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I've been reading Algebra The Easy Way, and there's a problem at the end of the chapter to prove that the product of two odd numbers is an odd number.

In this problem, an even number is defined as two multiplied by a natural number, so (2 * m). An odd number is defined as an even number plus one, so (2 * n + 1).

Therefore, two odd numbers multiplied by each other is

(2 * m + 1) * ( 2 * n + 1).

The answer says to use the distributive property to convert this to

(2 * m + 1) * 2 * n + (2 * m + 1)

My question is how does the distributive property allow (2 * m + 1) * ( 2 * n + 1) to be converted to (2 * m + 1) * 2 * n + (2 * m + 1)? Is this a typo in the book, or is it actually possible to do this conversion using the distributive property?

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    $\begingroup$ Because $2m + 1 = 1 \cdot (2m + 1)$. $\endgroup$ – user296602 Nov 13 '18 at 22:20
  • $\begingroup$ Yes. (2*n+1)=(2*n)+(1), so (2*m+1) can distribute. $\endgroup$ – herb steinberg Nov 13 '18 at 22:45
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To answer your specific question, let $\nu = (2m+1)$. Then, with substitution, $$(2m+1)(2n+1) = \\ \nu(2n+1) = \\ \nu2n + \nu $$Does that last line make sense? If not, please have a look over distributive multiplication. Going on, replace $\nu$ above with $(2m+1)$ and you get the answer in the book.

The overall proof may actually be easier to use FOIL, to obtain $4mn + 2m + 2n + 1$ then factor out a 2 to get $$2(2mn + m + n) + 1 = \\2(\mathrm{some\ integer})+1$$Since you've established that an odd number has the form, $2i+1$ for some integer $i$, and the above expansion has that form, then the above expansion is odd. Since the above expansion was created as a product of two odd numbers, then the product is an odd number.

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  • $\begingroup$ Using "v" as a substitution makes it perfectly clear. I also like your suggestion of using FOIL. Thank you. $\endgroup$ – Toby Artisan Nov 14 '18 at 14:59

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