible by 2.

Fact 4:  What all this means is that there after 5! + 1 there are four consecutive numbers that are composite.

This would also work for 100!  The number 100! + 100 would be composite.  For that matter, 100! + 37 would be composite also.  100! Plus all of the numbers up to and including 100 would be composite, (except possibly adding 1).

This means there is a gap of 99 after 100! + 1.

This goes on forever, arbitrarily large numbers, like 1,000,000,000,000!  There would be a gap of 1,000,000,000,000 – 1 numbers after this number that are composite.

We could write this in a general sense.  Let a and x be a whole numbers such that a is less than or equal to x.  (a x).

Then x! + a is composite.

Since x is a whole number and whole numbers are infinite, then there are infinitely large gaps between the large prime numbers, themselves being infinite.

Crazy, right?

So if the gaps between primes gets infinitely large, how can there be infinitely many prime numbers?

Well, there’s one more piece of information to be considered.  Twin primes are prime numbers that are just two numbers apart.  The primes 2 and 3 are only one apart, but all others are an even number apart, the smallest gap being a gap of two, like 5 and 7, or 11 and 13.

There’s a conjecture (not as strong as an axiom), that is yet unproven, but we’re getting closer, that states that there are an infinite number of twin primes.  The largest known pair of twin primes is below:

3,756,801,695,685 × 2666,689 – 1
and

3,756,801,695,685 × 2666,689 +1

Those numbers are too large to be written out!

While we do not yet know, with a proof, that there are infinitely many twin primes, we do know that there are infinitely many primes that have a maximum distance between them and it might be as low as a difference of sixteen.  This is all being discovered and explored and fought over at the moment.

So on one hand we have infinitely large gaps between prime numbers, but when they do pop up, they will do so in clumps and groups?

If all of this makes your head spin, then I have succeeded.  I am not trying to convince you that infinities do not exist, or that they do.  I am trying to show that math is contentious and changing.  As we learn and discover new things math is changing.  Math is just a language we use to describe the world around us.  So powerful is math that we are not even sure if it is a human invention at all or rather a discovery!

As always, thank you for your time. I hope this has stirred some thought, maybe even sparked a passion for mathematics!

At the time of the making of this video the world’s largest prime number is not the last one found by the GIMPS project.  However, they’re likely to find another even larger one, sometime soon.  There’s a video below (Largest prime number) that discusses that number and prints it out … it takes up as much paper as three large books!

For some fascinating and approachable treatment of prime numbers, consider the following videos:

Gaps between prime numbers: https://youtu.be/vkMXdShDdtY

If you found this helpful and would like to help make these videos possible, to help break down the obstacle that math presents itself as to young people, please consider visiting my patreon site:

www.patreon.com/beardedmathman

## The Problem with PEMDAS

The problem with PEMDAS

This problem has really stirred a lot of interest and created a buzz on the internet. I can see why, it’s an easy one to miss.  And yet, PEMDAS is such an easy thing to remember, the mnemonic devices offered make for a strong memory.  So people passionately defend their answers.

6 ÷ 2(2 + 1)

I am going to tell you the answer in just a moment, but before I do, please listen to why I think this is a worthy problem to explore.

There are two fundamental misconceptions with math that make math into a monster for so many people, and this problem touches on both.  In a sense, neither has anything to do with the order of operations specifically.

The first issue is understanding that spatial arrangements in math mean something.  The way we write the numbers and symbols has a meaning, very specific at that.  In this video by Mind Your Decisions, https://youtu.be/URcUvFIUIhQ, he shares where there was a moment in time when we used different conventions to write math.

And while math may or may not be a human invention, the symbols and arrangements and their meanings certainly are.  Just like the letter A is only a letter and with a specific sound because we all agree.  Just like a red light means stop, a green light means go and a yellow light means HURRY HURRY HURRY!

The second, and more over-arching issue here, is the misconception that addition and subtraction are different.  They are fundamentally the same thing.  Subtraction is really addition of opposite numbers.  Perhaps to shore this misconception negative numbers should be introduced instead of subtraction.

Now you might argue and say, Wait, addition has properties that subtraction lacks, like the commutative property.

