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The irrationality of the square root of 2

  1. Jan 18, 2005 #1
    I saw this proof in class today to prove the square root of 2 is irrational:

    1. Assume that √2 is a rational number. Meaning that there exists an integer a and b so that a / b = √2.
    2. Then √2 can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)^2 = 2.
    3. It follows that a^2 / b^2 = 2 and a^2 = 2 b^2.
    4. Therefore a^2 is even because it is equal to 2 b^2 which is obviously even.
    5. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
    6. Because a is even, there exists a k that fulfills: a = 2k.
    7. We insert the last equation of (3) in (6): 2b^2 = (2k)2 is equivalent to 2b^2 = 4k^2 is equivalent to b2 = 2k^2.
    8. Because 2k^2 is even it follows that b^2 is also even which means that b is even because only even numbers have even squares.
    9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

    This was copied from wikipedia by the way. I mean I know this works or it wouldn't have been around for as long as it has. I read this and it makes sense I understand why the square root of 2 is irrational. But it seems to me that you can use any number in there, take 4 for example, and reach the same conclusion when clearly the square root of 4 is rational. Can someone clarify this for me?
     
  2. jcsd
  3. Jan 18, 2005 #2

    AKG

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    4 won't work. Did you even try it? Everywhere where I've bolded the 2, you should replace it with 4. If you do that, then the underlined statement no longer holds. What you actually get, if we rewrite 7.

    7. We insert the last equation of (3) in (6): 4b^2 = (2k)2 is equivalent to 4b^2 = 4k^2 is equivalent to b2 = k^2. After this, the proof for [itex]\sqrt{2}[/itex] claims that b² is even since it would be equal to 2k². In this case, it is equal to 1k², so we can't claim that it's even, and so we don't get a contradiction from the original assumption that [itex]\sqrt{4} \in \mathbb{Q}[/itex].
     
  4. Jan 18, 2005 #3
    But for line 4, wouldn't you have a^2 is divisible by 4? And then in line 7 you would get 4b^2 = 16k^2?
     
  5. Jan 18, 2005 #4

    AKG

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    Look at line 6. For the case with [itex]\sqrt{2}[/itex], we have that a² is a multiple of 2, therefore a is a multiple of 2. In this case, you have that a² is a multiple of 4. What does that tell you about a? It does not tell you that a is a multiple of 4, therefore it does not tell you that there is a k such that a = 4k, therefore it does not tell you that there is a k such that 4b² = (4k)² = 16k².
     
  6. Jan 18, 2005 #5
    So what you are saying is that this proof only works with prime numbers?
     
  7. Jan 18, 2005 #6

    dextercioby

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    Nope,it works with any non perfect square numbers.24 is not prime,yet [itex] \sqrt{24} [/itex] is irrational.

    Daniel.
     
  8. Jan 18, 2005 #7

    AKG

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    No, that's not what I'm saying at all. When did I say that? I said exactly what I said, hopefully you see why the proof doesn't "work" if you use 4 instead of 2. In fact this proof on it's own, as far as I can tell, will only work for 2. If you plugged in 3, then you would get:

    a² = 3b²

    Whether or not a² is even depends on whether or not b² is even, whereas when a² = 2b², it obviously doesn't. You would need to do more than what is above to prove [itex]\sqrt{3}[/itex] to be irrational. However, the above proof is a sound proof that [itex]\sqrt{2} \in \overline{\mathbb{Q}}[/itex] and cannot be used to prove that [itex]\sqrt{4} \in \mathbb{Q}[/itex].
     
  9. Jan 18, 2005 #8

    AKG

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    It works with any non perfect square numbers? What do you mean? It doesn't work with square numbers at all.
     
  10. Jan 18, 2005 #9

    Tide

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    Assume [itex]\sqrt 2[/itex] is rational so that [itex]\sqrt 2 = \frac {p}{q}[/itex] where p and q are relatively prime integers. It follows that [itex]2 q^2 = p^2[/itex]. From the Fundamental Theorem of Arithmetic (Unique Factorization Theorem) [itex]p^2[/itex] and [itex]q^2[/itex] must each have only even powers of prime factors so it is impossible for [itex]2 q^2 = p^2[/itex] since the left side has an odd number of prime factors while the right side has an even number of prime factors. Hence, [itex]\sqrt 2[/itex] cannot be rational.
     
  11. Jan 18, 2005 #10

    Curious3141

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    He means numbers that are not perfect squares, i.e. non-(perfect square). :tongue2:
     
  12. Jan 19, 2005 #11

    AKG

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    Ah, I see. I thought he meant numbers that are square and non-perfect
     
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