anonymous
  • anonymous
can someone prove that root 2 is not a rational number
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
taramgrant0543664
  • taramgrant0543664
Here this link goes through how to prove it: http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.485329.html
freckles
  • 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 .
freckles
  • 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.

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freckles
  • 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
freckles
  • freckles
multiply both sides by b^2 *
xapproachesinfinity
  • 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
xapproachesinfinity
  • xapproachesinfinity
prove this a^2 implies a is even :)
xapproachesinfinity
  • xapproachesinfinity
a^2 is even implies a is even
xapproachesinfinity
  • xapproachesinfinity
can you prove it?
zzr0ck3r
  • 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.

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