Divisor Proof with ContrapositiveDate: 09/16/2002 at 13:59:01 From: Eric Subject: Proof of m^2 | n^2 => m | n I have been trying to prove that if n^2 divides m^2, then n divides m. m^2 = k*n^2 m,n,k all in Z. k = (m/n)^2 Is there a proof that says if a rational number squared is an integer, then it too is an integer? Since we know k is an integer, this would be sufficent to prove this, correct? Thank you for any help. Eric Date: 09/16/2002 at 18:08:37 From: Doctor Paul Subject: Re: Proof of m^2 | n^2 => m | n I think this proof is most easily accomplished by contraposition. Can you prove: n does not divide m => n^2 does not divide m^2 Proving this statement is equivalent to proving your statement, since a statement and its contrapositive are logically equivalent. To prove the statement, suppose that n does not divide m. Then there exists some prime p in the prime factorization of n that does not occur in the prime factorization of m. If p does not occur in the prime factorization of m, it won't occur in the prime factorization of m^2 either. Similarly, if p occurs in the prime factorization of n, it also occurs in the prime factorization of n^2. Thus there exists a prime (namely p) in the prime factorization of n^2 that does not occur in the prime factorization of m^2. Hence, n^2 does not divide m^2. This was to be shown... I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ Date: 09/16/2002 at 20:16:47 From: Eric Subject: Thank you (proof of m^2 | n^2 => m | n) AH! That makes sense. I wish I had thought of that. Once I read your response it became very clear. Thank you very much! Eric |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/