- #1
velouria131
- 13
- 0
I am coming across hiccups in my proof process. I am given this problem - Prove: if n is an integer and 3n + 2 is even that n is also even. I have to apply a contrapositive proof to this problem. The form is then [itex]\neg[/itex]q therefore [itex]\neg[/itex]p .The problem becomes - if n is odd, prove that 3n+2 is even.
Work:
Prove - if n is odd, prove that 3n+2 is even.
step 1 - if n is odd, n = 2k+1 for some integer k
step 2 - 3n + 2 = 3(2k+1) + 2 = 6k + 5
step 3 - This is my issue. A contrapositive proof for this problem would give not 'p', or, that 3n+2 is odd when n is odd. Do I now have to show that 6k + 5 is an odd number for any positive integer k? Or, should I just prove that 3n + 2 is odd when n is odd? If I take this route, could I choose another proof method, essentially having a 'proof within a proof'?
Work:
Prove - if n is odd, prove that 3n+2 is even.
step 1 - if n is odd, n = 2k+1 for some integer k
step 2 - 3n + 2 = 3(2k+1) + 2 = 6k + 5
step 3 - This is my issue. A contrapositive proof for this problem would give not 'p', or, that 3n+2 is odd when n is odd. Do I now have to show that 6k + 5 is an odd number for any positive integer k? Or, should I just prove that 3n + 2 is odd when n is odd? If I take this route, could I choose another proof method, essentially having a 'proof within a proof'?