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Proof that sqrt(2) is not a rational number

  1. Jun 14, 2014 #1
    1. Prove that [itex]\sqrt{2}[/itex] is not a rational number using the rational zero theorem



    2. Relevant equations: x2-2



    3. The attempt at a solution

    So, I was thinking to just show that the only rational numbers that have the form [itex]\frac{c}{d}[/itex] are [itex]\pm[/itex] 1, [itex]\pm[/itex] 2, but [itex]\sqrt{2}[/itex] is not any of those numbers, so it must be irrational? I feel like I'm supposed to have more of a justification as to why [itex]\pm 1, 2[/itex] are the only possible numbers
     
  2. jcsd
  3. Jun 14, 2014 #2
    I have no idea what you are saying in your attempted work. All rational numbers have the form c/d, by definition. What exactly do you mean?

    Try a proof by contradiction. That is, make the assumption that √2 can be expressed as a fraction a/b, where a and b are coprime integers (so that the fraction is in lowest terms). Can you find a contradiction?
     
  4. Jun 14, 2014 #3
    I was using the fact that any rational solutions of a polynomial equation must be an integer that divides the last coefficient c0, which in this case is -2. So the only rational solutions in the form [itex]\frac{c}{d}[/itex] where c divides -2 and d divides 1 are [itex]\pm 1,2 [/itex]. [itex]\sqrt{2}[/itex] solves the polynomial, but it does not divide [itex]\pm 1,2 [/itex] so it must be irrational.

    But trying to do a proof by contradiction...

    Assume [itex]x=\sqrt{2}[/itex] is rational. Then, that means [itex]\sqrt{2}[/itex] must be in the form of [itex]\frac{a}{b}[/itex]?

    Am I on the right track?
     
  5. Jun 14, 2014 #4

    Ray Vickson

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    You proof is OK if you are permitted to use that result about polynomial roots. However, there are other proofs that do not rely on that polynomial-root result, and maybe (or maybe not) the person posing the question wanted you to re-discover the old proof that the Greeks knew more than 2000 years ago. You started it, but did not finish. First, in the fraction ##a/b##, assume ##a,b## are positive integers having no common integer factors > 1; in other words, the fraction is in 'lowest' form. So, what does the equation ##a^2 = 2 b^2 ## tell you?
     
  6. Jun 14, 2014 #5
    Okay so that would mean that a is an even number, so it technically be written in the form 2a? And if it is written like that, then the equation can take the form 4a2 = 2b2. Thus, b is also an even number.Since they are both even numbers, this means that they have not been factored to their lowest possible form.

    I don't know what to do from here
     
  7. Jun 14, 2014 #6
    Can I show that a/b cant equal 1?
     
    Last edited: Jun 14, 2014
  8. Jun 14, 2014 #7
    Hi Jef123.

    The square root of 2 being proved to be irrational is usually done through proof by contradiction i.e. proving that the square root of 2 can not be rational.

    Suppose that √2 were a rational number. Then we can write it √2 = a/b where a,b are whole numbers, b not zero and that a/b is in lowest terms. From the equality √2 = a/b it follows that 2 = a^2/b^2, or a^2 = 2 * b^2. From this we can know that a itself is also an even number. Then a is 2 times some other whole number, or a = 2k where k is this other number. Then, substitute that into the original equation and see what happens!
     
    Last edited: Jun 14, 2014
  9. Jun 14, 2014 #8

    mfb

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    ##\frac{a}{b} \neq 1## is both true and irrelevant.

    You found a contradiction, done. You shouldn't use "a" twice with two different values, however. Use a' or something else.
     
  10. Jun 14, 2014 #9
    Hi SpecialPerson,

    So after substituting a = 2k into the equation, I get 4k^2=2b^2. This means that a and b are still divisible and not in lowest terms. Is that the contradiction? Because both a and b are even numbers, meaning they can still be factored
     
  11. Jun 14, 2014 #10
    Hi Jef123.

    If we substitute a = 2k into the original equation 2 = a^2/b^2, this is what we get:
    2 = (2k)^2/b^2
    2 = 4k^2/b^2
    2*b^2 = 4k^2
    b^2 = 2k^2

    This means b^2 is even, from which follows again that b itself is an even number! This is a contradiction because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.
     
  12. Jun 14, 2014 #11
    Yes, this is the contradiction to find. You have found that a fraction representation of √2 would be a ratio of even integers in any representation, but that is impossible, so such a representation does not exist.

    The fancy name is "infinite descent."

    Though I want to iterate what mfb said. Do not write 2a for a, pick a new letter, writing 2a is both false and confusing to someone reading your proof.
     
  13. Jun 14, 2014 #12

    AlephZero

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    Well, the way I read the question, it doesn't say its "OK" to use it, it tells you to use it.

    Rediscovering Euclid's proof is educational, but if the question was in an exam you would deserve to get zero marks for not doing what you were asked to do!

    The fact that the first response apparently didn't know what the rational roots theorem was, is just "stuff that happens", not a reason to get sidetracked.
     
  14. Jun 14, 2014 #13
    Wow, it does explicitly tell you to use it! I think we all missed that.
     
  15. Jun 14, 2014 #14
    I missed that completely.

    Since sqrt 2 is a root of the polynomial f(x) = x^2 - 2, the rational root theorem tells us that the rational roots of f(x) are of the form + or - 1.2/1. It is easy to check that none of these are roots of f(x), hence has no rational roots. Thus, is not rational.
     
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