Is the Square Root of 2 an Irrational Number?

[ltkach]

SquareRoot 2 is Irrational?

\sqrt{}2 I've attached an image of what I'm talking about. Tell me what you think.

 

[micromass]

Yes, that is the standard proof.

 

[DrClaude]

Aren't you basically saying that assuming that $\sqrt{2}$ can be written as a rational number, then it is a rational number?

 

[micromass]

Aren't you basically saying that assuming that $\sqrt{2}$ can be written as a rational number, then it is a rational number?

No. http://en.wikipedia.org/wiki/Proof_by_contradiction

 

[DrClaude]

No. http://en.wikipedia.org/wiki/Proof_by_contradiction

Thanks. The last part, namely showing the contradiction, was missing from the OP.

 

[rbj]

Thanks. The last part, namely showing the contradiction, was missing from the OP.

i didn't think that the proof was complete until you express m and n as products of prime numbers. because of the squaring, there will always be an even number of any prime factor in both m² and n². but with the extra 2 (or whatever the prime number) on one side, you can show that equality is not possible, thus the proof by contradiction.

it's the same for the square root of any prime number. it cannot be rational.

 

[ltkach]

I have no experience with upper level math. I am barely in ODE. Anyways, someone showed me this and I thought it was amazing. Basically everything I learned is wrong.

 

[ltkach]

what are it's implications in math?

 

[DrClaude]

Basically everything I learned is wrong.

Had you learned that $\sqrt{2}$ was rational?

 

[Number Nine]

it's the same for the square root of any prime number. it cannot be rational.

It's the same for any natural number that is not a perfect square (actually, the n'th root of any natural number that is not a perfect n'th power is irrational).

FYI. An easier proof uses the rational zero theorem. Consider the possible rational roots of the polynomial $x^2 - 2 = 0$.

 

[ltkach]

wow i cannot believe myself. yeah ignore me.