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Author
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
I forgot to define "even." A number is even iff it can be written in the form 2k, where k is an integer.

Nice proof MHajduk, it's more elegant than mine. (Which is to be expected; I haven't taken a proof class yet.)

Mine:

Assume x is odd, which implies x=2n+1 for some integer n. By the substitution property of equality, we can say
x^2-x=2(2n^2-n).
Given that the integers are closed under addition and multiplication, it follows that 2n^2-n is an integer and that 2(2n^2-n) is of the form 2k, where k is an integer. Given that an integer is even iff it can be written in the form 2k, we can say 2(2n^2-n) is even. Since, by assuming x is odd, we are able to rewrite it as an even number, we have proven it to be even for all odd x.

EDIT: 1000th post

Last edited by Tyler on 21 May 2011, 02:59; edited 1 time in total
20 May 2011, 21:20
neville

Joined: 13 Jul 2008
Posts: 507
Location: New Zealand
neville
gabiz_ro wrote:
But also who is tied to brackets: (hope you understand)
6:2(2+1)=6:2(3)=6:6 (because of brackets) =1 final answer
There is no valid mathematical construct called "tied to brackets". Implied multiplication is still just multiplication. After evaluating expressions inside brackets, calculation is performed from left to right.
So 6:2(3) = 6:2*3 = 3*3 = 9
9 is the ONLY correct answer

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20 May 2011, 23:08
typedef

Joined: 25 Jul 2010
Posts: 2913
Location: 0x77760000
typedef
6/(2(3)) = 1
6/2(3) = 9
20 May 2011, 23:50
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