You’re correct, 5 + 3 = 3 + 5, while 5 – 3 does not equal 3 – 5.  However, 5 – 3 is really five plus the opposite of three, like written below.

5 + - 3

And that is the same as this expression below.

-3 + 5

So the AS at the end of PEMDAS is really just A, or S, whichever leads to the better nursey rhyme type device to improve recall.

Since we believe that addition and subtraction are different, we also come away with the belief that multiplication and division are different.  Sorry, they’re not.  Division is multiplication of the reciprocal.  Remember that whole phrase from your school days? (How was that for a mnemonic device?)

And while division does not have the commutative property, that again is a consequence of the way we write math.  If we only wrote division as multiplication of the reciprocal, we would see that multiplication and division are in fact the same.

So, back to the problem.  The most common wrong answer is 1.  The correct answer is 9.  Here’s a great video on the order of operations, super catchy and articulates the importance of left to right as written for multiplication and addition.

Last thing:  Now, in creative writing the intent of the author must be considered, should it also be considered here?

Let me know what about this you like, dislike or disagree with.  Let me know what is helpful.  I really want to promote success through making math transparent.  It’s my mission.  You can help support my mission by just sharing and liking this.  Subscribe to my blog if you’re a teacher as I will be populating it with lots of teacher advice, not all math related.

## The Square Root Club

If you’re a teacher, I have a short story that you can share, adapted to fit your own style, that you can use to address the biggest issue with teaching … students learn what they want to learn.  Creating interest in mathematics for teenagers can sometimes be a challenge.  One of the easiest ways to do so is with humor.  The following story is actually true, but humorous, and I think will create some curiosity and thus learning opportunities for students.

I believe the appropriate audience would be pre-algebra students learning about square roots up to algebra students learning about square roots.  Anyhow, if you find this helpful, please let me know.

# The Square Root Club

My daughter, a senior at the University of Arizona, called and said she’d uncovered an issue in math that is both absolutely impossible and yet, true.  My interest piqued, I listened attentively as she asked if I’d ever heard of the square root club.

The square root club, I was informed, is a club of dubious membership.  To become a member the square root of your GPA must exceed your GPA.  What a delightful treat this was…and to think, I’d never heard of such a thing!

She continues to tell me that she met someone who was a member.  I asked her how she knew, because certainly her friend would be ignorant of his membership.  Surely, someone in the club would not be smart enough to be aware of the fact, right?

That’s what she said was the funniest part, the part that was seemingly impossible!  He knew about it, even made up the name of the club himself.  He was no longer a member, just graduated with his bachelor’s degree with a 3.0 GPA.

Note:  GPA (grade point average) is calculated by assigning a numerical value to letter grades.  An A is 4, B is 3, C is 2, D is 1 and an F is zero.

The moral of the story is that grades don’t reflect potential, they reflect what you show you know.  Many high school students get by with intelligence but never work.  Upon arriving in college they are overwhelmed, never having had to work hard or apply themselves.  Before they know it, they’re buried and there’s no quick fix like there can be in high school.

To that point, nobody cares about someone’s potential, not even your mother. Imagine your mom told you to clean your room.  Because she told you to do it, she believes you have the potential.  However, if you do not clean it, she will be satisfied by the fact that you could have cleaned it.

Now of course, the question being begged here is, what could his GPA have been?

## How Math Fixed Music

### Rational Exponents Sound GREAT

Before we dive in, music is primarily defined by what we hear, not by the analysis and insight provided by math.  For example, an octave is a note whose frequency is double that of its parent note.  The mathematical relationship was discovered after the fact.  The following is an exploration of how math is used in music, but I don’t want to put the cart before the horse here.  The math supports the music, makes it work.  But the math is really fine-tuning what we hear.

Pythagoras developed a musical system that over the years evolved into what we have today. (At least Pythagoras is often credited for it.)  Not until “recently,” however, has one of the major problems with music been resolved (see what I did there with resolve?).

The problem the ancients had is that their octaves didn’t line up.  An octave, as I mentioned early, is a note that has twice (or half) the frequency of another note.  Octaves, in modern western music, share the same names, too.  The note A, at 440 Hz, has an octave at 880 Hz, and also 220 Hz.   (There are infinitely many octaves, in theory, though our ears have a limited range of things we can hear.)  The ancients, however, had a problem because after a few octaves, well, they were no longer octaves.

