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

In summary, to prove that √(1+a) is irrational when a is irrational, we can use a proof by contradiction by assuming that √(1+a) is rational and showing that this would imply that a is rational, which contradicts our initial assumption. Therefore, if a is irrational, √(1+a) must also be irrational.
  • #1
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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
 
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  • #2
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.
 
  • #3
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!
 
  • #4
math-help-me said:
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 [itex]\sqrt{1+a}[/itex] is rational, then [itex]a[/itex] is rational. This is exactly what you just proved.

And therefore:
The contrapositive is equivalent.
Therefore if a is irrational, √(1+a) is irrational.!
 
  • #5
suppose that a was irrational, but √(1+a) = r was rational.

what can you say, then, about r2?
 

1. What is the definition of an irrational number?

An irrational number is a real number that cannot be expressed as a ratio of two integers. This means that its decimal representation is non-terminating and non-repeating.

2. How is the square root of an irrational number defined?

The square root of a number is a value that, when multiplied by itself, gives the original number. In the case of irrational numbers, the square root cannot be expressed as a rational number and is therefore also irrational.

3. How can we prove that the square root of an irrational number is also irrational?

One way to prove this is by contradiction. We assume that the square root of an irrational number is rational, and then show that this leads to a contradiction. This contradiction proves that our initial assumption was false, and therefore the square root must be irrational.

4. What is the proof for the statement "Suppose a is irrational, prove √(1+a) is irrational"?

We can prove this statement through contradiction. Suppose √(1+a) is rational. This means that there exist two integers, p and q, such that √(1+a) = p/q. We can then manipulate this equation to show that a must be rational, which contradicts our initial assumption that a is irrational. Therefore, √(1+a) must be irrational.

5. Can this proof be applied to similar statements involving other irrational numbers?

Yes, this proof technique can be applied to similar statements involving other irrational numbers. As long as the initial assumption is that the irrational number is rational, and this leads to a contradiction, the proof is valid.

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