Question

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

Prove that `sqrt(3)` is irrational by contradiction. E.g. show that it is not rational

Answer 1

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 `a/b`, meaning `3b^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.

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Suppose, to the contrary, that `sqrt(3)` is rational. Then there exist integers `a` and `b` such that `sqrt(3) = 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) = a/b`

`=> 3 = (a/b)^2`

`=> 3 = a^2/b^2`

`=> 3b^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 = a/3^(k_a)` is an integer not divisible by `3`. Similarly, let `b = 3^(k_b)b_0` where `b_0 = b/3^(k_b)` is an integer not divisible by `3`.

Substituting those in, we get

`3(3^(k_b)b_0)^2 = (3^(k_a)a_0)^2`

`=> 3^(2k_b+1)b_0^2 = 3^(2k_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

`2k_b+1 = 2k_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. ∎

Answer 2

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 (`1/2` instead of `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 `1/2`)

We know that 5 is rational because it can be expressed as `10/2` or `5/1`. 4.333333333(...) is rational because it can be expressed as `13/3`.

So, let's prove that `sqrt(3) = 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) = 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`

`3b^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`.

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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.

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Let's keep going here. Since we know that `a` is divisble by `3`, we can make this equation.
`a = 3c`, 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 = 9c^2`

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

Our new one is:
`3b^2 =9c^2`

Let's divide both sides by 3 to get:
`b^2 =3c^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.

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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 `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!!!!