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Suppose the quotient of two odd integers is an integer. Make and prove a conjecture about whether the quotient is either even or odd.

If you had an even (2n) divided by and odd (2m+1), it wont work. so my odd integers would be a= 2n+1 and b=2m+1

a/b = c = (2n+1)/(2m+1) which is also odd so c = (2w + 1)

c|a, so a = [(2n+1)/(2m+1)] * k for some integer k  

????? or I have

Let a = 2n +1 and b = 2m + 1. From the definition a/b = c and c|a, then we get a/b = c . Thus a = b*c = (2m+1)(c) = 4mc + c = c(4m + 1). Then, we have an equation that is (c)*(odd) making the final result odd.

examples: 9/3 = 3 21/7 = 3 81/9 = 9 49/7 = 7 35/5 = 7

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    How does it look now? I'm not sure where to go from that.2012-09-09

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Let $a=2n+1$ and $b = 2m+1$ be odd integers. I see now you have a conjecture that $c=\frac{a}{b}$ is odd i.e $c=2k+1.$

Let's prove it:

Now assume the opposite, i.e assume $c$ is even, $c=2k.$ What happens in this case? Use the fact that if $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ then $x_1y_2=x_2y_1$.

You will find that $a=cb=2kb=2(kb)=2l$ (for some integer $l=kb$), but this mean $a$ is even which contradicts our original hypothesis that $a$ is odd. Hence our assumption $c$ is even is wrong, this implies $c$ must be an odd integer. QED

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Since the OP has mentioned being new to proof writing, I'll try to help him/her get started by providing a example.
It looks like you(the OP) are having difficulty moving from examples you have tried ( assuming you have done an extensive trial ) I'm going to help you get started a generalization of ideas.

Dennis Gulko gave a starting point. After reading what's written here, go back to his comment and think about it.

Ask yourself: what do I have and what is asked for.
As data ( hypotheses ) you have $a$ and $b$ being odd.
Now the question is find out whether $\frac{a}{b} = c$ is either odd or even.

To conjecture means you try as many instances of the problem as you can then formulate a conjecture ( a statement you assume to be true but you haven't proved yet ).

Example of a conjecture is If a is an odd integer and b is an odd integer then a+b is even. We don't know for sure but $3+3$ is even and we can try with others. ( In your case, do multiple trials. )

You see, the value of the conjecture is that it allows you to have something to prove. A sentence like If a is an odd integer and b is an odd integer then a+b is even or odd. is not a useful conjecture because it is always true. There is no third possibility.

After you have a conjecture it helps to know the definition of key concepts involved. E.g what is an odd number, an even number? The good news in mathematics is that definition are written not to be ambiguous.

In the conjecture i provided, we need to know exactly what an odd and an even numbers are.

And we say: let a=2n+1 and b=2m+1 be two odd numbers. We have to prove that a+b is equal to a third number c=2k which is even.

Then we have:

We have a=2n+1 and b=2m+1, then c=a+b=2n+1+2m+1=2(n+m)+2=2(n+m+1)=2k. qed

That is a flow you might use to get started proving things.

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    Shew, I'm so drained $f$rom this problem. $c$(2m+1) = 2n + 1 right?2012-09-09