/sci/ - product of two consecutive integers is even

Name: Anonymous 2007-01-25 15:39

Can someone prove for me that n(n + 1) where n is an arbitrarily chosen integer can be represented as 2k where k is an integer? It was on the board today, and apparently I need to know it, but I was sleeping.

Name: Anonymous 2007-01-25 15:47

nevermind I'm retarded, 2n(2n + 1) = 2(2n^2 + n)

Name: Anonymous 2007-01-25 19:05

Indeed.

Name: Anonymous 2007-01-25 22:08

>>2
You didn't answer your own question...

if n is even, n(n+1) = 2*((n/2)(n+1)). if n is odd, n(n+1) = 2*(n((n+1)/2)).

Name: Anonymous 2007-01-25 23:30

>>4
noob
GTFO my science and math

Name: Anonymous 2007-01-26 1:26 (sage)

>>5
Nice rebuttal

Name: Anonymous 2007-01-26 12:47

this thread is full of fucking morons.

Name: 4tran 2007-01-28 0:01

product of n consecutive integers is divisible by n

Name: Anonymous 2007-01-28 0:05

>>8
what if n=0?!

Name: Anonymous 2007-01-28 0:38

use induction

Name: Anonymous 2007-01-28 1:30

>>10
no need since >>4 already directly proved it

Name: Anonymous 2007-01-28 2:29

>>2 and >>4 have already shown what was required but for something different let
S = n + n-1 + n-2 + . . . + 1
S = 1 + 2 + 3 + . . . + n
S is an element of the integers since integers are closed under addtion. Adding related components gives
2S = (n+1) + (n+1) + (n+1) + . . . + (n+1)
= n(n+1)
=> n(n+1) is an even integer since 2*integer is always even.

Name: Anonymous 2007-01-28 2:33

>>12
Opps forgot to point out n is an integer as well otherwise fail.

Name: Anonymous 2007-01-28 2:44

>>11

Actually, >>10 was right - you do need induction. What >>4 was doing was "plug in the integer into this eqn if it's even, and plug it in the other if it's odd"

You need to prove it's true for the entire set of integers, which is where induction comes into play.

Name: Anonymous 2007-01-28 8:02

test test

Name: Anonymous 2007-01-28 8:05

um you guys are all morons

if you have two numbers
number and number+1
one of them is even, because the other is odd.

every even number is divisible by two, i.e. its factorisaton is like 2 * (otherstuff)
so 2 * (otherstuff) * (number + 1)

OH LOOK 2 IS IN THE FACTORISATION OF THE PRODUCT OF THE TWO NUMBERS
WEL FUUUUUUUUUUUCK MEEEEEEE THAT WAS DIFFIFUCKLT.

Name: Anonymous 2007-01-28 14:38

>>14
As >>16 said, integers are either even or odd, and if n is not even than n+1 is. This is the basis of >>4's non-inductive proof.

product of two consecutive integers is even

1 Name: Anonymous 2007-01-25 15:39

2 Name: Anonymous 2007-01-25 15:47

3 Name: Anonymous 2007-01-25 19:05

4 Name: Anonymous 2007-01-25 22:08

5 Name: Anonymous 2007-01-25 23:30

6 Name: Anonymous 2007-01-26 1:26 (sage)

7 Name: Anonymous 2007-01-26 12:47

8 Name: 4tran 2007-01-28 0:01

9 Name: Anonymous 2007-01-28 0:05

10 Name: Anonymous 2007-01-28 0:38

11 Name: Anonymous 2007-01-28 1:30

12 Name: Anonymous 2007-01-28 2:29

13 Name: Anonymous 2007-01-28 2:33

14 Name: Anonymous 2007-01-28 2:44

15 Name: Anonymous 2007-01-28 8:02

16 Name: Anonymous 2007-01-28 8:05

17 Name: Anonymous 2007-01-28 14:38