Hello, somestudent2!

Prove: .

where are integers and .

Definition: .

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 isrational.

Then: . .for some integers and[1]

Since is rational: . for some integers and

Then equation [1] 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 saiddirectproof, didn't you? . . . Never mind.]