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anonymous
 4 years ago
Prove that if n is an integer then 3n + 2 is even, then n is even
anonymous
 4 years ago
Prove that if n is an integer then 3n + 2 is even, then n is even

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mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0we have to prove two things? : 1) 3n+2 is even , 2) n is even

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, it's prove n is even GIVEN 3n + 2 is even.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 3n + 2 is even, then 3n is even

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0if 3n is even then n is even.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Given : 3n+2 is even To prove : n is even 3n+2 = 2k 3n = 2k2 3n = 2(k1)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0not true for n=1 ? 3+2=5<not even

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hartnn GIVEN that 3+2 is even, n is even

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0IF 3n+2 is even , prove that is even. @hartnn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But it's not so no guarentee is made.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0lets clear it from @lgbasallote what the exact question is...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just out of interest whats the general way to prove something is even, divide by 2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we prove a number is of form 2k that proves it is even

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw any1 here mind taking a look at my question? http://openstudy.com/updates/5078b9d7e4b02f109be44f95

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i just came back....anyway...i'd llike to see how proving by contradiction is done.. direct proof is too easy

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0but whats the question ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the statement in the blue box

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forgot to mention by the way....that the direct proof done here was wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they took n as the condition... 3n + 2 is supposed to be the condition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 3n + 2 = (a different n)?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0if 3n+2 is even , n is even. thats the question and u need to prove that using contradiction, right ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you use direct proof... it should be 3n + 2 = 2x then prove n is even

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but like i said...should be contradiction though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well okay thats easy enough, too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm then let's see you try

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0All 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
 4 years ago
Best ResponseYou've already chosen the best response.0Given : 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 = 2k2 3n = 2(k1) so we got the right side as even number, but we assumed n is odd, so left 3n becomes odd  contradiction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sara12345 you cannot say "assume that n is odd" and then equate it to 2k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sara12345 how is that contradiction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we do that always in proof by contradiction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS = even , LHS = odd => contradiction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was referring to your solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How can you prove that it equals 2k, though?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so in proof by contradiction...you still substitute back huh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As @JamesWolf showed for you,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you take the contradiction, show that it is not internally consistent, and therefore it cannot be true.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ive been an idiot int he first step

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@JamesWolf no, you inadvertently put up my proof.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in proof by contradiction, we assum e the opposite of what we need to prove as true, and proceed, not the opposite of given conditions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0@sara12345 yes you're right, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh so you take the negation of q then proceed from there?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0anyway...is my proof right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lgba ur proof is more correct as it shows the assumption n is odd as well by letting n =2k+1
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