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

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

Prove that

##### 2 Answers

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

The same technique can be used to show that

Suppose, to the contrary, that

Let

Substituting those in, we get

As neither

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

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 (

Remember an integer is a whole number (1, 2, 3, etc) and ** not** a fraction of any sort (so not 3.592 or

We know that 5 is rational because it can be expressed as

So, let's prove that *never* be a rational number, so this proof should have a contradiction in it.

Let's start with the statement in the previous paragraph.

And now, we divide both sides by 3.

Ok so now we need to use some logic. We know both

Now, this also means that

So, since

an integer

We figured that out through simple substitution of "an integer" and

Ok, so back to the problem. We just figured out that

**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 *prime numbers* that

Here's a pretty simple way to look at it. Let's say that

So, now we know that

Let's keep going here. Since we know that

*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:

Now, let's substitute the equation we just made above into the original equation. The original is:

Our new one is:

Let's divide both sides by 3 to get:

And divide both sides by 3 again:

And suddenly we find ourselves repeating ourselves. Remember when we proved that

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

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,

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

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