Alexmsi wrote:

@Bunuel:

I don't understand your approach for the first argument:

\(2x-1=odd\)

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?

Several things:

1. \(2x-1=odd\) --> \(2x=odd+1=odd+odd=even\) --> so \(2x=even\) --> \(x=\frac{even}{2}=integer\). Hence \(2x-1=odd\) just means that \(x\) is an integer (it can be even as well as odd).

2. If \(x=1\)the result wont be zero, it'l be 1, so odd: \(2*1-1=2-1=1=odd\).

3.

Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an

integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an

integer of the form \(n=2k\), where \(k\) is an integer. So for \(k=0\) --> \(n=2*0=0\).

Hope it's clear.

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