In western music we have 12 semi-tones, A, A# (or B-flat), B, C, C#, D, D#, E, F, F#, G, G# and then A again.  It’s cyclic, repeated infinitely both higher and lower. Each semi-tone in the next series of 12 notes is an octave of our first series of notes.  And the relationship between notes is what makes them, well, musical, not just sounds.

The problem is defining that relationship.  You see, because each note is slightly higher (has a higher frequency), and each note’s octave is double that frequency, what happens is the notes get further and further apart (the differences in their frequencies increases).

Let’s take a look at the frequencies:

 Note Frequency A 220.00 A# 233.08 B 246.94 C 261.63 C# 277.18 D 293.66 D# 311.13 E 329.63 F 349.23 F# 369.99 G 392.00 G# 415.30 A 440.00 A# 466.16 B 493.88 C 523.25 C# 554.37 D 587.33 D# 622.25 E 659.25 F 698.46 F# 739.99 G 783.99 G# 830.61 A 880.00

As you can see, the differences between consecutive notes is increasing, at an increasing rate!  This is not a linear relationship.  Because of this, the ancients had a very hard time defining what was an A and what was a D, especially when you started moving around between octaves.  Things got jumbled, and out of tune.

It is tricky to find the proportion and rate of change between consecutive notes, any two consecutive notes that is.  That’s where math comes in to save the day.  Let’s build the rate of change, shall we.

First, note that the rate is increasing, at an increasing rate, so we cannot add.  I show that in the video below.  We have to multiply.  When we repeatedly multiply, we can use exponents.  Since we need a note and it’s octave to be doubles, our base number is 2.

Since there are twelve notes between a note and its octave, we need to break the multiple of two into twelve equal, multiplicative parts.  That’s a rational exponent, 1/12.

The number we need to multiply each note by is 21/12.  Each note is one-twelfth of the way to the octave.  It is pretty cool indeed.

For a great read on this topic, consider the book Harmonograph.

## Cube Roots and Higher Order Roots

other roots

Cube Roots
and

Square roots ask what squared is the radicand. A geometric explanation is that given the area of a square, what’s the side length? A geometric explanation of a cube root is given the volume of a cube, what’s the side length. The way you find the volume of a cube is multiply the length by itself three times (cube it).

The way we write cube root is similar to square roots, with one very big difference, the index.

$\begin{array}{l}\sqrt{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\\ \sqrt[3]{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}cube\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\end{array}$

There actually is an index for a square root, but we don’t write the two. It is just assumed to be there.

Warning: When writing cube roots, or other roots, be careful to write the index in the proper place. If not, what you will write will look like multiplication and you can confuse yourself. When writing by hand, this is an easy thing to do.

$3\sqrt{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{8}$

To simplify a square root you factor the radicand and look for the largest perfect square. To simplify a cubed root you factor the radicand and find the largest perfect cube. A perfect cube is a number times itself three times. The first ten are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1,000.

Let’s see an example:

Simplify:

$\sqrt[3]{16}$

Factor the radicand, 16, find the largest perfect cube, which is 8.

$\sqrt[3]{8}×\sqrt[3]{2}$

The cube root of eight is just two.

$2\sqrt[3]{2}$

The following is true,

$\sqrt[3]{16}=2\sqrt[3]{2}$,

only if

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Arithmetic with other radicals, like cube roots, work the same as they do with square roots. We will multiply the rational numbers together, then the irrational numbers together, and then see if simplification can occur.

${\left(2\sqrt[3]{2}\right)}^{3}={2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}$

Two cubed is just eight and the cube root of two cubed is the cube root of eight.

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{2\cdot 2\cdot 2}$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{8}$

The cube root of eight is just two.

$8×\sqrt[3]{8}=8×2$

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Negatives and cube roots: The square root of a negative number is imagery. There isn’t a real number times itself that is negative because, well a negative squared is positive. Cubed numbers, though, can be negative.

$-3×-3×-3=-27$

So the cube root of a negative number is, well, a negative number.

$\sqrt[3]{-27}=-3$

Other indices (plural of index): The index tells you what power of a base to look for. For example, the 6th root is looking for a perfect 6th number, like 64. Sixty four is two to the sixth power.

