Proof about an irrational number.

[cragar]

Homework Statement

Prove that [itex] \sqrt{3} [/itex] is irrational.

The Attempt at a Solution

SO I will start by assuming that [itex] \sqrt{3} [/itex] is rational and i can represent this as
[itex] 3=\frac{b^2}{a^2} [/itex] and I assume that a and b have no common factors.
so now I have [itex] 3b^2=a^2 [/itex]
but this is not possible because if a and b have no common factors.
I probably need to add more to this, what do you guys think?

 

[jedishrfu]

First you must state let's assume that its rational then that √3 = a / b
And that a / b is reduced to lowest form then square both sides.

Next You need to reach a contradiction about a and b.

Check out the sqroot of 2 proofs online to get an idea.

 

[Slats18]

You're doing great at the moment, there's just a few more steps to the proof. Consider two cases: a,b are even, and a,b are odd. a,b being even is very similar to the irrationality proof of sqrt(2), just as jedishrfu said. So, now consider when they are odd, and substitute simple expressions for a,b to show they are odd numbers. See where you go from there =)

 

[cragar]

ok I looked at the [itex] \sqrt{2} [/itex] proofs. And I saw how they reached a contradiction about both if of them are even and that would imply they shared a common factor. But as for [itex] \sqrt{3} [/itex] a and b could both be odd .
What about this. Since 3 is prime and a and b are both integers. the only way to divide 2 integers to get a prime number is to have a and b share common factors, therefore this is a contradiction. Will this work. maybe its recursive

 

[Dick]

ok I looked at the [itex] \sqrt{2} [/itex] proofs. And I saw how they reached a contradiction about both if of them are even and that would imply they shared a common factor. But as for [itex] \sqrt{3} [/itex] a and b could both be odd .
What about this. Since 3 is prime and a and b are both integers. the only way to divide 2 integers to get a prime number is to have a and b share common factors, therefore this is a contradiction. Will this work. maybe its recursive

If b^2=3*a^2 then b is divisible by 3. Why? Keep going from there.

 

[cragar]

thanks for everyones help.
ok so b is divisible by 3 because b*b=3a^2 . so now I let b=3r where r is an integer.
so now 9r^2=3a^2 , and then 3r^2=a^2, and now this is saying that a is divisible by 3, which is a contradiction because we assumed at the start that a and b shared no common factors.

 

[Dick]

thanks for everyones help.
ok so b is divisible by 3 because b*b=3a^2 . so now I let b=3r where r is an integer.
so now 9r^2=3a^2 , and then 3r^2=a^2, and now this is saying that a is divisible by 3, which is a contradiction because we assumed at the start that a and b shared no common factors.

That's it alright.