Question

can someone prove that root 2 is not a rational number

Answer 1

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

Answer 2

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 .

Answer 3

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.

Answer 4

\[\sqrt{2}=\frac{a}{b} \]

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

Answer 5

multiply both sides by b^2 *

Answer 6

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

Answer 7

prove this a^2 implies a is even :)

Answer 8

a^2 is even implies a is even

Answer 9

can you prove it?

Answer 10

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.