Question

Proof by using contrapositive. If n is an integer such that n^2 + 5 is odd, then n is even.

Answer 1

the contra-positve is ; if n is an odd integer, then n^2 + 5 is even

Answer 2

prove this and you have proved the original

Answer 3

Is that all? Dont I need to solve it with (k+1)?

Answer 4

contrapositive is ~q -> ~p so it should be

if n is an even integer then n^2 +5 is odd

Answer 5

if n is an even integer then n^2 +5 is odd
... this is q->p (inverse)

Answer 6

Oh ok. Do I need to solve it with (k+1)?

(k+1)^2 + 5?

Answer 7

p= n^2+5 is odd
q= n is even

(Assuming n is an integer)

Answer 8

right so contrapositve is ~q --> ~p

Answer 9

i don't know why you'd need k+1

Answer 10

to see if I can solve it to odd or even integer then it would make the statement true

Answer 11

1. n is an odd integer .... {Assumption}
2. n^2 is odd .... {odd x odd = odd }
3 n^2+5 is even .... { odd + odd = even}

Answer 12

oh you mean 2k+1 ... ???

Answer 13

Ah ok. I see it now. Thanks.

Answer 14

yes I meant 2k + 1.. sorry lol

Answer 15

2k+1 should work too

Answer 16

Im going to post another question. Can you please help me with that, too?

Answer 17

sure