1
$\begingroup$

I know that the fundamental theorem of arithmetic gives a unique (except, for ordering) prime factorization of any natural $\gt 1$. Now, any odd number will not have 2 as a prime in that factorization. If I have to prove using the fundamental theorem of arithmetic, that product of two odd integers is odd; does it suffice to say that the product's prime-factorization does not have a prime value of 2.

$\endgroup$
  • 1
    $\begingroup$ You are right. And it is not necessary to write neither the factorization nor the form $2n+1$. A proof is also to say that the product does not contains the prime $2$. $\endgroup$ – Piquito Nov 10 '17 at 12:25
1
$\begingroup$

The FTA does indeed suffice, but the claim follows perhaps more easily by considering $$(2n+1)(2k+1)= 2n \cdot 2k +2k +2n +1 \\ = 4nk+2k+2n+1 \\ = 2(2nk+k+n)+1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.