Question

Show that (n^2+ n) is divisible by 2 for all natural numbers n.

Answer 1

you can factor n^2+n

Answer 2

And guess what must be true about one of the factors of n^2+n?

Answer 3

or you can also do this by induction

Answer 4

well, yes.... for even n, (n) is even and for odd n (n+1) is even.

and Even times Odd = Even
(and that means that any output is divisible by 2)

(myininaya already posted that)

Answer 5

yeh by induction is what im looking for

Answer 6

could you help me out ?

Answer 7

so we are proving this for natural numbers

Answer 8

base case is n=1

Answer 9

Do you have 2 divides 1^2+1?

Answer 10

If so move on to the assumption part of the proof.

Answer 11

Suppose that for some integer k>=1 we have 2 divides k^2+k

Answer 12

Now the trick is to use that assumption to prove 2 divides (k+1)^2+(k+1)

Answer 13

hint: You will need 2a=k^2+k for some integer a. And you will need to multiple out the (k+1)^2 of the (k+1)^2+(k+1) expression .

Answer 14

oh ok let me see

Answer 15

base case :
\[k^{2}+k\]
now for k+1:
\[(k^2+1+(k+1))\]

Answer 16

you mean (k+1)^2+(k+1)?

Answer 17

base case is proving it for the first case which is n=1

Answer 18

\[k^2 +2k+1+(k+1)\]

Answer 19

ok now you are given something about k^2+k

Answer 20

\[k^2+3k+2\] now what Do i have to do .. ?

Answer 21

\[k^2+2k+1+(k+1) \\ =k^2+k+2k+1+1=k^2+k+2k+2=k^2+k+2(k+1)\]

Answer 22

what are you given about k^2+k

Answer 23

Remember
you made the assumption that 2a=k^2+k for some integer a.

Answer 24

so 2a +2(k+1) ?

Answer 25

yep yep
factor the 2 out :)

Answer 26

wow i never thought that my answer would be right lol

Answer 27

So you have
Base case: n=1:
2 divides 1^2+1

Assumption part:
Assume for some integer k>=1 we have 2a=k^2+k for some integer a.

Inductive case:
(k+1)^2+(k+1)=k^2+k+2(k+1)=2a+2(k+1)
=2(a+k+1)
which means what?

Answer 28

That 2 divides ?

Answer 29

is it beacuase of the coeficient two ?

Answer 30

(k+1)^2+(k+1) has the factor 2 and the factor (a+k+1)
where a+k+1 is also an integer.
having 2 as a factor means that 2 divides the (k+1)^2+(k+1)

Answer 31

wow you're so great at induction

Answer 32

thanks a lot you're amazing

Answer 33

:)

Answer 34

The earlier proof was to factor n^2+n and get n(n+1).
Knowing n is natural then that means that n or (n+1) will be even since n and (n+1) are consecutive natural numbers.
Example
2,3 you have 2 is even

11,10 you have 10 is even

so n(n+1) will always have an even factor which means it is divisible by 2

Answer 35

any integer (a +k+1) multyplied by 2 is an even number that's why is divisible by two as well

Answer 36

meant to list 10,11 in order oh well
9,10 you have 10 is even

Answer 37

many thanks really really i apreciate your help