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

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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0Here this link goes through how to prove it: http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.485329.html

freckles
 one year ago
Best ResponseYou've already chosen the best response.1You 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 .

freckles
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{2}=\frac{a}{b} \] 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1multiply both sides by b^2 *

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0first 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0prove this a^2 implies a is even :)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0a^2 is even implies a is even

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0can you prove it?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Suppose \(a\) 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.
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