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

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we have to prove two things? : 1) 3n+2 is even , 2) n is even
no, it's prove n is even GIVEN 3n + 2 is even.
ok , thanks

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Other answers:

If 3n + 2 is even, then 3n is even
if 3n is even then n is even.
Given : 3n+2 is even To prove : n is even 3n+2 = 2k 3n = 2k-2 3n = 2(k-1)
not true for n=1 ? 3+2=5<---not even
@hartnn GIVEN that 3+2 is even, n is even
IF 3n+2 is even , prove that is even. @hartnn
But it's not so no guarentee is made.
lets clear it from @lgbasallote what the exact question is...
Just out of interest whats the general way to prove something is even, divide by 2?
if we prove a number is of form 2k that proves it is even
right I see
btw any1 here mind taking a look at my question?
sorry i just came back....anyway...i'd llike to see how proving by contradiction is done.. direct proof is too easy
but whats the question ?
the statement in the blue box
forgot to mention by the way....that the direct proof done here was wrong
they took n as the condition... 3n + 2 is supposed to be the condition
so 3n + 2 = (a different n)?
if 3n+2 is even , n is even. thats the question and u need to prove that using contradiction, right ?
if you use direct proof... it should be 3n + 2 = 2x then prove n is even
but like i said...should be contradiction though
well okay thats easy enough, too
hmmm then let's see you try
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
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
@sara12345 you cannot say "assume that n is odd" and then equate it to 2k
@sara12345 how is that contradiction
we do that always in proof by contradiction
RHS = even , LHS = odd => contradiction
i was referring to your solution
How can you prove that it equals 2k, though?
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
hmm seems legit
so in proof by still substitute back huh
As @JamesWolf showed for you,
you take the contradiction, show that it is not internally consistent, and therefore it cannot be true.
no its not legit
ive been an idiot int he first step
@JamesWolf no, you inadvertently put up my proof.
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
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 contradiction i suppose that works as well
@sara12345 yes you're right, sorry.
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
oh so you take the negation of q then proceed from there? my proof right?
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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