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We prove `p`(`x`, `y`) : <=> `x`+`y`
is even is an equivalence relation on **N**.

`p`is clearly a binary relation on**N**.`p`is reflexive on**N**: Take arbitrary`x`in**N**. We have to show`x`+`x`is even, i.e, 2`x`is even.`p`is symmetric on**N**: Take`x`in**N**,`y`in**N**. We assume`x`+`y`is even. Then`y`+`x`is even.`p`is transitive on**N**: Take arbitrary`x`in**N**,`y`in**N**, and`z`in**N**. We assume(1)

We have to show (2)`x`+`y`is even /\`y`+`z`is even.`x`+`z`is even. From (1), we have some`a`in**N**and`b`in**N**such that(3) 2

Thus we know (2) because of`a`=`x`+`y`/\ 2`b`=`y`+`z`.`x`+`z`= (`x`+`y`) + (`y`+`z`) - 2`y`= 2`a`+ 2`b`- 2`y`= 2(`a`+`b`-`y`).

Author: Wolfgang Schreiner

Last Modification: January 12, 2000