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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.)
Assume x is odd, which implies x=2n+1 for some integer n. By the substitution property of equality, we can say
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||
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.
But also who is tied to brackets: (hope you understand)
So 6:2(3) = 6:2*3 = 3*3 = 9
9 is the ONLY correct answer
FAMOS - the first memory operating system
|20 May 2011, 23:08||
6/(2(3)) = 1
6/2(3) = 9
|20 May 2011, 23:50||
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