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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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    Did you try some examples?2012-09-09
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    If this is HW, please tag it as such. Additionally, how would you expand an odd integer such that you can use the distributive rule?2012-09-09
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    Observe that if $a/b=c$ is an integer, then $c$ divides $a$. Can an even integer divide an odd one?2012-09-09
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    Sorry I am new to this website, It is a homework question. I am not sure how to even start this. I am very new to the solving proofs to begin with.2012-09-09
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    The way to start making a conjecture is to try some examples, as Robert Israel said. Have you tried some examples?2012-09-09
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    To complement what Dennis Gulko said, when attempting a proof, it helps to remember the definitions of keys concepts involved. In your case namely: what it means $a$ divides $b$ and the definition of odd and even numbers. Then think about the comment of Mr Gulko.2012-09-09
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    I have tried a few examples like 9/3 = 3 but 6/3 = 2 so i'm confused. Am I doing this right?2012-09-09
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    @Christene: $6$ isn't odd, so that has no bearing on what your conjecture should be.2012-09-09
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    Notice that $6$ is not odd. Keep trying with odd numbers only...2012-09-09
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    No, you wrote "quotient of two odd integers" but $6$ is not odd, it is even2012-09-09
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    Oh, sorry i didn't even realize it. 21/3 = 72012-09-09
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    Yes, keep trying and do not limit yourself to $3$ as denominator. Try other numbers. Then think about what Dennis Gulko said. After come back to edit your post with regard to your progress.2012-09-09
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    Correct me if I'm wrong but its more that I'm proving why it can't be even?2012-09-09
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    How does it look now? I'm not sure where to go from that.2012-09-09

2 Answers 2

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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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    Where I edited... is this more one right track? I'm not sure what to do next.2012-09-09
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    Almost but you assumed that $c$ is odd and that is not given. You are supposed to figure out if $c$ will always be odd or even. Follow 3 simple steps: 1/ provide examples 2/ make a conjecture ( like `if a and b are odd integers the c=a/b is an ...` ) 3/ Then prove then conjecture. You are already on the right track, just use $a$ and $b$ to figure out $c$. So congratulations for trying harder. Just one more step!2012-09-09
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    I'm not sure where to go.2012-09-09
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    First, tell what is your conjecture, then we'll move from there. Do like I did in the example in my answer.2012-09-09
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    My conjecture would be : if a and b are odd integers, then the c=a/b is also odd. c = (2n+1)/(2m+1).2012-09-09
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    Good! Now you have to show that $c=\frac{2*n+1}{2*m+1}$ is indeed an odd number. Just perform a few multiplications and additions and deduce the value of $c$. **Hint:** You'll find something you might not like about $c$, but okay. Then update your question! Don't stop trying, that is how we all learn, by trying harder!2012-09-09
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    No, no my dear!!! The fraction is $c=frac{2n+1}{2m+1}$. Remember in elementary calculus they taught that if $\frac{a}{b}=\frac{c}{d}$ then $ad=cb$! Just do the same there and get the value of $c$ ( a second time if you wish ) but without fractions.2012-09-09
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    Shew, I'm so drained from this problem. c(2m+1) = 2n + 1 right?2012-09-09