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Homework Help: Square Root of an Irrational Number is Irrational

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Let a be a positive real number. Prove that if a is irrational, then √a is
    irrational. Is the converse true?


    2. Relevant equations
    So, an irrational number is one in which m=q/p does not exist. I understand that part, but then trying to show that the square root of an irrational number is irrational is giving me problems. I would think this needs to be done by contradicition

    Thanks!!!
     
  2. jcsd
  3. Feb 23, 2012 #2

    LCKurtz

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    More carefully, what you mean is that there are not integers p and q such that m = q/p.

    I would think so to. So try it...
     
  4. Feb 23, 2012 #3
    Here's what I have so far:

    Proceed by way of contradiction. Assume a is rational, so a=x/y. So

    √a=p/q
    √(x/y)=p/q
    x/y=p^2/q^2
    q^2(x/y)=p^2

    This is where I don't know where to go from here. Normally, I would take advaadvantagetnage of knowing something about a, that is was prime or even for example, but I don't know what I an say to show that there is indeed a contradiction.

    Thanks
     
  5. Feb 23, 2012 #4

    Dick

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    Which part are you trying prove?
     
  6. Feb 23, 2012 #5
    I'm trying to show that if a is a rational number then √a can not be rational (the contradiction). Like I said in the OP, I'm pretty lost with this one.
     
  7. Feb 23, 2012 #6
    Could I say a=√2, which is an irrational number?
     
  8. Feb 23, 2012 #7

    Dick

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    That's not really the contradiction. And you can't prove it because it's not true. Can't you think of a simple counterexample? What's the correct contradiction (or contrapositive)?
     
  9. Feb 23, 2012 #8

    Dick

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    4 is rational, so is √4=2. The contradiction of "if a is irrational, then √a is
    irrational" is "if √a is NOT irrational then a is NOT irrational".
     
  10. Feb 23, 2012 #9
    You mean contrapositive, right? A proof by contradiction would be like "if √a is not irrational and a is rational, then (deductive logic), and thus (clearly false statement), but (point out why the statement is false) and thus contradiction."

    But yeah, use the contrapositive.
     
  11. Feb 23, 2012 #10

    Dick

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    Yes, of course, I meant contrapositive. I said that in post 7. I was trying to follow the OP's words and may have muddled this up more than I helped. Thanks.
     
  12. Feb 24, 2012 #11
    It could be like this. First you prove that something like √2 is irrational.
    Then let, on contrary, √(√2) = p/q where p and q are co-prime
    Square both sides, √2= p^2/q^2=(p/q)^2
    But you know that the square of any fraction which contains co-prime can't be irrational or something with an under root.
    So, this contradiction has arise because our assumption was wrong.
     
  13. Feb 24, 2012 #12

    SammyS

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    By contradiction.

    Suppose that a is irrational but that √(a) is rational.

    Then √(a) that [itex]\displaystyle \sqrt{a}=\frac{p}{q}\,,[/itex] for some integers p and q .

    Now consider what all this says about [itex](\sqrt{a})^2\,.[/itex]
     
  14. Jun 12, 2013 #13
    let √a be rational
    √a=p/q
    this √a,being a rational no. must be between 2 integers,so
    I< p/q <I+1
    =Iq<p<q(I+1)
    =0<p-Iq<q

    now,as p and q are integers , therefore (p-Iq) is also an integer,so
    when it is multiplied by a rational no. ,it will be a rational no i.e.
    k= (p-Iq)*(p/q)........(integer)*(rational no), this will be a rational no.
    now,k=(p^2)/q -Ip
    also,k={(p^2)/(q^2)}q - Ip
    then,k=aq - Ip..................................as (p^2)/(q^2)=a
    now,k=(irrational)*integer - integer
    hence k is also irrational,thereby arriving at a contradiction,where before it was
    supposed to be rational .. ..
    so ,therefore √a is also irrational when a is rational
     
  15. Jun 13, 2013 #14

    HallsofIvy

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    This
    Is NOT true. If √a= 3 it does not lie between I and I+1 for any integer I.

    Why not simply "if [itex]\sqrt{a}= m/n[/tex]" then a= what?
     
  16. Jun 14, 2013 #15
    let √a be rational,(including integers)
    √a=p/q
    this √a,being a rational no. must be between any 2 consecutive integers,so
    I≤ p/q <I+1
    =Iq≤p<q(I+1)
    =0≤p-Iq<q
    now let us take, p- Iq=0
    so,I=p/q=√a
    also,I^2=a ,which makes "a"(which is irrational) ,an integer.hence a contradiction.
    so,excluding the equality case,we have
    =0<p-Iq<q........................(A)
    now,as p and q are integers , therefore (p-Iq) is also an integer(lets say I1),so
    when it is multiplied by a rational no. ,it will be a rational no i.e.
    k= (p-Iq)*(p/q)........{integer(I1)}*(a rational no,whose denominator is an integer less than I1)..........................from (A)
    this "k" will be a rational no.(since q cannot be a factor of I1)
    now,k=(p^2)/q -Ip
    also,k={(p^2)/(q^2)}q - Ip
    then,k=aq - Ip..................................as (p^2)/(q^2)=a
    now,k=(irrational)*integer - integer
    hence k is also irrational,thereby arriving at a contradiction,where before it was
    supposed to be rational .. ..
    so ,therefore √a is also irrational when a is rational

    hope ,this time its correct.........
     
  17. Jun 14, 2013 #16

    Dick

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    Why are you making this so complicated? If sqrt(a)=p/q, then sure a=p^2/q^2. Which is rational. So if a is irrational, then mustn't it be that sqrt(a) is irrational? It's just a simple proof by contradiction. I don't know what all the extra complication is about.
     
  18. Jun 15, 2013 #17

    D H

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    That's proof by contraposition: Prove ⌉q→⌉p and you have proved p→q. Proof by contradiction would involve showing that the statement "there exists an irrational number a whose square root is rational" results in a contradiction.
     
  19. Jun 15, 2013 #18

    Dick

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    I think it's a fine distinction. "there exists an irrational number a whose square root is rational" implies a is rational. Isn't that a contradiction?
     
  20. Jun 15, 2013 #19

    D H

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    Sure. But you don't need to contradiction to prove this. All one has to prove is that if √a is rational then a is a rational. By contraposition, if a is irrational then √a is irrational. QED. There's no need to look at the inverse of "if a is irrational then √a is irrational" and find a contradiction because the two statements "if √a is rational then a is rational" and "if a is irrational then √a is irrational" are one and the same, just worded differently.
     
  21. Jun 15, 2013 #20
    The way I would do this is prove that the square root of any non-perfect square is irrational. This is easily done with prime factorization in a couple of lines.

    Then, since the definition of a perfect square requires that it is a natural number, the proof is complete.
     
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