lgbasallote
  • lgbasallote
Prove that if n is an integer then 3n + 2 is even, then n is even
Mathematics
schrodinger
  • schrodinger
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mathslover
  • mathslover
we have to prove two things? : 1) 3n+2 is even , 2) n is even
anonymous
  • anonymous
no, it's prove n is even GIVEN 3n + 2 is even.
mathslover
  • mathslover
ok , thanks

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anonymous
  • anonymous
If 3n + 2 is even, then 3n is even
mathslover
  • mathslover
if 3n is even then n is even.
anonymous
  • anonymous
Given : 3n+2 is even To prove : n is even 3n+2 = 2k 3n = 2k-2 3n = 2(k-1)
hartnn
  • hartnn
not true for n=1 ? 3+2=5<---not even
anonymous
  • anonymous
@hartnn GIVEN that 3+2 is even, n is even
mathslover
  • mathslover
IF 3n+2 is even , prove that is even. @hartnn
anonymous
  • anonymous
But it's not so no guarentee is made.
hartnn
  • hartnn
lets clear it from @lgbasallote what the exact question is...
mathslover
  • mathslover
anonymous
  • anonymous
Just out of interest whats the general way to prove something is even, divide by 2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
if we prove a number is of form 2k that proves it is even
anonymous
  • anonymous
right I see
anonymous
  • anonymous
btw any1 here mind taking a look at my question? http://openstudy.com/updates/5078b9d7e4b02f109be44f95
lgbasallote
  • lgbasallote
sorry i just came back....anyway...i'd llike to see how proving by contradiction is done.. direct proof is too easy
hartnn
  • hartnn
but whats the question ?
lgbasallote
  • lgbasallote
the statement in the blue box
lgbasallote
  • lgbasallote
forgot to mention by the way....that the direct proof done here was wrong
lgbasallote
  • lgbasallote
they took n as the condition... 3n + 2 is supposed to be the condition
anonymous
  • anonymous
so 3n + 2 = (a different n)?
hartnn
  • hartnn
if 3n+2 is even , n is even. thats the question and u need to prove that using contradiction, right ?
lgbasallote
  • lgbasallote
if you use direct proof... it should be 3n + 2 = 2x then prove n is even
lgbasallote
  • lgbasallote
but like i said...should be contradiction though
anonymous
  • anonymous
well okay thats easy enough, too
lgbasallote
  • lgbasallote
hmmm then let's see you try
anonymous
  • 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
  • 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
  • anonymous
@sara12345 you cannot say "assume that n is odd" and then equate it to 2k
lgbasallote
  • lgbasallote
@sara12345 how is that contradiction
anonymous
  • anonymous
we do that always in proof by contradiction
anonymous
  • anonymous
RHS = even , LHS = odd => contradiction
lgbasallote
  • lgbasallote
i was referring to your solution
anonymous
  • anonymous
How can you prove that it equals 2k, though?
anonymous
  • 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
  • lgbasallote
hmm seems legit
lgbasallote
  • lgbasallote
so in proof by contradiction...you still substitute back huh
anonymous
  • anonymous
As @JamesWolf showed for you,
anonymous
  • anonymous
you take the contradiction, show that it is not internally consistent, and therefore it cannot be true.
anonymous
  • anonymous
no its not legit
anonymous
  • anonymous
ive been an idiot int he first step
anonymous
  • anonymous
@JamesWolf no, you inadvertently put up my proof.
anonymous
  • 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
  • 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
  • anonymous
@sara12345 yes you're right, sorry.
anonymous
  • 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
  • lgbasallote
oh so you take the negation of q then proceed from there?
lgbasallote
  • lgbasallote
anyway...is my proof right?
anonymous
  • 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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