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Proof that the Square Root of 2 is Irrational.

  1. Jan 9, 2014 #1
    I am trying to prove that √2 is irrational using proof by contradiction. Here is my work so far:

    √2 = p/q where p & q are in their lowest terms. Where q is non-zero.

    2=p2/q2

    2q2 = p2

    Which tells me that p2 is an even number, using the definition of an even number. We can use this definition because we know that q is an integer and an integer squared is also an integer. Hence 2q2 is even and as 2q=p2 we see that p2 is also even. Can we use the knowledge that p is an integer to say that p is also an even number? Or is there another rule that I'm not aware of?

    If we can use this knowledge, then we can say that p=2n, where n is some integer, using the definition of an even number.

    2q2=(2n)2
    2q2=4n2
    q2=2n2 where n2 is an integer, as n was an integer from the definition of an even number.

    ∴ q2 is also an even number. If both p and q are even numbers then the fraction p/q can be simplified further and p/q is not in its lowest terms. This shows that √2 is in fact not a rational number and proves that it is irrational.
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2

    Dick

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    All you need is that the square of an even number is even and the square of an odd number is odd. You can prove that. So if p^2 is even, p must be even since it can't be odd.
     
  4. Jan 9, 2014 #3
    If I take p2 = 6, p = √6 which is why I thought the definition of p as an integer was important, as √6 isn't even.
     
  5. Jan 9, 2014 #4

    Dick

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    Of course, it's important that p is an integer. Now I'm not sure what your question actually is.
     
  6. Jan 9, 2014 #5
    It was really just whether knowing p was an integer was important. Now I know it is, so thank you.
     
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