$\sqrt[6]{64}=2\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}because\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{2}^{6}=64.$

A few points to make clear.

·         If the index is even and the radicand is negative, the number is irrational.

·         If the radicand does not contain a factor that is a perfect power of the index, the number is irrational

·         All operations, including rationalizing the denominator, work just as they do with square roots.

Rationalizing the Denominator:

Consider the following:

$\frac{9}{\sqrt[3]{3}}$

If we multiply by the cube root of three, we get this:

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}}=\frac{9\sqrt[3]{3}}{\sqrt[3]{9}}$

Since 9 is not a perfect cube, the denominator is still irrational. Instead, we need to multiply by the cube root of nine.

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}}=\frac{9\sqrt[3]{9}}{\sqrt[3]{27}}$

Since twenty seven is a perfect cube, this can be simplified.

$\frac{9}{\sqrt[3]{3}}=\frac{9\sqrt[3]{9}}{3}$

And always make sure to reduce if possible.

$\frac{9}{\sqrt[3]{3}}=3\sqrt[3]{9}$

This is a bit tricky, to be sure. The way the math is written does not offer us a clear insight into how to manage the situation. However, the topic we will see next, rational exponents, will make this much clearer.

Practice Problems:

Simplify or perform the indicated operations:

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[4]{64}\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}+4\sqrt[3]{9}\\ \\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}×4\sqrt[3]{9}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[5]{64}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{7}}{\sqrt[3]{7}}\end{array}$

## Multiplying and Dividing Square Roots, Rationalizing the Denominator

1.7.2 square root operations continued

Square Roots
Multiplication and Division

At some point square roots should no longer be considered an operation but rather the most efficient way to express a number. For example, the best way to write one hundred trillion is $1×{10}^{14}$. The best way to express the number times itself that is two is as $\sqrt{2}.$

That provides insight when we consider multiplying a rational number and an irrational number together. It is not confusing for some irrational numbers, like π. Nobody confused 3π because we understand that symbol is the best way to write the number. There’s not a way to rewrite multiples of π other than by writing the multiple in front.

However, $3\sqrt{2}$ is often written as $\sqrt{6}$. There are reasons explained by the order of operations which tell us why this is false, but understanding what the square root of two is perhaps offers the simplest insight.

$\sqrt{2}\approx 1.414$

$3\sqrt{2}=\sqrt{2}+\sqrt{2}+\sqrt{2}$

$3\sqrt{2}\approx 1.414+1.414+1.414$

4.242

The square root of six is approximately 2.449. Not the same thing at all.

The following, however, is true:

$\sqrt{2}×\sqrt{3}=\sqrt{6}$

and

$\sqrt{2×3}=\sqrt{6}$.

The following generalization can be used. Sometimes it is best to write things one way versus another, and it is up to you to decide if rewriting an expression offers insight.

$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$

If two numbers are both square roots you can multiply their radicands together. But you cannot multiply the radicand of a square root with rational number like we saw above.

Division is a little more nuanced, but only when your denominator is a fraction.

This generalization is true for division:

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

For example:

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{2}.$

This can be calculated two ways.

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}.$

or

$\frac{\sqrt{8}}{\sqrt{4}}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$

But you cannot divide rational numbers into the radicand, or the radicand of a square root into a rational number. Remember, square roots, when simplified, are the most efficient way of writing irrational numbers. If we used k to represent the square root of two, these types of confusing things would not be happening.

Nobody would confuse what is happening with
$\frac{6}{k}$. We simply cannot evaluate that because 6 and k do not have common factors. When k is written as the square root of two, sometimes people just see a 2 and reduce.

The only issue with division of square roots occurs if you end up with a square root in the denominator.

$\frac{5}{\sqrt{2}}$

Denominators must be rational and the square root of two is irrational. However, there’s an easy fix. Remember that $\sqrt{2}×\sqrt{2}=\sqrt{4},$ and $\sqrt{4}=2.$ It is also true that:

$\frac{\sqrt{2}}{\sqrt{2}}=1$ .

