Prove that if n is an integer then 3n + 2 is even, then n is even

- lgbasallote

Prove that if n is an integer then 3n + 2 is even, then n is even

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- mathslover

we have to prove two things? : 1) 3n+2 is even , 2) n is even

- anonymous

no, it's prove n is even GIVEN 3n + 2 is even.

- mathslover

ok , thanks

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## More answers

- anonymous

If 3n + 2 is even, then 3n is even

- mathslover

if 3n is even then n is even.

- anonymous

Given : 3n+2 is even
To prove : n is even
3n+2 = 2k
3n = 2k-2
3n = 2(k-1)

- hartnn

not true for n=1 ?
3+2=5<---not even

- anonymous

@hartnn GIVEN that 3+2 is even, n is even

- mathslover

IF 3n+2 is even , prove that is even. @hartnn

- anonymous

But it's not so no guarentee is made.

- hartnn

lets clear it from @lgbasallote what the exact question is...

- mathslover

@lgbasallote ?

- anonymous

Just out of interest whats the general way to prove something is even, divide by 2?

- anonymous

yes

- anonymous

if we prove a number is of form 2k that proves it is even

- anonymous

right I see

- anonymous

btw any1 here mind taking a look at my question?
http://openstudy.com/updates/5078b9d7e4b02f109be44f95

- lgbasallote

sorry i just came back....anyway...i'd llike to see how proving by contradiction is done.. direct proof is too easy

- hartnn

but whats the question ?

- lgbasallote

the statement in the blue box

- lgbasallote

forgot to mention by the way....that the direct proof done here was wrong

- lgbasallote

they took n as the condition... 3n + 2 is supposed to be the condition

- anonymous

so 3n + 2 = (a different n)?

- hartnn

if 3n+2 is even , n is even.
thats the question and u need to prove that using contradiction, right ?

- lgbasallote

if you use direct proof...
it should be 3n + 2 = 2x
then prove n is even

- lgbasallote

but like i said...should be contradiction though

- anonymous

well okay thats easy enough, too

- lgbasallote

hmmm then let's see you try

- anonymous

All you have to say is that assume that given 3n + 2 is even, n is not even
Then n has to be odd. Thus, 3n + 2 can be represented as 2k + 1 for some k

- anonymous

Given : 3n+2 is even
To prove : n is even
to prove by contradiction, lets assume the opposite -
lets assume n is odd,
3n+2 = 2k
3n = 2k-2
3n = 2(k-1)
so we got the right side as even number,
but we assumed n is odd, so left 3n becomes odd - contradiction

- anonymous

@sara12345 you cannot say "assume that n is odd" and then equate it to 2k

- lgbasallote

@sara12345 how is that contradiction

- anonymous

we do that always in proof by contradiction

- anonymous

RHS = even , LHS = odd => contradiction

- lgbasallote

i was referring to your solution

- anonymous

How can you prove that it equals 2k, though?

- anonymous

if 3n + 2 is odd, then it can be expressed 3n + 2 = 2k + 1. solving for n gives \[n = \frac{2k}{3} - 1\] substituting back for n gives
\[3 (\times \frac{2k}{3} - 1) + 2 = 2k + 1so\]
so \[2k - 1 = 2k + 1 \]
which is absurd

- lgbasallote

hmm seems legit

- lgbasallote

so in proof by contradiction...you still substitute back huh

- anonymous

As @JamesWolf showed for you,

- anonymous

you take the contradiction, show that it is not internally consistent, and therefore it cannot be true.

- anonymous

no its not legit

- anonymous

ive been an idiot int he first step

- anonymous

@JamesWolf no, you inadvertently put up my proof.

- anonymous

in proof by contradiction, we assum e the opposite of what we need to prove as true, and proceed, not the opposite of given conditions

- lgbasallote

if you assume n is odd and 3n + 2 is even...
3(2k + 1) + 2
6k + 3 + 2
6k + 2 + 3
2(3k + 1) + 3
so even + odd would be odd...so contradiction
i suppose that works as well

- anonymous

@sara12345 yes you're right, sorry.

- anonymous

yes its right actually i managed to not divide by 3 at the start, but luckily i messed up by not multiplying - 1by 3 at the end

- lgbasallote

oh so you take the negation of q then proceed from there?

- lgbasallote

anyway...is my proof right?

- anonymous

lgba ur proof is more correct as it shows the assumption n is odd as well by letting n =2k+1

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