discrete math - proof

[Ikenna]

Show that if n3 + 5 is odd, then n is even, where n is an integer.

In your proof method, what will you assume?
(a) n is even
(b) n = 2k + 1, where k is an integer
(c) n3 + 5 is odd
(d) n3 + 5 = 2k, where k is an integer
(e) n = 2k, where k is an integer

Note: n3 is n to the power 3

 

[lookagain]

Show that if n3 + 5 is odd, then n is even, where n is an integer.

In your proof method, what will you assume?
(a) n is even
(b) n = 2k + 1, where k is an integer
(c) n3 + 5 is odd
(d) n3 + 5 = 2k, where k is an integer
(e) n = 2k, where k is an integer

Note: n3 is n to the power 3

How many correct choices are there to be?

It depends on the method of proof: direct, by contradiction, by contrapositive, inductive, etc.

 

[Ikenna]

Method of proof is by Contraposition

 

[Deleted member 4993]

Show that if n3 + 5 is odd, then n is even, where n is an integer.

In your proof method, what will you assume?
(a) n is even
(b) n = 2k + 1, where k is an integer
(c) n3 + 5 is odd
(d) n3 + 5 = 2k, where k is an integer
(e) n = 2k, where k is an integer

Note: n3 is n to the power 3

Is that n*3 (=3*n = 3n) or n[sup:11phtdyt]3[/sup:11phtdyt]?