# SquareRoot 2 is Irrational?

1. Mar 30, 2013

### ltkach

SquareRoot 2 is Irrational???

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

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2. Mar 30, 2013

### micromass

Staff Emeritus
Yes, that is the standard proof.

3. Mar 30, 2013

### Staff: Mentor

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

4. Mar 30, 2013

### micromass

Staff Emeritus
5. Mar 30, 2013

### Staff: Mentor

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

6. Mar 30, 2013

### rbj

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.

7. Mar 30, 2013

### 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.

8. Mar 30, 2013

### ltkach

what are it's implications in math?

9. Mar 30, 2013

### Staff: Mentor

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

10. Mar 30, 2013

### Number Nine

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 $x^2 - 2 = 0$.

11. Mar 30, 2013

### ltkach

wow i cannot believe myself. yeah ignore me.