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How can I prove that the integer $3n+2$ is odd if and only if the integer $9n+5$ is even, where n is an integer?

I suppose I could set $9n+5 = 2k$, to prove it's even, and then do it again as $9n+5=2k+1$

Would this work?

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@Arturo: the tag logic sounds equally correct as well – user17762 May 30 '11 at 2:49
@Sivaram: I'm not so sure; it touches on "logic" as much as it touches on "algebra-precalculus". If you look at the info page, I think this does not rise to the level (even though george is asking how to prove something)... – Arturo Magidin May 30 '11 at 2:53
Probably there should be a tag named "elementary logic" – user17762 May 30 '11 at 2:57
If you are familiar with modluar arithmetic, probably the fact that $3n+2 \cong n$ modulo 2 and $9n+5 \cong n+1$ mod 2 is the fastest way to do it. – N. S. May 30 '11 at 17:22
up vote 4 down vote accepted

No, you should not set $9n+5$ equal to $2k$ in order to prove it is even; that is tantamount to assuming what you want to prove.

This is an "if and only if"; there are two standard ways of proving it:

  1. By proving two implications: prove that if $3n+2$ is odd, then $9n+5$ is even; so, assume $3n+2 = 2k+1$ for some integer $k$; try to conclude that $9n+5$ must be equal to $2m$ for some integer $m$. Then prove that if $9n+5$ is even, then $3n+2$ is odd. So, assume $9n+5=2m$ for some integer $m$, try to use this to conclude that $3n+2=2k+1$ for some integer $k$. Each of this implications can be proven in any of the usual ways (directly, by contradiction, by contrapositive, etc).

  2. Starting with one of the two statements, say "$3n+2$ is odd", construct a chain of statements, each of them equivalent to the previous one, that ends with "$9n+5$ is even". For example, $3n+2$ is odd if and only if there is a $k$ such that $3n+2 = 2k+1$, which happens if and only if there is a $k$ with $3n+1=2k$, which happens if and only .... and keep going until you manage to get to "9n+5 is even".

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@ShreevatsaR: Thanks. I've corrected is. – Arturo Magidin May 30 '11 at 12:38

HINT $\rm\ \ 3\ (3\:n+2)\ -\ (9\:n+5)\:\ =\:\ 1$

Alternatively note that their sum $\rm\:12\:n + 7\:$ is odd, so they have opposite parity.

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Hint: Prove that $3n+2$ is odd if and only if $n$ is odd and $9n+5$ is even if and only if $n$ is odd

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It's enough to look at 3n, of course: 3n+2 is odd <=> 3n is odd <=> 9n is odd <=> 9n+5 is even. – ShreevatsaR May 30 '11 at 7:22
@ShreevatsaR: True. Your argument is way easier. – user17762 May 30 '11 at 17:07

(This bit of writing is organized by notes and proofs, as noted by headings and X)s)

Proof 1

1) Assuming an even plus an even is an odd, odd plus an odd is an even, and even plus an odd is an odd... -- (See end proofs)

2) Assuming n is even...

3) Anything even can be stated as 2x, where x equals any non-decimal number, therefore the product of 3*2x would be divisible by two, and even. Adding two to that would produce an even according to our original assumption.

4) Now assuming n is odd...

5) 3n would always result in an odd, due to the fact that any value for n that is even would result in an even number, we would only need to add 3 to n-1 to get 3n. Since we're assuming n is odd, n-1 would be even, and an even plus an odd is an odd.

6) Now adding two to this odd would result in another odd, due to an odd plus an even being an odd.

Proof 2

Now to prove that 9n + 5 is an even if 3n + 2 is an odd...

1) N must be an odd for 3n + 2 to be an odd

2)Plugging any odd number into the equation 9n would result in an odd number(see note five of previous proof)

3) An odd plus an odd is an even (see note 1 of previous proof), five is an odd, 9n is an odd, so 9n + 5 is an even.

Now, a quick proof on odd plus odd being an even and etc.

Definitions: Even - 2n or an odd +- 1 where n is any complete number Odd - 2n +- 1 where n is any complete number

Proof 3

1) an odd plus an odd would be (2n +- 1) + (2x +- 1).

2) After simplification, this can become 3 things. 1 - 2n + 2x 2 - 2n + 2x + 2 3 - 2n + 2x - 2

3) To check even-ness, divide by two and check for decimals. Doing so in the above three results results in no decimals, therefore, all even.

So an odd plus an odd is always an even.

Proof 4

1) An even plus an odd would be 2x + (2n + 1)

2) Dividing by two gives us x + n + .5. .5 is a decimal, therefore, the result is odd.

So an even plus an odd is an odd.

Proof 5

1) an even plus an even would be 2x + 2n

2) Divide by two - x + n. Since there are no decimals in this result, it is an even.

Therefore, an even plus an even is always an even.

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