Proving \sqrt{3} is Irrational

[Mentallic]

Homework Statement

I need to prove that \sqrt{3} is irrational.

The Attempt at a Solution

The prior problem was to show \sqrt{2} is irrational and the solution had to do with a contradiction that each number must be even or something (frankly, I didn't understand it too well). But I don't see how I can apply the same idea to this one, since evenness is not applicable here.

All i can really do so far is assume \sqrt{3}=\frac{p}{q} where p and q are integers.
Now, by squaring, 3q^2=p^2.

I am completely lost here though...

Some help please?

 

[CompuChip]

The proof of case with 2 depends on the fact that either p is divisible by 2 (and then you can show that q is too, which contradicts the assumption that you choose them to be "minimal"), or it is not (and since 2q^2 = p^2 tells you that it is, you have a contradiction).

The proof for 3 is similar: p is either divisible by 3, and then so is q contradicting our assumption; or p is not divisible by 3 and then it must be divisible by 3 which is also impossible.

Perhaps it is a good idea to go through the proof of the irrationality of \sqrt{2} once more, and try to find out what part confuses you. Can you try and explain it to us?

 

[Mentallic]

With the \sqrt{2} question the solution says that if 2q^2=p^2 then p^2 is even, thus p is even, so we can put p=2r for integer r, so now we have 2q^2=4r^2 which also implies that q is even but this is the contradiction.

Actually.. while writing this I tested to see what the difference is for rational cases like \sqrt{4} and \sqrt{9}. Yes, now I understand it. It wasn't the evenness of \sqrt{2}'s example that was special, it was the fact that it was divisible by 2 (the same thing I know ).

 

[CompuChip]

It wasn't the evenness of \sqrt{2}'s example that was special, it was the fact that it was divisible by 2 (the same thing I know ).

Yes, in this case it is the same.
But in the general case \sqrt{n}, you want to say something about the divisibility by n.

(Note that in passing you have also shown that the square of an even number is always divisible by not just 2, but by 4).

 

[icystrike]

Homework Statement

I need to prove that \sqrt{3} is irrational.

The Attempt at a Solution

The prior problem was to show \sqrt{2} is irrational and the solution had to do with a contradiction that each number must be even or something (frankly, I didn't understand it too well). But I don't see how I can apply the same idea to this one, since evenness is not applicable here.

All i can really do so far is assume \sqrt{3}=\frac{p}{q} where p and q are integers.
Now, by squaring, 3q^2=p^2.

I am completely lost here though...

Some help please?

If you add an additional information that gcd(p,q)=1 , which also means the greatest common divisor of p and q is 1 . However for \sqrt{2} you can draw the conclusion that the gcd(p,q) is atleast 2 and for \sqrt{3} gcd(p,q) is atleast 3. In fact you can prove a generalised one analogously which states the \sqrt{prime} will lead you to gcd(p,q) is atleast the prime number , yet definition of prime states it cannot be 1 , will leads to the irrationality of all prime.