Question

Prove that: For all n that are elements of the integers, (n^2 + n) / 2.

Answer 1

There is no resultant proposition in this statement.

Answer 2

didn't finish it! sorry! it ends with ...is an integer.

Answer 3

induction

Answer 4

prove by induction

Answer 5

1) check if it's true for n=1\[\frac{1^2+1}2=1\checkmark\]

Answer 6

2)assume it is true for some n\[\frac{n^2+n}2=k;k\in\mathbb N\]

Answer 7

\[\frac{(n+1)^2+(n+1)}2=\frac{n^2+n}2+n+1=k+n+1\]there we go

Answer 8

since k and n are integers, so is k+n+1
QED

Answer 9

Hooray for relearning induction!

Answer 10

would you be able to explain each of your steps in words so i can follow through with your thinking process? i really want to understand this, and i'm currently struggling with grasping the concept.

Answer 11

ok, induction involves three steps:
1) prove that the statement is true for some particular n
(often n=1 is used, as in this case)
2) assume that the statement is true for some n
(in other words we assume the statement is true if you just put in a variable n)
3)try to find if those assumptions imply that the statement is true for n+1

Answer 12

this works is because of the the axioms of numbers, in particular, Peano's 6th axiom
http://en.wikipedia.org/wiki/Peano_axioms

Answer 13

anyway, so we first check that this is true for some n
n=1 is an obvious choice...

Answer 14

1)is it true for n=1 ?\[\frac{1^2+1}2=1\in\mathbb Z\]the last two symbols mean 'is an integer'
hence we have shown that our statement is true for at least one n:
n=1
good so far?

Answer 15

yup...i wasn't aware you could specify an integer in the proof. was under the impression you had to keep it generalized. i missed the class in which we were lectured about induction so it's a new concept.

Answer 16

yes, that is a critical step

Answer 17

Apparently, I need to relearn algebra... I messed up a basic step when setting up after induction. XD

Answer 18

now that we know it is true for at least one integer, we can write it with a variable and assume that it is true
(if nothing else, we have already shown that it is true for at least one number, so this is a legal assumption)\[\frac{n^2+n}2=k\in\mathbb Z\]we assume the above to be true, which means that this unknown value k is assumed to be an integer

now finally, we prove that this implies that it is true for n+1...

Answer 19

you always use n + 1?

Answer 20

in this case we can do that by plugging in n+1 to the equation
in other cases you may have to express the successive term differently
for example look at the proof of the sum of n squares:
http://pirate.shu.edu/~wachsmut/ira/infinity/answers/sm_sq_cb.html
the hardes part of induction is deciding how you can formulate the n+1 term to prove your statement
anyhow, in this case we can prove it by putting in n+1\[\frac{(n+1)^2+(n+1)}2=\frac{n^2+2n+1+n+1}2=\frac{n^2+n}2+\frac{2n+2}2\]\[=\frac{n^2+n}2+n+1\]now because of our previous assumption we know that\[\frac{n^2+n}2=k\in\mathbb Z\]so since this first term that we call k is an integer, and n is an integer, we know that\[k+n+1\in\mathbb Z\]because the sum of integers is always an integer
thus we have proven our statement

Answer 21

yes, we always check if n+1 is true
how we represent that varies at times though

Answer 22

it's starting to become a bit more clear now. can you look at my previous question? i started a way of solving it and want to know if i am on the right track with my solution.