Thread: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd"

1. "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd"

I am having trouble understanding how to "prove" the following statement: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd". What I am aware of currently is that the definition of an odd number is "n = 2k + 1", where n is the number you are trying to prove is negative and k is any integer. However, if you place 2, or 4 into the formula as x, the result is an odd, negative number, which cannot be proven odd using the definition of a negative number since there is no integer than when multiplied by 2 results in a negative number. If you place a large, even x value into the equation, then the result is a positive, odd number which can be proven odd using the definition of an odd number, but if the result of the formula is negative and odd, how do you prove that?

2. Re: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd"

Originally Posted by ProtoflareX
I am having trouble understanding how to "prove" the following statement: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd". What I am aware of currently is that the definition of an odd number is "n = 2k + 1", where n is the number you are trying to prove is negative and k is any integer. However, if you place 2, or 4 into the formula as x, the result is an odd, negative number, which cannot be proven odd using the definition of a negative number since there is no integer than when multiplied by 2 results in a negative number. If you place a large, even x value into the equation, then the result is a positive, odd number which can be proven odd using the definition of an odd number, but if the result of the formula is negative and odd, how do you prove that?
I don't know what you mean by "direct method" but if x is even....
Is $\displaystyle x^2$ even or odd
Is 6x even or odd
Is 5 even or odd

-Dan

3. Re: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd"

So, x is an even integer. To put this in arbitrary terms, x = 2k, where k is any integer.

So, we have (2k)2 - 6(2k) + 5.

Going further, we have 4k2 - 12k + 5.

Now assess these terms as even or odd. What do we know when we add two evens? An even and odd? Two odds? It should be fairly simple from here.

4. Re: "Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd"

Originally Posted by ProtoflareX
"Prove by Direct Method: If x is an even integer, then x^2 - 6x + 5 is odd".
$\large x^2-6x+5=(x-1)(x-5)$