Calculus theory proof- Suppose a is irrational, prove√(1+a) is irrational.

[math-help-me]

Homework Statement

Suppose a is irrational, prove√(1+a) is irrational.

Homework Equations

A number is rational if it can be expressed as p/q, p,q integers with q≠0

The Attempt at a Solution

I can reason through it intuitively but not sure how to demonstrate it formally. Any help or advice would be greatly appreciated

 

[Ansatz7]

I think you can use a proof by contradiction here. Assume you can write sqrt(1 + a) as p/q with p and q integers and see what happens.

 

[math-help-me]

that was sort of my attempt at a solution. This is what I've got now.
Assume a irrational, but √(1+a) is rational.
Then √(1+a) = p/q for p,q integers q≠0.
1+a = p^2/q^2 → a = p^2/q^2 - 1 → (p^2- q^2)/q^2
Since p,q integers, p^2-q^2 and q^2 must be integers.
Thus a must also be rational by definition, a contradiction.
Thus if a is rational, √(1+a) is rational. The contrapositive is equivalent.
Therefore if a is irrational, √(1+a) is irrational.

Is this all valid and a valid conclusion? Thanks again!

 

[jbunniii]

that was sort of my attempt at a solution. This is what I've got now.
Assume a irrational, but √(1+a) is rational.
Then √(1+a) = p/q for p,q integers q≠0.
1+a = p^2/q^2 → a = p^2/q^2 - 1 → (p^2- q^2)/q^2
Since p,q integers, p^2-q^2 and q^2 must be integers.
Thus a must also be rational by definition, a contradiction.

Correct up to this point.

Thus if a is rational, √(1+a) is rational.

No, it's the other way around. If $\sqrt{1+a}$ is rational, then $a$ is rational. This is exactly what you just proved.

And therefore:

The contrapositive is equivalent.
Therefore if a is irrational, √(1+a) is irrational.!

 

[Deveno]

suppose that a was irrational, but √(1+a) = r was rational.

what can you say, then, about r²?