Mathematical Induction (1 Viewer)

[jchoi]

Hi, I've got a question which asks if n(n+1) is even for all n>0.

I'm just wondering if "even" just means divisible by 2.

so I got on my paper
Showing that statement is true for n=1:
1(2)=2 which is divisible by 2
Assuming that statement is true for n=k:
k(k+1)==2a where a is an interger.
Proving that statement is true for n=k+1:
(k+1)(k+2)=
=2a +2k+2
=2(a+k+1)
which is divisible by 2 as a+k+1 is an interger.

[Then the mouthful about if statement is tru for n=k, it is also.... blah]

This is correct, right?

 

[youngminii]

I guess.. The working doesn't seem very elegant to me and so in lieu of a better reply by someone smarter, I'll just say you'll get the mark.

 

[Iruka]

Does the question specifically tell you to use induction?

 

[kurt.physics]

Hi, I've got a question which asks if n(n+1) is even for all n>0

if the question doesnt specifically reference induction, then you could just simply prove it by cases.

 

[kurt.physics]

so I got on my paper
Showing that statement is true for n=1:
1(2)=2 which is divisible by 2
Assuming that statement is true for n=k:
k(k+1)==2a where a is an interger.
Proving that statement is true for n=k+1:
(k+1)(k+2)=
=2a +2k+2
=2(a+k+1)
which is divisible by 2 as a+k+1 is an interger.

[Then the mouthful about if statement is tru for n=k, it is also.... blah]

This is correct, right?

Yeh, this seams like a good way to go about it. Yeh, in a nutshell, even basically means divisible by 2 ie in the form 2m (2 times whatever )

 

[addikaye03]

Step three, Prove true for n=k+2 instead of n=k+1, assume that we begin on a even term

 

[Trebla]

k doesn't have to be even...it can be any positive integer. It's assumed k(k + 1) is even, not k.

Notice that n(n + 1) is the product of two consecutive integers, so one must be odd and the other must be even. Since the even number always has a factor of 2, then this factor will always make the number even. i.e. n(n + 1) = 2a.b for some integers a and b, where b is the odd part and 2a is the even part of the sequence {n, n + 1} and it is obvious that there is divisibility by 2...if that makes any sense lol