$x$ is odd if and only if $3x+6$ is odd

Question

Prove the following proposition. Let $x\in\Bbb Z$. Then $x$ is odd if and only if $3x+6$ is odd.

I'm currently not seeing a way to transform $3x+6$ into the format of $2k+1$ in order to prove odd. This is my first time dealing with discrete mathematics and proofs and I'm trying to get a feel for it.

To prove $3x+6$ is odd thus $x$ is odd could be easily proven if you prove the contrapositive of the statement. — InsigMath, Jun 24 '16 at 22:14
If it helps, the even integers are defined as ${n\mid n/2\in\Bbb{Z}}$ and the odd integers are defined as ${n\mid (n+1)/2\in\Bbb{Z}}$ — gen-ℤ ready to perish, Jun 25 '16 at 00:05
A formal proof is below. But a quick glance should convince you the statement is true: because 6 is an even number, adding 6 does not change the parity (evenness or oddness). So the parity of $3x+6$ is the same as the parity of $3x$. It should be clear $3x$ is odd if and only if $x$ is odd because 3 is an odd integer. — Ashwin Ganesan, Jun 25 '16 at 14:22

score 5 · Answer 1 · answered Jun 24 '16 at 02:27

5

Hint: Write $3x+6=x+2(x+3)$ .

answered Jun 24 '16 at 02:27

lhf

216,483

score 1 · Answer 2 · answered Jun 24 '16 at 02:27

1

Hints:

(1) Adding $2$ to an integer doesn't change its parity.

(2) Multiplying an integer by an odd number doesn't change its parity.

(3) You can write $3x+6=3(x+ 2)$

answered Jun 24 '16 at 02:27

MPW

43,638

score 1 · Answer 3 · answered Jun 24 '16 at 02:45

If $x$ is odd then it is of the form $2k+1$ for some integer $k$. … Thus $3x+6$ is of the form $2(\underline{\qquad})+1$ for some integer $(\underline{\qquad})$; meaning it is odd.

If $x$ is even then it is of the form $2k$ for some integer $k$. … Thus $3x+6$ is of the form $2(\underline{\qquad})$ for some integer $(\underline{\qquad})$; meaning that it is even.

Therefore $3x+6$ is odd if and only if $x$ is odd.

score 0 · Answer 4 · answered Jun 24 '16 at 02:42

0

$3x+6$ is odd,we know $3x$ is odd ,so set $3x=2k+1$,$k=1,2,3...$,

so $$x=\frac{2k}{3}+1,k=1,2,3...$$

then $\frac{2k}{3}$ is integer so $k=3n$,and $$x=2n+1,n=3k,k=1,2,3$$

and $x$ is odd

answered Jun 24 '16 at 02:42

Young1997

198

score 0 · Answer 5 · answered Jun 24 '16 at 03:00

Hints:

(1) An "if and only" statement can be split into two statements. You can prove "A if and only if B" by proving "if A then B" and "if B then A".

(2) A conditional statement is logically equivalent to its contrapositve. For example "if A then B" is logically equivalent to "if (not B) then (not A)".

(3) Using hints (1) and (2), one way to prove the statement you are asking about is to prove the following two statements:

Let $x\in\Bbb Z$. If $x$ is odd, then $3x+6$ is odd.
Let $x\in\Bbb Z$. If $x$ is even, then $3x+6$ is even.

score 0 · Answer 6 · answered Jun 24 '16 at 21:53

0

Hint $ $ negate $\ 2\mid 3(x\!+\!2)\iff 2\mid x\!+\!2\iff 2\mid x$

answered Jun 24 '16 at 21:53

Bill Dubuque

272,048

score 0 · Answer 7 · answered Jun 25 '16 at 14:21

We want to prove that $x$ is odd iff $3x+6$ is odd. For many of these kinds of problems, the proof is easier if we start by assuming that $x$ is of the form $x=2k$ or $x=2k+1$ (these exhaust all possibilities for $x$) and substitute these expressions into $3x+6$.

Suppose $x$ is even, say $x=2k$ for some integer $k$. Then $3x+6 = 3(2k)+6=6x+6=2(3x+3)$, which is even.

Suppose $x$ is odd, say $x=2k+1$ for some integer $k$. Then $3x+6 = 3(2k+1)+6 = 6k+9 = 6k+8+1 = 2(3k+4)+1$, which is odd.

$x$ is odd if and only if $3x+6$ is odd

7 Answers7