Here's the question you clicked on:
lgbasallote
Prove that if n is an integer then 3n + 2 is even, then n is even
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.
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
btw any1 here mind taking a look at my question? http://openstudy.com/updates/5078b9d7e4b02f109be44f95
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
so in proof by contradiction...you 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.
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 odd...so 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?
anyway...is my proof right?
lgba ur proof is more correct as it shows the assumption n is odd as well by letting n =2k+1