prove or disprove that no integer is both even and odd?

Answer 1

Every even integer can be expressed as, 2n, where n is an integer. Likewise, every odd integer can be expressed as, 2n + 1, where n is an integer.
Assume, m is an integer, which is both odd and even. Then there exists integers, a & b, such that:
m = 2a & m = 2b + 1.
2a = 2b + 1
2(a - b) = 1
a - b = (1/2)
This is a contradiction, since a & b are integers. Therefore, m can not be both even and odd.

Answer 2

If an integer, say, X is divisible by 2, then it is even.
If u add 1 to it (now it becomes odd), n then divide it by 2, the result won't be an integer, it'll only be a decimal number of the form " X.5 "
Hence proved, that no integer can be both even n odd at the same tme.

Answer 3

I would say that is pretty much true by definition. But lets go for definitions that possibly give a little leeway:
n is even if it belongs to the set 2Z = { ... -4,-2,0, 2, 4 ... }
n is odd if it belongs to the set 2Z+1 = { ... -3, -1, 1, 3, 5 ... }
(where Z is the set of all integers)

now let n belong to 2Z (is even). then n = 2x for some x in Z
assume n also belongs to 2Z+1 (is odd). then n=2y + 1 for some y in Z
so 2x= 2y+1
x = y+0.5
BUT y is a whole number, so x cannot be an integer which is a contradiction.

So n is not odd.

Answer 4

An even integer could be divided by 2 exactly while odd leaves a remainder of 1. It's logic! Ok you think of a number that could be divided by 2 exactly but still leaving a remainder 1. There's none right?

Answer 5

You make the decomposition of the integer into its prime factors.
( You know that each integer is or prime or has a unique decomposition into prime factors)
If the prime factor 2 appears it is even .If it doesn´t appear it is odd.Both things can´t happen

Answer 6

let the number be m,such that there exists a,b members of Z such that m=2a(even) and m=2b+1(odd)
so, 2a=2b+1
2(a-b)=1
1=2k where k=(a-b)...this is a contradiction since 1 is odd.
@wiz