You are watching: Are the odd numbers a closed set under addition

Let $a = 2n + 1$ and also let $b = 2m + 1$ whereby $n, m geq 0$.

We desire to show that the collection of natural odd numbers space closed under the defined operation $*$.

So: $$a*b = a + b + ab$$$$= (2n + 1) + (2m + 1) + (2n + 1)(2m + 1)$$$$= (2n + 2m + 2) + (4nm + 2n + 2m + 1)$$$$= (4n + 4m + 4mn + 2) + 1$$

Thus $*$ is closeup of the door under the defined operation.

You might use the fact that $$a*b=(a+1)(b+1)-1$$ We have actually that $$(2a+1)*(2b+1)=(2a+2)(2b+2)-1$$ do you watch why the number should be odd?

If you want to view it inmediately, if $a$ and $b$ room odd, $a + b$ is even and also $acdot b$ is odd; and also odd plus also is odd.

$ ewcommandodd<1>#1 ext is odd ewcommandeven<1>#1 ext is even$Just for fun, below is a slightly different (a "logical") strategy compared come the currently answers.

"The collection of odd organic number is close up door under $;*;$" means that if any type of $;a;$ and $;b;$ space odd natural numbers, then likewise $;a * b;$ is an odd natural number.

Therefore us ask ourselves: as soon as is $;a * b;$ one *odd natural* number? First, indigenous the definition of $;*;$ that is clear that if $;a,b;$ are natural numbers, climate $;a * b;$ additionally is a *natural* number.

So, what around the *oddness* the $;a * b;$? Let"s calculate:

eginalign& odda * b \equiv & qquad ext"definition that $;*;$" \& odda + b + a imes b \equiv & qquad ext"sum is strange if exactly one is odd" \& odda + b ;
otequiv; odda imes b \equiv & qquad ext"sum is odd if specifically one is odd; product is odd if both space odd" \& odda ;
otequiv; oddb ;
otequiv; odda ;land; oddb \equiv & qquad ext"logic: simplify by removing double negation" \& odda ;equiv; oddb ;equiv; odda ;land; oddb \equiv & qquad ext"logic: gold rule" \& odda ;lor; oddb \endalignSo $;a * b;$ is strange iff one of two people $;a;$ or $;b;$ is odd, so certainly if *both* are odd.

This completes the proof.

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Note how both $; otequiv;$ and $;equiv;$ space associative, so that we could safely leave out the clip in the above proof. The gold rule mentioned over is$$P ;equiv; Q ;equiv; p land Q ;equiv; p lor Q$$for any boolean expression $;P,Q;$.