# How do I show that the set of odd natural numbers is closed under the operation * defined by a*b=a+b+ab?

I really need help with this question. I am required to show that the set of odd natural numbers is closed under the operation * defined by a*b=a+b+ab, and I'm not quite sure how. Any work/help is greatly appreciated.

## 7 Answers

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

We want to show that the set of natural odd numbers are 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 closed under the defined operation.

show a+b+ab is odd whenever a and b are odd

• Thank you, but this method will not definitively prove that the operation is closed as it is not a generalised statement. – Alex Cross Mar 25 '14 at 2:39
• This may be a summary rather than a complete detailed answer, but it is a generalised statement and it is perfectly correct. +1. – David Mar 25 '14 at 3:06
• @AlexCross There's nothing wrong with this approach. You can solve the problem easily at a high level: "If $a$ and $b$ are odd, then $a+b$ is even and $ab$ is odd, hence $a+b+ab$ is odd." This proof is rigorous, and you don't have to mess around with a bunch of error-prone algebra in order to write it or review it. – Chris Culter Mar 25 '14 at 3:11
• Indeed, no matter what $a$ and $b$ are, you’re adding three odd numbers. Result: odd. This argument is fully general. – Lubin Mar 25 '14 at 4:26
• @AlexCross: What do you mean here when you say 'generalized statement'? – ColinK Mar 25 '14 at 12:12

You could use the fact that $$a*b=(a+1)(b+1)-1$$ We have that $$(2a+1)*(2b+1)=(2a+2)(2b+2)-1$$ Do you see why that number must be odd?

• I am not sure how you reached the second part above. Do you mind clarifying? – Alex Cross Mar 25 '14 at 2:43
• Since the numbers we are dealing with are all odd, they will have the form $2n+1$ for some number $n$. When you "multiply" two such numbers, you get a product of two even numbers ($2a+2$ and $2b+2$) minus 1. Even X Even = Even, and Even - 1 = Odd, so you always get an odd number out. In other words, the operation always sends odd numbers to other odd numbers. It is "closed" under the set of odd numbers. – user137500 Mar 25 '14 at 2:49
• Your idea is correct but the writing is not quite clear :) – Silviu Burcea Mar 25 '14 at 11:32

If you want to see it inmediately, if $a$ and $b$ are odd, $a + b$ is even and $a\cdot b$ is odd; and odd plus even is odd.


"The set of odd natural number is closed under $\;*\;$" means that if any $\;a\;$ and $\;b\;$ are odd natural numbers, then also $\;a * b\;$ is an odd natural number.

Therefore we ask ourselves: when is $\;a * b\;$ an odd natural number? First, from the definition of $\;*\;$ it is clear that if $\;a,b\;$ are natural numbers, then $\;a * b\;$ also is a natural number.

So, what about the oddness of $\;a * b\;$? Let's calculate:

\begin{align} & \odd{a * b} \\ \equiv & \qquad \text{"definition of $\;*\;$"} \\ & \odd{a + b + a \times b} \\ \equiv & \qquad \text{"sum is odd if exactly one is odd"} \\ & \odd{a + b} \;\not\equiv\; \odd{a \times b} \\ \equiv & \qquad \text{"sum is odd if exactly one is odd; product is odd if both are odd"} \\ & \odd{a} \;\not\equiv\; \odd{b} \;\not\equiv\; \odd{a} \;\land\; \odd{b} \\ \equiv & \qquad \text{"logic: simplify by removing double negation"} \\ & \odd{a} \;\equiv\; \odd{b} \;\equiv\; \odd{a} \;\land\; \odd{b} \\ \equiv & \qquad \text{"logic: golden rule"} \\ & \odd{a} \;\lor\; \odd{b} \\ \end{align} So $\;a * b\;$ is odd iff either $\;a\;$ or $\;b\;$ is odd, so certainly if both are odd.

This completes the proof.

Note how both $\;\not\equiv\;$ and $\;\equiv\;$ are associative, so that we could safely leave out the parentheses in the above proof. The golden rule mentioned above is $$P \;\equiv\; Q \;\equiv\; P \land Q \;\equiv\; P \lor Q$$ for any boolean expressions $\;P,Q\;$.

Definition: A number is odd if and only if it can get written in the form [(2*n)+1] where n is an integer.

Now, "a" and "b" both qualify as odd. But, you don't want to write them in the same form for this question, since they might not equal each other. Thus, select one letter to go in the blank space of [(2*_)+1] for "a" and another letter to go in the blank space for "b". Now put those things in place of "a" and "b" in a+b+ab. Then expand a+b+ab. After expanding a+b+ab, select letters to represent certain equations. Eventually, you should end up with something of the form [(2*n)+1].

For instance, say I wanted to show that if x, and y are even, that x+y is even. I would first let x=2a, and y=2b. Then we can see that x+y=2a+2b=2(a+b). Letting a+b=z we then have x+y=2z. But all the variables are arbitrary (within the set of even numbers), and thus z is arbitrary also. Consequently, in "2z" z indicates an arbitrary variable, which means that z means the same thing as an arbitrary variable in the definition of an even number. Therefore, x+y is even.

a+b*ab=(2k+1)+(2j+1)+(2k+1)(2j+1)=(2k+2j)+2+(4kj+2k+2j+1)=2(k+j+1+2kj+k+j)+1=2z+1.

• @AlexCross See the last line as updated. – Doug Spoonwood Mar 25 '14 at 3:02

To just answer the question, simplest is to observe that $a+b+ab$ is the sum of three numbers that are each positive and odd (if $a,b$ are so), and hence itself positive and odd.

To better understand what this operation does, one can compute $$(x-1)*(y-1) = (x-1)+(y-1)+(x-1)(y-1)=xy-1$$ for any $x,y$, which would be in practice taken to be positive even numbers. This shows the operation is just the multiplication of positive even numbers in disguise, the disguise consisting of representing each such number systematically (on input and on output) by the odd number before it. Thus one sees for instance immediately that '$*$' is (commutative and) associative, which would otherwise require a bit of computation to verify. Multiplication of even numbers has the property that only numbers divisible by$~4$ ever occur as value of the operation, and more generally that combining $n$ numbers results in a number divisible by$~2^n$; correspondingly, combining $n$ odd numbers by '$*$' result in a number that is congruent to$~{-}1$ modulo$~2^n$. Finally just as the even positive numbers remain closed under multiplication if one adds $0$ to the set (which acts as absorbing element: $0x=x0=0$ for all$~x$), one could add $-1$ to the set of odd positive numbers, which would become an absorbing element for '$*$'.