To Rationalize the Denominator, which means make the denominator a rational number, we just multiply as follows:

$\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$

Sometimes we end up with something like this:

$\frac{5}{3\sqrt{2}}$

Three is a rational number and is perfectly okay in the denominator. If you multiply by the fraction $\frac{3\sqrt{2}}{3\sqrt{2}},$ you can still get the simplified equivalent, but you’ll have extra reducing to do at the end. Instead, just multiply by the irrational portion.

$\frac{5}{3\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{6}$.

In summary, to divide or multiply with square roots, you can multiply or divide the radicands. However, if you’re multiplying or dividing rational numbers and square roots, you cannot combine the radicands and the rational numbers.

Practice Problems:

Perform the indicated operations:

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(5\sqrt{7}\right)\left(3\sqrt{14}\right)\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\sqrt{15}\right)\left(\sqrt{3}\right)\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3\sqrt{2}}{\sqrt{8}}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{5}}{\sqrt{3}}\\ \\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{\sqrt{8}}\cdot \frac{\sqrt{2}}{6}\end{array}$

## Addition and Subtraction of Square Roots

Mathematical Operations and Square Roots

Part 1

In this section we will see why we can add things like $5\sqrt{2}+3\sqrt{2}$ but cannot add things like $2\sqrt{5}+2\sqrt{3}$. Later we will see how multiplication and division work when radicals (square roots and such) are involved.

Addition and Subtraction: Addition is just repeated counting. The expression $5\sqrt{2}$ means $\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}$, and the expression So if we add those two expressions, $5\sqrt{2}+3\sqrt{2},$ we get $8\sqrt{2}$ . Subtraction works the same way.

Consider the expression $2\sqrt{5}+2\sqrt{3}$. This means $\sqrt{5}+\sqrt{5}+\sqrt{3}+\sqrt{3}.$ The square root of five and the square root of three are different things, so the simplest we can write that sum is $2\sqrt{5}+2\sqrt{3}$.

A common way to describe when square roots can or cannot be added (or subtracted) is, “If the radicands are the same you add/subtract the number in front.” This is not a bad rule of thumb, but it treats square roots as something other than numbers.

$5×3+4×3=9×3$

The above statement is true. Five groups of three and four groups of three is nine groups of three.

$5\sqrt{3}+4\sqrt{3}=9\sqrt{3}$

The above statement is also true because five groups of the numbers squared that is three, plus four more groups of the same number would be nine groups of that number.

However, the following cannot be combined in such a fashion.

$3×8+5×2$

While this can be calculated, we cannot add the two terms together because the first portion is three $–$ eights and the second is five $–$ twos.

$3\sqrt{8}+5\sqrt{2}$

The same situation is happening here.

Common Mistake: The following is obviously wrong. A student learning this level of math would be highly unlikely to make such a mistake.

$7×2+9×2=16×4$

Seven $–$ twos and nine $–$ twos makes a total of sixteen $–$ twos, not sixteen $–$ fours. You’re adding the number of twos you have together, not the twos themselves. And yet, this is a common thing done with square roots.

$7\sqrt{2}+9\sqrt{2}=16\sqrt{4}$

This is incorrect for the same reason. The thing you are counting does not change by counting it.

Explanation: Why can you add $5\sqrt{2}+3\sqrt{2}$? Is that a violation of the order of operations (PEMDAS)? Clearly, the five and square root of two are multiplying, as are the three and the square root of two. Why does this work?

Multiplication is a short-cut for repeated addition of one particular number. Since both terms are repeatedly adding the same thing, we can combine them.

But if the things we are repeatedly adding are not the same, we cannot add them together before multiplying.

What About Something Like This: $3\sqrt{40}-9\sqrt{90}$?

Before claiming that this expression cannot be simplified you must make sure the square roots are fully simplified. It turns out that both of these can be simplified.

$3\sqrt{40}-9\sqrt{90}$

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

The dot symbol for multiplication is written here to remind us that all of these numbers are being multiplied.

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

$3\cdot 2\cdot \sqrt{10}-9\cdot 3\cdot \sqrt{10}$

$6\sqrt{10}-27\sqrt{10}$

$-21\sqrt{10}$

What About Something Like This: $\sqrt{7+7}$ versus $\sqrt{7}+\sqrt{7}.$

Notice that in the first expression there is a group, the radical symbol groups the sevens together. Since the operation is adding, this becomes:

$\sqrt{7+7}=\sqrt{14}$.

