# Can you prove that sqrt(3) is irrational?

## Prove that $\sqrt{3}$ is irrational by contradiction. E.g. show that it is not rational

Oct 22, 2016

The idea behind the proof is pretty simple, although depending on how rigorous you want to be, the setup can make it a little lengthy. The idea is that you suppose it's of the form $\frac{a}{b}$, meaning $3 {b}^{2} = {a}^{2}$, and then show that the left hand side is divisible by $3$ an odd number of times, whereas the right hand side is divisible by $3$ an even number of times.

The same technique can be used to show that $\sqrt{n}$ is irrational for any integer which is not a perfect square.

Suppose, to the contrary, that $\sqrt{3}$ is rational. Then there exist integers $a$ and $b$ such that $\sqrt{3} = \frac{a}{b}$. Without loss of generality, we will assume that $a$ and $b$ share no common factors, as if they did, we could cancel them and obtain two new integers with the desired properties.

$\sqrt{3} = \frac{a}{b}$

$\implies 3 = {\left(\frac{a}{b}\right)}^{2}$

$\implies 3 = {a}^{2} / {b}^{2}$

$\implies 3 {b}^{2} = {a}^{2}$

Let ${k}_{a}$ be the greatest power of $3$ by which $a$ is divisible. Then $a = {3}^{{k}_{a}} {a}_{0}$, where ${a}_{0} = \frac{a}{3} ^ \left({k}_{a}\right)$ is an integer not divisible by $3$. Similarly, let $b = {3}^{{k}_{b}} {b}_{0}$ where ${b}_{0} = \frac{b}{3} ^ \left({k}_{b}\right)$ is an integer not divisible by $3$.

Substituting those in, we get

$3 {\left({3}^{{k}_{b}} {b}_{0}\right)}^{2} = {\left({3}^{{k}_{a}} {a}_{0}\right)}^{2}$

$\implies {3}^{2 {k}_{b} + 1} {b}_{0}^{2} = {3}^{2 {k}_{a}} {a}_{0}^{2}$

As neither ${a}_{0}$ nor ${b}_{0}$ is divisible by $3$, then by the uniqueness of the prime factorization of integers, we have

$2 {k}_{b} + 1 = 2 {k}_{a}$.

However, as the left hand side of this is an odd number and the right is even, we have reached a contradiction. Thus our initial premise was false, meaning $\sqrt{3}$ is not rational, i.e., $\sqrt{3}$ is irrational. ∎

Oct 22, 2016

Well, we'll prove it by proving that it's not a rational number. I'm really sorry this is long but hopefully I explain every step thoroughly.

#### Explanation:

So, let's start with the definition of a rational number:
Rational Number: a real number that can be expressed through a fraction of REAL, INTEGER, NONZERO numbers. The fraction has to be in it's lowest form ($\frac{1}{2}$ instead of $\frac{2}{4}$)

Remember an integer is a whole number (1, 2, 3, etc) and not a fraction of any sort (so not 3.592 or $\frac{1}{2}$)

We know that 5 is rational because it can be expressed as $\frac{10}{2}$ or $\frac{5}{1}$. 4.333333333(...) is rational because it can be expressed as $\frac{13}{3}$.

So, let's prove that $\sqrt{3} = \frac{a}{b}$. We're doing this because we want to show that it can never be a rational number, so this proof should have a contradiction in it.

Let's start with the statement in the previous paragraph.
$\sqrt{3} = \frac{a}{b}$. Let's square both sides to start.
$3 = {a}^{2} / {b}^{2}$. We just squared each number, getting rid of the root. Next, we'll multiply both sides by ${b}^{2}$

$3 {b}^{2} = {a}^{2}$

And now, we divide both sides by 3.

${b}^{2} = {a}^{2} / 3$

Ok so now we need to use some logic. We know both $a$ and $b$ are integers (whole numbers) because they have to be for $\sqrt{3}$ to be a rational number. So, this means that ${a}^{2}$ is also an integer.

Now, this also means that $b$ and ${b}^{2}$ are both integers. Keep that in mind.

So, since $b$, ${b}^{2}$, $a$, and ${a}^{2}$ are all integers, ${a}^{2} / 3$ should be an integer as well. We know this because ${b}^{2}$ is an integer, and ${a}^{2} / 3$ is equal to ${b}^{2}$, so whatever ${a}^{2} / 3$ is, it has to be an integer since its equal to an integer. Here's a quick visualization of that.

${b}^{2}$ is an integer
${b}^{2} = {a}^{2} / 3$
an integer $= {a}^{2} / 3$

We figured that out through simple substitution of "an integer" and ${b}^{2}$.

Ok, so back to the problem. We just figured out that ${a}^{2} / 3$ is an integer. So, a number squared divided by three is a whole number. That means that ${a}^{2}$ is divisible by three. This may not seen revolutionary, but it helps a lot.

Mathematical Rule: A square of an integer is divisible by the same prime numbers that it's square root is divisible by.

This rule means that ${a}^{2}$ should be divisible by the same prime numbers that $a$ is. And, 3 is a prime number. So, $a$ should also be divisible by three.
Here's a pretty simple way to look at it. Let's say that $a = 14$. This means that ${a}^{2} = 196$. We know that $146$ is divisible by $7$ and $2$ because $14$ is divisible by $7$ and $2$

So, now we know that $a$ and ${a}^{2}$ are divisible by 3.

Let's keep going here. Since we know that $a$ is divisble by $3$, we can make this equation.
$a = 3 c$, where $c$ is an integer. This makes sense because 3 times some number equals a. But the key here is that C is an integer.
If we square both sides in the equation we made, we get that:
${a}^{2} = 9 {c}^{2}$

Now, let's substitute the equation we just made above into the original equation. The original is:
$3 {b}^{2} = {a}^{2}$

Our new one is:
$3 {b}^{2} = 9 {c}^{2}$

Let's divide both sides by 3 to get:
${b}^{2} = 3 {c}^{2}$

And divide both sides by 3 again:
${b}^{2} / 3 = {c}^{2}$

And suddenly we find ourselves repeating ourselves. Remember when we proved that $a$ was divisible by 3? We can do that again here.

We can prove that b is divisible by 3 since c is an integer just like we proved that $a$ was divisible by 3.

So, I'm going to skip over that step but re-read the part where we proved a was divisible by 3 if you want to jog your memory.

Ok, we're getting close to finishing. Let's examine the definition of a rational number again:
Rational Number: a real number that can be expressed through a fraction of REAL, INTEGER, NONZERO numbers. The fraction has to be in it's lowest form (1/2 instead of 2/4).

Based on what we have proven, $a$ and $b$ are both divisible by 3. If we had the fraction $\frac{a}{b}$, it would not be in it's simplest form because we can divide both sides by 3. Therefor, $\sqrt{3}$ cannot be rational since it's proposed fraction isn't in simplest form.

Since it's not rational, it has to be irrational.

Thanks for reading, sorry for it being so long!!!!