A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

can someone prove that root 2 is not a rational number

  • This Question is Open
  1. taramgrant0543664
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  5. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    multiply both sides by b^2 *

  6. xapproachesinfinity
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    prove this a^2 implies a is even :)

  8. xapproachesinfinity
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a^2 is even implies a is even

  9. xapproachesinfinity
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you prove it?

  10. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.