Since the square root of fourteen cannot be simplified, we are done.

The other expression becomes:

$\sqrt{7}+\sqrt{7}=2\sqrt{7}.$

Summary: If the radicals are the same number, the number in front just describes how many of them there are. You can combine (add/subtract) them if they are the same number. You are finished when you have combined all of the like terms together and all square roots are simplified.

Practice Problems: Perform the indicated operation.

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{25}-5\sqrt{5}+5\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}+3\sqrt{3}\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\sqrt{75}+8\sqrt{24}+\sqrt{75}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{200}+8\sqrt{8}-2\sqrt{32}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\sqrt{98}+16\sqrt{2}\end{array}$

## Square Roots Part 1

Note:  Square roots are pretty tricky to teach and learn because the tendency is to seek answer-getting methods.  Patience at the onset, allowing for full development of conceptual understanding is key.  Do not revert to tricks and quick "gets" when first learning square roots.  Always revert back to the question they ask and how you know if you've answered that question.

square roots part 1

Square Roots

Part 1

Introduction: Square roots are consistently among the most misunderstood topics in developmental math. Similar to exponents, students must possess both procedural fluency but also a solid conceptual foundation and the ability to read and understand what square roots mean, in order to be proficient with them. It is often the case that problems with square roots do not lend themselves to a correct first step, but rather, offer many equally viable methods of approach.

Square Roots Ask a Question: What number squared is equal to the radicand? The radicand is the number inside the square root symbol (radical). This expression asks, what number times itself (squared) is 11?

$\sqrt{11}$

This is a number. It is not 11. It turns out this number is irrational and we can never actually write what it is more accurately than this.

Big Idea: The area of a square is calculated by squaring a side (multiplying it by itself). Since all sides of a square are equal, this about as easy of an area to calculate as possible. A square root is giving us the area of a square and asking us to find out how long a side is.

$\overline{)42}$

For example, this square has an area of forty-two. Instead of writing out the question, “How long is the side of a square whose area is forty-two?” we simply write, $\sqrt{42}.$

The majority of the confusion with square roots comes back to this definition of what a square root is. To make it as clear as possible, please consider the following table.

 English Math How long is the side of a square that has an area of 100? $\sqrt{100}$ How long is the side of a square that has an area of 10? $\sqrt{10}$

These two numbers were chosen because students inevitably write $\sqrt{100}=\sqrt{10}.$

 English Math Answer How long is the side of a square that has an area of 100? $\sqrt{100}$ 10 How long is the side of a square that has an area of 10? $\sqrt{10}$ 3.162277660168379…

Key Knowledge: In order to be proficient with square roots we need to know about perfect squares. A perfect square is a number that is the product of a number squared. Sixteen is a perfect because four times four is sixteen.

The reason you need to know perfect squares is because square roots are asking for numbers squared that equal the radicand. So if the radicand is a perfect square, we have an easy ‘get,’ that is, simplification.

For example, since 42 = 16, and the square root of sixteen $\left(\sqrt{16}\right)$ asks for what number squared is 16, the answer is just four.

Let’s take a look at the first twenty perfect squares and what number has been squared to arrive at the perfect square, which we will call the parent.

 Perfect Square “Parent” 1 1 4 2 9 3 16 4 25 5 36 6 49 7 64 8 81 9 100 10 121 11 144 12 169 13 196 14 225 15 256 16 289 17 324 18 361 19 400 20

You should recognize these numbers as perfect squares as that is a key piece of knowledge required!

Pro-Tip: When dealing with square roots it is wise to have a list of perfect squares handy to help you familiarize yourself with them.

How to Simplify a Square Root: To simplify a square root all you do is answer the question it is asking.

The best way to go about that is to see if the radicand is a perfect square. If so, then just answer the question. For example:

Simplify $\sqrt{256}$

Since this is asking, “What number squared is 256?” and 256 is a perfect square, 162, the answer to the question is just 16.

$\sqrt{256}=16$

What if we had something like this:

$\sqrt{{x}^{2}}$

If you’re confused by this, revert back to the question it is asking. This is asking, “What squared is x $–$ squared?” All you have to do is answer it.

$\sqrt{{x}^{2}}=x$

What if the radicand was not a perfect square?

