Question

"if x and y are odd integers, then x + y is even"

Give a proof by contradiction of this theorem.

Answer 1

the statement "if x and y are odd integers, then x + y is even" is true, and a proof by contradiction is not applicable. Therefore, a proof by contradiction is not possible in this case. The sum of two odd integers will always be even.

This can be proven using the definition of even and odd integers:

An even integer is any whole number that can be expressed as a multiple of 2 (e.g., 2, 4, 6, 8).

An odd integer is any whole number that leaves a remainder of 1 when divided by 2 (e.g., 1, 3, 5, 7).

Now, let x and y be any two odd integers. We can represent them mathematically as:

x = 2a + 1, where a is an integer (since x is an odd multiple of 2 plus 1).

y = 2b + 1, where b is an integer (similarly for y).

Adding these two equations:

x + y = (2a + 1) + (2b + 1)

x + y = 2a + 2b + 2

Combining like terms:

x + y = 2(a + b + 1)

Since (a + b + 1) is clearly an integer, this represents an even multiple of 2, making x + y even.

Therefore, the statement "if x and y are odd integers, then x + y is even" is true, and a proof by contradiction is not applicable.

Answer 2

Hello!

The statement is "if x and y are odd integers, then x + y is even"

and we want to prove it by contradiction.

Suppose that we have x and y odd numbers, and suppose that his addition is odd.

We know that an odd number can be writen as (2n +1) (and a even number can be written as 2n) where n is an integer number; then:

x = (2k + 1) and y = (2m + 1)

and x + y = j, where j is also a odd number, then j = (2h + 1)

then:

2k + 1 + 2m + 1 = 2h + 1

2(k + m) + 2 = 2h + 1

2(k + m) +1 = 2h

if k and m are integers, then k + m is also an integer, suppose that k + m = g

then 2g + 1 = 2h

this says that in odd number is equal to an even number, then we have a contradiction, and the addition of two odd numbers cant be an odd number.