where are integers and .
In baby-talk: and are congruent (mod ) if their difference is a multiple of
Since , a multiple of
. . then: .
2) If is any rational number and is any irrational, then is irrational.
Proof by contradiction: .Assume is rational.
Then: . .for some integers and 
Since is rational: . for some integers and
Then equation  becomes: .
Since and are integers (integers are closed under multiplication),
. . is a rational number. .But we are told that is irrational.
We have reached a contradiction, hence our assumption is incorrect.
. . Therefore: . is irrational.
[Oh-oh! .You said direct proof, didn't you? . . . Never mind.]