If you end up with an ugly square root, like $\sqrt{48},$ all you have to do is factor the radicand to find the largest perfect square.

List all factors, not just the prime factors. In fact, the prime factors are of little use because prime numbers are not perfect squares. And again, we are looking for perfect squares because they help us answer the question posed by the square root.

48

1, 48

2, 24

3, 16

4, 12

6, 8

Pro-Tip: When factoring, do not skip around. Check divisibility by all of the numbers in order until you get a turn around. For example, after 6, check 7. Seven doesn’t divide into 48, but 8 does. Eight times six is forty eight, but you already have that pair. That’s how you know you’re done!

In our list we need to find the largest perfect square. While four is a perfect square, sixteen is larger. So we need to use three and sixteen like shown below.

$\sqrt{48}=\sqrt{16}\cdot \sqrt{3}$

The square root of three is irrational (square roots of prime numbers are all irrational), but the square root of sixteen is four. So rewriting this we get:

$\begin{array}{l}\sqrt{48}=\sqrt{16}\cdot \sqrt{3}\\ \sqrt{48}=4\cdot \sqrt{3}\end{array}$

See Note 1 and Note 2 below for an explanation of why the above works.

Fact:

That means that Let’s see if it is true.

${\left(4\sqrt{3}\right)}^{2}=4\sqrt{3}×4\sqrt{3}$

Because we can change the order in which we multiply, we can rearrange this and multiply the rational numbers together first and the irrational numbers together first.

$4\sqrt{3}×4\sqrt{3}=4×4×\sqrt{3}×\sqrt{3}$

The square root of three times itself is the square root of nine.

$4×4×\sqrt{3}×\sqrt{3}=16×\sqrt{9}$

The square root of nine asks, what squared is nine. The answer to that is three.

$16×\sqrt{9}=16×3$

So,

Note 1: $4\sqrt{3}$ cannot be simplified because the square root of three is irrational. That means we cannot write it more accurately than this. Also, the product of rational number and an irrational number is irrational. So, $4\sqrt{3}$ is just written as “four root three.”

Note 2: We can separate square roots into the product of two different square roots like this:

$\sqrt{75}=\sqrt{25\cdot 3}$ figure a.

or

$\sqrt{75}=\sqrt{25}\cdot \sqrt{3}$ figure b.

If we consider the question being asked, what number squared is seventy five, we can see why this works. What number squared is seventy five is the same as what number squared is twenty five times three,” (figure a). The number squared that is twenty five times the number squared that is three is the same as the number times itself that is twenty five times three.

For example:

But also: $\sqrt{64}=\sqrt{4}\cdot \sqrt{16}$

And this simplifies to:

$\sqrt{64}=2\cdot 4=8$

What we will see in a future section is that square roots are actually exponents, exponents are repeated multiplication and the order in which you multiply does not matter. This allows us to manipulate square root expressions in such a fashion.

Let us work through two examples. Before we do, let us define what simplify means in the context of square roots. Simplify with square roots means that the radicand does not contain a factor that is a perfect square and that all terms are multiplied together.

Simplify: $9\sqrt{8}$

What is the nine doing with the square root of eight? It is multiplying by it. We cannot carry out that operation. However, eight, the radicand, does contain a perfect square, four. Do not allow the fact that 9 is also a perfect square confuse you. This is just 9, as in 1, 2, 3, 4, 5, 6, 7, 8, 9. The square root of eight cannot be counted. It is asking a question, remember?

$9\sqrt{8}=9\sqrt{4}\cdot \sqrt{2}$

Pro-Tip: When rewriting radical expressions (square roots), write the perfect square first as it is easier to manipulate (you won’t mess up as easily).

