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

1. mathslover

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

2. anonymous

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

3. mathslover

ok , thanks

4. anonymous

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

5. mathslover

if 3n is even then n is even.

6. anonymous

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

7. hartnn

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

8. anonymous

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

9. mathslover

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

10. anonymous

But it's not so no guarentee is made.

11. hartnn

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

12. mathslover

@lgbasallote ?

13. anonymous

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

14. anonymous

yes

15. anonymous

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

16. anonymous

right I see

17. anonymous

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

18. anonymous

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

19. hartnn

but whats the question ?

20. anonymous

the statement in the blue box

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. hartnn

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

25. anonymous

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

26. anonymous

but like i said...should be contradiction though

27. anonymous

well okay thats easy enough, too

28. anonymous

hmmm then let's see you try

29. 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

30. 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

31. anonymous

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

32. anonymous

33. anonymous

we do that always in proof by contradiction

34. anonymous

RHS = even , LHS = odd => contradiction

35. anonymous

i was referring to your solution

36. anonymous

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

37. 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

38. anonymous

hmm seems legit

39. anonymous

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

40. anonymous

As @JamesWolf showed for you,

41. anonymous

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

42. anonymous

no its not legit

43. anonymous

ive been an idiot int he first step

44. anonymous

@JamesWolf no, you inadvertently put up my proof.

45. 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

46. anonymous

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

47. anonymous

@sara12345 yes you're right, sorry.

48. 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

49. anonymous

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

50. anonymous

anyway...is my proof right?

51. anonymous

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