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Proof of irrationality of sqrt(2)

  1. Jun 4, 2015 #1
    There's something I don't understand in the popular "proof by contradiction" of sqrt(2) after the following step:

    a2/b2 = 2

    The above equation implies that a2 is even. Fair enough. But the way I see it, the above equation also makes it impossible for a to be even, as


    This implies that a should actually be an irrational number as it is a multiple of sqrt(2). Though it seems obvious to me that if some number n^2 is even, then n must also be even, but in the context of the equation a2=2b2, it doesn't seem possible.

    Shouldn't this be the end of the proof? We started by assuming that a and b are whole numbers and a/b can be used to represent sqrt(2), but found that a, at least, is actually irrational. Doesn't that count as a contradiction?

    I'm really confused. :(
  2. jcsd
  3. Jun 4, 2015 #2
    It does, but you don't know that a priori. At this stage of the proof, you don't know yet that ##\sqrt{2}## is irrational.
  4. Jun 4, 2015 #3
    Ohhhhh! I'm an utter idiot! I can't help laughing at myself now. Literally. :p

    Thank you so much, micromass! I really appreciate it!
  5. Jun 4, 2015 #4
    One another thing I would like to ask. When we say that we are assuming an irreducible fraction, are we just saying it? I'm trying to say that our math should be aware of the assumptions we are making so when we write down the premise of the proof, we should be translating that assumption into mathematical form.

    How does the math know that a/b is not supposed to be a reducible fraction?
  6. Jun 4, 2015 #5
    The irreducibility of a/b is assumed by the prover. At a later stage in the proof we show that a and b must have a common factor - there lies the contadiction that is the heart of the proof.
  7. Jun 4, 2015 #6


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    The proof goes like this: Assume that [itex](\frac{a}{b})^{2} = 2[/itex] where a and b are integers (and b≠0). We can also assume that the fraction is irreducible (a and b have no common factors). Then [itex] a^{2}=2\cdot b^{2}[/itex], which implies that a2 is even. But the only way a2 can be even is that a is even. Therefore you can write a = 2⋅p, where p is an integer. This gives [itex](\frac{2p}{b})^{2}=2 [/itex] or [itex] 4\cdot p^{2} = 2\cdot b^{2}[/itex] which gives [itex]b^{2}=2\cdot p^{2} [/itex] which implies that b2 is even. But the only way b2 can be even is that b is even. Therefore you can write b = 2⋅q, where q is an integer. But this again says that [itex]\frac{a}{b}=\frac{2p}{2q} [/itex] which means that a and b have a common factor (2) which is contrary to the assumption.
  8. Jun 4, 2015 #7
    Thanks I understand pretty much all of the proof now except the use of assumption which is troubling me.

    The first step of the proof is:

    sqrt(2) = a/b.

    Then we just tell ourselves that a/b is an irreducible fraction. There should be a way of letting the math know precisely what assumptions we are making because the statement (a/b)^2 = 2 could mean anything, so if it gives us a fraction with common factors, it's not really a contradiction because we only made that assumption verbally.

    I'm sorry if I sound obnoxious. I'm just a bit of a dimwit.
  9. Jun 4, 2015 #8
    No we don't, we make sure it is an irreducible fraction. How? You have given the answer yourself...
    then what do you think you should do?
  10. Jun 4, 2015 #9
    I'm sorry I don't get it. How do we make sure it is an irreducible fraction?
  11. Jun 5, 2015 #10


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    Basis knowledge of fractions: If there is a common factor in the numerator and denominator, we can remove it in both places (if the common factor is c, we have [itex]\frac{a}{b}=\frac{c\cdot p}{c\cdot q}= \frac{c}{c}\cdot\frac{p}{q}=1\cdot \frac{p}{q} [/itex].
  12. Jun 5, 2015 #11
    I think the thing I'm having trouble grasping is how we are to make sure that a/b is already in its reduced form when we start the proof. Sure we could cancel it out later (even a and b can be reduced when we find they are both even), but I think it is important to the proof that a/b be already written in its reduced form. For example, if I write x = p/q, does that guarantee that I will get a value of x that is already reduced to it lowest form?
  13. Jun 5, 2015 #12


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    No. But you know that if it is not, you can reduce it.

    You have to make some assumption when it comes to fractions, otherwise you will have to deal with an infinity of fractions all representing the same number. For example 1 = n/n for all n.
  14. Jun 5, 2015 #13
    Right. Thank you very much!
  15. Jun 5, 2015 #14


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    Well, maybe there's one number for which this isn't true... All is a very general term.
  16. Jun 5, 2015 #15


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    Yes. I was a bit sloppy there. I should have said (∀n∈ℕ), but that is a bit unreadable.
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