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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? –  Robert Israel Sep 9 '12 at 18:31
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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? –  soandos Sep 9 '12 at 18:32
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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? –  Dennis Gulko Sep 9 '12 at 18:33
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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. –  Christene Sep 9 '12 at 18:45
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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? –  Qiaochu Yuan Sep 9 '12 at 18:51
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3 Answers

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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Im just going to write down all the stuff the other people have written. Ok so lets say you have numbers a and b. We will expand the number into its prime factorization. So $a=2^{a_1}*3^{a_2}....p^{a_p}$ and $b=a=2^{b_1}*3^{b_2}....p^{b_p}$. So then $a/b=2^{(a_1-b_1)}*3^{(a_2-b_2)}....p^{(a_p-b_p)}$-

The number is an integer because for any $i:a_i\geq b_i$ However we know that a number is even only if it has at least one 2 in its prime factorization but since $a_1= 0$ (because a is odd) it is impossible for a/b to be even.

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The notion of a prime number is not needed for the problem though, much less unique factorization. –  Trevor Wilson Sep 9 '12 at 19:06
    
Well, im a noob in math but this is still a valid solution right?It might not be a very nice one but it gets the job done.(I think) –  user4140 Sep 9 '12 at 19:09
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You have practically given an answer! The OP is supposed to reach any conclusion him/herself. Only by trying harder can the OP learn not by reading someone else's reasoning. –  nt.bas Sep 9 '12 at 19:09
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You lost me, we haven't gone over prime factorization. –  Christene Sep 9 '12 at 19:09
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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. –  Christene Sep 9 '12 at 20:57
    
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! –  nt.bas Sep 9 '12 at 21:07
    
I'm not sure where to go. –  Christene Sep 9 '12 at 21:36
    
First, tell what is your conjecture, then we'll move from there. Do like I did in the example in my answer. –  nt.bas Sep 9 '12 at 21:44
    
My conjecture would be : if a and b are odd integers, then the c=a/b is also odd. c = (2n+1)/(2m+1). –  Christene Sep 9 '12 at 21:48
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