Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof help please!

  1. Oct 5, 2005 #1
    Hi can anyone help me check if I've approached this question correctly and offer any help on part b) of the question? Thanks! :smile:

    a) Prove that if n is an integer and n^3 is a multiple of 2 then n is a multiple of 2.

    Let n^3 be a multiple of 2 but suppose n is not a multiple of 2.

    then n= 2k+1

    => n^3 = (2k+1)^3

    = (4k^3+4k+1)(2k+1)
    = 8k^3+12k^2+6k+1
    = 2(4k^3+6k^2+3k)+1
    = 2m + 1 where m = 4k^3+6k^2+3k

    ==> n^3 is not a multiple of 2

    therefore by contradiction n^3 is a multiple of 2 and n must also be a multiple of 2.

    Is that correct? :confused: plz correct me if im wrong, im not too good at proof.

    b) Deduce that 3 (sqrt 2) is irrational.
  2. jcsd
  3. Oct 5, 2005 #2


    User Avatar
    Science Advisor

    a) proof is correct (except for typo in the middle, but it doesn't really matter).

    b) if you have proved sqrt(2) is irrational, then it is trivial to show 3 times is also irrational - you can use the contradiction argument by showing that if 3*sqrt(2) is rational, then sqrt(2) is rational.
  4. Oct 5, 2005 #3
    oh thanks for your help! I'll give the 3(sqrt)2 one a go :smile:

    I've also got another problem trying to get my head round the following type of question. It's so basic but :confused:

    Question :-

    Decide whether this condition is necessary, sufficient or both.

    If statement A: n is a positive integer and n is divisible by 6

    statement B: n is divisible by 12.

    My Answer

    B => A but A does not imply B as n = 18 is a counter example.

    So B => A therefore:

    => n = 12a
    => n = 6b where b is 2a
    => n is divisible by 6

    therefore A is necessary but not sufficient for B.

    Is that right? :uhh:

    Last edited: Oct 5, 2005
  5. Oct 5, 2005 #4


    User Avatar
    Homework Helper


    You understand how proof by contradiction works. Practice.

    Just for fun, here is a direct proof:

    Consider the quantity: n^3 - n = n(n^2-1) = (n-1)n(n+1) , which is the product of 3 consequective integers and is therefore even (for at least one of said integers is even). If n^3 is even, then n must also be even, as their difference is even. What you proved above also follows from this: if n^3 is odd, then so must n be odd, as their difference is even.

    Likely 3 (sqrt 2) was a typo and should be 2^(1/3) (i.e., the cube root of 2) or else it wouldn't be deduced from the above.

    Here is a proof that 3 (sqrt 2) is irrational:

    Suppose not. Then 3*sqrt2 is rational, i.e., there exist positive integers p and q having no common factors such that p/q = 3*sqrt2. Then reason as follows:

    p/q = 3*sqrt2
    => (p/q)^2 = 18
    => p^2 = 18q^2 (eqn #1)
    => p^2 is even
    => p is even

    so put p = 2k (for some positive integer k) in eqn #1 to get

    (2r)^2 = 18q^2
    => 2r^2 = 9q^2
    => q^2 is even
    => q is even, which is a contradiction since p and q have no common factors

    Therefore 3*sqrt2 is irrational.

    Your proof that that 2^(1/3) is irrational will involve something like p^3 is even => p is even :wink: .
  6. Oct 5, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    It looks like the answer is "neither."
    This proves that B => n is divisible by 6, but A is the statement "n is divisible by 6 and n is a positive integer". What if n = -24?
  7. Oct 5, 2005 #6


    User Avatar
    Homework Helper

    Certianly A does not imply B as n = 18 is a valid counter example.

    B => A is true only if B includes "n is a positive integer," since n = -24 would be a counter example if it didn't.

    But were it generally assumed that n is a positive integer, then what you had worked.
  8. Oct 5, 2005 #7


    User Avatar
    Homework Helper

    seems you type faster than I
  9. Oct 5, 2005 #8
    proving that sq(2) is irrational is very simple: 2 is divisible only by 1 and 2. there is no n^2 that can = 0 for n<10 so decimal values squared will never form integer values. 1^2 does not equal 2. 2^2 does not equal 2 no decimal number ^ 2 can equal 2 so there you go.

    0^2=0 but 0 after decimal is neglected

    (considering strictly the last decimal)
  10. Oct 5, 2005 #9
    ahh that was very helpful benorin and AKG, thanks! I think I'm starting to get the idea now :smile:

    Can I ask just one more question? :biggrin:

    Decide whether this condition is necessary, sufficient or both.

    If statement A: n is a positive integer and n is divisible by 6

    statement B: n = 384

    My Answer

    Again I think only B => A because 384 is divisible by 6, A does not necessarily imply B since n divisible by 6 does not imply that n is 384.

    Is there any way I can make this proof better rather than just stating the obvious? Would providing a counter-example be helpful? :confused: And also I'm not sure whether it is necessary or sufficient :uhh:

    Help very much appreciated
  11. Oct 5, 2005 #10
    sufficient. 3840 is divisible by 6 also.
  12. Oct 5, 2005 #11


    User Avatar
    Science Advisor
    Homework Helper

    To be honest, the question seems oddly stated. It says "decide whether this condition is necessary or sufficient." Which condition is "this" condition? It would be nice if the unambiguously asked, "Decide whether statement A is necessary, sufficient, both necessary and sufficient, or neither necessary nor sufficient for statement B." Assuming that is what the meant, then you have to decide, is A necessary for B. That is, can B be true without A being true? If A is necessary for B, then whenever B is true, A is true, so B => A. If A is sufficient for B, then as long as A is true, B will be true, so A => B. So does A => B? You should provide a counterexample (and it's easy enough to do it) to show that A => B is false. Does B => A? Yes it does, and you may want to explicitly say that 6x64 = 384 and that 384 is a positive integer (yeah, it sounds stupid, but the question itself is stupid). So is A necessary, insufficient condition for B.
  13. Oct 6, 2005 #12
    oh thanks a lot guys! ur replies were very helpful ~ hmm Im feeling slightly confident now with all this proof but maybe not for long... :uhh:

    I've given the question: 'Prove that 2^(1/3) is irrational' a go, and would appreciate if you can help me check if its right. Thank you!!

    2^(1/3) = p/q

    2 = p^3/q^3

    p^3 = 2q^3 (equation 1)

    => p^3 is divisible by 2
    => p is divisible by 2
    => p = 2k (for some +ve integer)

    => p^3 = (2k)^3
    => p^3 = 8k^3 = 2q^3 (from equation 1)
    => 2q^3 = 8k^3
    => q^3 = 4k^3
    => q^3 = 2(2k^3)
    => q^3 is divisible by 2
    => q is divisible by 2

    => p, q are both divisible by 2

    Hence by contradicition 2^(1/3) is irrational
  14. Oct 6, 2005 #13


    User Avatar
    Science Advisor
    Homework Helper

    Yes, it looks good. Note that you should probably state that if 21/3 were rational, then there would exist p, q co-prime such that 21/3 = p/q. Assuming p and q are such numbers, you deduce that 2 divides both p and q, hence p and q are not co-prime. There is your actual contradiction, so you know that there DOES NOT exist a pair p, q of co-prime numbers such that 21/3 = p/q, and that's why it is irrational.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook