## anonymous one year ago can someone prove that root 2 is not a rational number

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1. taramgrant0543664

Here this link goes through how to prove it: http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.485329.html

2. freckles

You can suppose it is rational. That means it will be of the form a/b where a and b are integers . You should wind up with a contradiction which will prove it is irrational .

3. freckles

The key is using a/b is already a reduced fraction like a and b should have no common factors except one (or - one) of course.

4. freckles

$\sqrt{2}=\frac{a}{b}$ So hint: square both sides and multiply both sides by b say something about a factor of a and then you will say something about a factor of b eventually and you should see that a and b actually do have a common factor and that is where you have your contradiction

5. freckles

multiply both sides by b^2 *

6. xapproachesinfinity

first we suppose it is rational so we can write as $$\sqrt{2}=\frac{a}{b}$$ where a and b are integers in reduced form means cannot be factored in any form we square both sides $$2=\frac{a^2}{b^2}$$ then we get $$a^2=2b^2$$ this implies that a^2 is even hence a must be even (we said a cannot be reduced but this says the opposite already) since a is even then a=2k so 4k^2=2b^2 ===> b^2=2k^2 hence b^2 is even then it must be that b is even aha but we said b is irreducible! so according to this a and b have a common factor! contradiction! hence root2 is not rational

7. xapproachesinfinity

prove this a^2 implies a is even :)

8. xapproachesinfinity

a^2 is even implies a is even

9. xapproachesinfinity

can you prove it?

10. zzr0ck3r

Suppose $$a$$ is odd. Then $$a=2k+1$$ for some $$k\in \mathbb{Z}$$. So $$a^2 = (2k+1)^2=4k^2+4k+1 = 2(2k^2+2k)+1$$. Since $$2k^2+2k\in \mathbb{Z}$$ we have that $$2(2k^2+2k)+1$$ is odd. So by contrapositive we are done.