$\begin{array}{l}9\sqrt{4}\cdot \sqrt{2}\\ 9\cdot 2\cdot \sqrt{2}\\ 18\sqrt{2}\end{array}$

Example 2:

Simplify $8\cdot \sqrt{\frac{32}{{x}^{4}}}$

The eight is multiplying with the radical expression. Just like we could separate the multiplication of square roots, we can also separate the division, provided it is written as multiplication by the reciprocal. So, let’s consider these separately, to break this down into smaller pieces that are easier to manage.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{32}}{\sqrt{{x}^{4}}}$

Let’s factor each square root, looking for a perfect square. Note that x2 times x2 is x4.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{16}×\sqrt{2}}{\sqrt{{x}^{4}}}$

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{4×\sqrt{2}}{{x}^{2}}$

Notice that 8 is a fraction 8/1.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{8}{1}×\frac{4×\sqrt{2}}{{x}^{2}}$

Multiplication of fractions is easy as π.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{32\sqrt{2}}{{x}^{2}}$

Summary: Square roots ask a question: What number squared is the radicand? This comes from the area of a square. Given the area of a square, how long is the side?

To answer the question you factor the radicand and find the largest perfect square.

Time for some practice problems:

1.7 Square Roots Part 1 Practice Set 1

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{125}=\\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{27}=\\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{162}=\\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{75}=\\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{45}=\end{array}$ $\begin{array}{l}6.\text{\hspace{0.17em}}\sqrt{4{x}^{2}}=\\ 7.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4{x}^{4}}=\\ 8.\text{\hspace{0.17em}}\sqrt{\frac{4}{25}}=\\ 9.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{98}=\\ 10.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}=\end{array}$ $\begin{array}{l}11.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4}\cdot \sqrt{8}=\\ 12.\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\sqrt{8}=\\ 13.\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\sqrt{9}=\\ 14.\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\sqrt{\frac{1}{4}}=\\ 15.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{300}=\end{array}$

1.7 Square Roots Part 1, Practice 2

Simplify problems 1 through 8.

1.      $\sqrt{24}$ 2. $\sqrt{8{x}^{3}}$

3. $\sqrt{200}$ 4. $\sqrt{27}$

5. $\sqrt{7x}$ 6. $4\sqrt{{a}^{2}b}$

7. $3x\sqrt{98{x}^{2}}$ 8. $\frac{1}{3}\cdot \sqrt{\frac{9}{{x}^{2}}}$

9. Show that ${a}^{2}b=\sqrt{{a}^{4}{b}^{2}}$ 10. Why is finding perfect squares appropriate

when simplifying square roots?

## Where the First “Law” of Logarithms Originates – Wednesday’s Why E6

rule 1

Wednesday’s Why $–$ Episode 6

In this week’s episode of Wednesday’s Why we will tackle why the following is “law” of exponents is true in a way that hopefully will promote mathematical fluency and confidence. It is my hope that through these Wednesday’s Why episodes that you are empowered to seek deeper understanding by seeing that math is a written language and that by substituting equivalent expressions we can manipulate things to find truths.

${\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N$

Now of course the base of 2 is arbitrary, but we will use a base of two to explore this.

The first thing to be aware of is that exponents and logarithms deal with the same issue of repeated multiplication. There connection between the properties of each are tightly related. What we will see here is that the property of logarithms above and this property of exponents below are both at play here. But it is not so easy to see, so let’s do a little exploration.

Just to be sure of how exponents and logarithms are written with the same meaning, consider the following.

${a}^{b}=c↔{\mathrm{log}}_{a}c=b$

Let us begin with statement 1: ${2}^{A}=M$

We can rewrite this as a logarithm, statement 1.1: ${\mathrm{log}}_{2}M=A$

Statement 2:
${2}^{B}=N$

We can rewrite this as a logarithm, statement 2.1: ${\mathrm{log}}_{2}N=B$

If we take the product of M and N, we would get
2A ·2B. Since exponents are repeated multiplication,

2A ·2B = 2A + B

This gives us statement 3: $M\cdot N={2}^{A+B}$

Let us rewrite statement 3 as a logarithmic equation.

$M\cdot N={2}^{A+B}\to {\mathrm{log}}_{2}\left(M\cdot N\right)=A+B$

In statements 1.1 and 2.1 we see what A and B equal. So let’s substitute those now.

$\begin{array}{l}{\mathrm{log}}_{2}\left(M\cdot N\right)=A+B\\ {\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N\end{array}$

It took a little algebraic-juggling to get it done, but hopefully you can now see that this is not a law or a rule, but a property of repeated multiplication, just like all of the properties of exponents are consequences of repeated multiplication.

Let me know what worked for you here and what did not. Leave me a comment.

Thank you again.