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Prove the Square Root of 2 is irrational

  1. Sep 8, 2005 #1
    This is Algebra 2 question...

    I have to prove that the square root of 2 is irrational...

    First we must assume that

    sqrt (2) = a/b

    I never took geometry and i dont know proofs...

    Please help me.

    Thank you.
     
  2. jcsd
  3. Sep 8, 2005 #2
    a rational number is of form a/b where a and be are mutually prime. I will give you a hint: you must prove that and and b cannot possibly be mutually prime.

    and what does this have to do with geometry?
     
  4. Sep 8, 2005 #3
    You're off to a good start. Let "a" and "b" be natural numbers.

    [tex]\sqrt{2}=\frac{a}{b}\implies b\sqrt{2}=a[/tex]

    Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. This in turn, would mean that "b" isn't a natural number.

    Can you see where this is going? You need to prove this by contradiction.
     
    Last edited: Sep 8, 2005
  5. Sep 9, 2005 #4

    HallsofIvy

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    "Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. This in turn, would mean that "b" isn't a natural number."
    How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that [itex]\sqrt{2}[/itex] is irrational.

    Better to note that if [itex]\frac{a}{b}= \sqrt{2}[/itex] then, squaring both sides, [itex]\frac{a^2}{b^2}= 2[/itex] so that a2= 2b2 showing that a2 is even.

    Crucial point: the square of an odd integer is always odd:

    If p is an odd integer, then it can be written 2n+ 1 where n is any integer.

    p2= (2n+1)2= 4n2+ 4n+ 1= 2(2n2+2n)+1.

    Since 2n2+ 2n is an integer, p2 is of the form 2m+1 (m= 2n2+2n) and so is odd.

    Do you see why knowing that tells us that a must be even?
     
    Last edited: Sep 9, 2005
  6. Sep 9, 2005 #5
    Wow!! You are so good at teaching! Thank you everybody!
     
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