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Proof Check! (irrational number)

  1. Sep 16, 2006 #1
    Hi, can somebody please check my proof(s). I am pretty sure they are right...but i just feel they are too elelmentary.

    1) Prove that 12 is irrational

    -Let x be a rational number
    -Therefore, x = {a/b: m in Z and n in N} n not 0.
    -m and n have to be in the LOWEST FORMS...therefore, m and n cannot be divided any further and have no divisors

    let x^2 =12
    then (a/b)^2 = 12
    then a^2/b^2=12
    then (12)(b^2) = a^2
    Contradiction since b^12 is divisiable by 12, hence b is divisable by 12

    2. Prove that |x + y|^2 + |x - y|^2 = 2|x|^2 + 2|y|^2

    |x + y|^2 + |x - y|^2
    =(x+y)(x+y) + (x-y)(x-y)
    =xx + 2xy + yy + xx -2xy + yy
    =2xx +2yy
    =2|x|^2 + 2|y|^2

    Are the above correct?
  2. jcsd
  3. Sep 16, 2006 #2

    matt grime

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    1. you mean sqrt(12), and no, because b^12 is divislbe by 12 does not mean b is divisible by 12, though you mean a^2, not b^12 anyway, and from where you've got the only thing you can say is that 12 divides a^2. Note 12 divides 6^2 in case you still don't believe me on the first one.

    Clearly you only need to show that sqrt(3) is irrational anyway.

    2. Why are there absolute value signs in there? They are not necessary, perhaps you don't mean absolute value of real numbers, but something to do with complex numbers?
  4. Sep 16, 2006 #3
    oops... I made a typo!

    for 1. I actually meant (12)(b^2) is divisiable by 12 hence b is divisable by 12 and so 12 is irrational.

    for 2... the question in the book actually asked "Prove that |x + y|^2 + |x - y|^2 = 2|x|^2 + 2|y|^2 with x,y in R^k (R = real numbers). Interpret this geometrically, as a statement about parallelograms."

    I have no idea what all that means as I am jumping really far ahead of lectures...so i won't get into all that now.
  5. Sep 16, 2006 #4

    matt grime

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    1.How did you conclude that 12 divides b^2 from the assertion that 12 divides 12b^2? b=1 as a counter example for instance?

    2. You can't just multiply vectors. You have to use dot products. The proof though, is identical once you state what the terms are (and put the dots in).
  6. Sep 16, 2006 #5
    1. Ohh! I was confused over this :uhh: but it should be a is divisible by 12.

    a^2 = 12b^2 means a is divisible by 12 and not b being divisiable by 12!

    2. We didn't get this far in lecture...I was trying to be smart and jumping ahead too far :cool: I will re-read your above posts during the week and then it will make more sense to me
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