Question

Let x,y,z e Z. Prove that if x+y= 2, then at least one of , y, and z must be even.

Answer 1

It is given that, .

2, 4,2 EZ

Therefore, x can be either an odd integer or an even integer. Similarly, y can be either an odd integer or an even integer.

We are given, $x+y=z$

Since x can take 2 types of values (odd/even), and y can also take 2 types of values (odd/even), we have to check 4 ($2\times 2=4$) cases, as follows.

CASE 1: When x is an odd integer, and y is an odd integer.

ODD INTEGER(x) + ODD INTEGER(y) = EVEN INTEGER(z). Here z, has to be an even integer, because the sum of two odd integers, is always an even integer. Therefore, at least one of x,y, and, z must be even. Hence proved.

CASE 2: When x is an odd integer, and y is an even integer.

Here, y is already an even integer. Therefore, at least one of x,y, and, z must be even. Hence proved.

CASE 3: When x is an even integer, and y is an odd integer.

Here, x is already an even integer. Therefore, at least one of x,y, and, z must be even. Hence proved.

CASE 4: When x is an even integer, and y is an even integer.

Here, x & y are already an even integer. Therefore, at least one of x,y, and, z must be even. Hence proved.

Hence, we can conclude, that, at least one of x,y, and, z must be even. Hence proved.