# Irrationality homework problem help

1. Jan 5, 2008

### lkh1986

1. The problem statement, all variables and given/known data
Prove that for any n, sqrt n + sqrt (n+1) is irrational.

2. Relevant equations

3. The attempt at a solution

Well, I know that sqrt 2 + sqrt 3 is irrational. How about sqrt n + sqrt (n+1)?

Let sqrt n + sqrt (n+1) be rational.
[sqrt n + sqrt (n+1)]^2 = (m^2/n^2)
n + (n+1)^2 + 2 sqrt(n^2 + n) = (m^2/n^2)
n + (n+1)^2 is rational.
Then, I don't know how to continue...

I read from a book that to prove the sqrt of a composite number is irrational, we should use one method. Then if we were to prove the sqrt of a prime number is irrational, we should use another method. Then I read from another website that in Real Analsysis, there are many approaches to solve a problem.

I take this Introductory Analysis this semester. This course is so different from the other maths courses. Does memorising all of the axioms help in this case? Since all the proof need them. Thanks.

2. Jan 5, 2008

### morphism

This is only true if n is a positive integer. Let's use what you have so far to prove it: 2 sqrt(n^2 + n) = (m^2/n^2) - 2n - 1 [note: your (n+1)^2 should have been (n+1)]. The right side is rational, so if we manage to show that the left side is irrational, we will get our desired contradiction. Here it's helpful to know that sqrt(k) is rational if and only if k is a perfect square. We can then finish things off by showing that n^2+n=n(n+1) is a perfect square iff n=0. I'll leave this to you.

3. Jan 5, 2008

### Rainbow Child

I think that the first step should be

$$(\sqrt{n}+\sqrt{n+1})^2=(\frac{m}{k})^2\Rightarrow n=(\frac{m^2-k^2}{2\,m\,k})^2$$

why you can write the rational as $$\frac{m}{n}$$, i.e. use the same symbol n?

Am I loosing something?

Last edited: Jan 5, 2008
4. Jan 5, 2008

### morphism

No, you're absolutely right. It shouldn't be the same n.

5. Jan 5, 2008

### Rainbow Child

Ok!
First of all, since n>1 we must have m>k, thus $$\frac{k}{m}<1$$.
Let's now, investigate the fraction $$\frac{m^2-k^2}{2\,m\,k}$$.
The positive integer m must divides either m-k or m+k, but

$$\frac{m-k}{m}=1-\frac{k}{m}, \quad \frac{m+k}{m}=1+\frac{k}{m}$$

either of which is integer.

6. Jan 5, 2008

### Shooting Star

All this is all right but not necessary. It is enough to show that n(n+1) cannot be a perfect square, as mentioned by morphism.

7. Jan 5, 2008

### Rainbow Child

But that was wrong because the OP wrote the rational as $$\frac{m}{n}$$ where n was the original positive integer. See post #3

8. Jan 5, 2008

### morphism

No, there's nothing wrong with what I posted (aside from copying down the OP's m^2/n^2). I didn't use the second n at all, only the fact that the right side is rational.

9. Jan 5, 2008

### Rainbow Child

Ok! I was stuck with the second n and I didn't notice what you said!
You are both right!

10. Jan 5, 2008

### lkh1986

Let see. Well, the product of any 2 succecive integers cannot be a perfect square. Unless it is 0x1=0. Can this statement be used? Or we should write something like n(n+1) = k^2?

11. Jan 5, 2008

### morphism

You just translated it into words. Try to prove it: why can't the product of two successive integers be a perfect square?

12. Jan 5, 2008

### Shooting Star

That's the way to proceed.

13. Jan 5, 2008

### Gib Z

Would it be correct to instead say; Since perfect squares are never consecutive integers ( I guess we could prove that if we had to), we have 2 cases: One of the terms is a perfect square, and the other is irrational - the addition of a rational number to an irrational is once again irrational. Or neither of the terms are perfect squares, and are both irrational. Now my conjecture is that the sum of 2 irrational numbers is another irrational number provided one of the numbers is the additive inverse of the other. There may be other conditions I don't see right now, but with further work could this be made into another method of proof?

14. Jan 5, 2008

### morphism

If you add a number to its additive inverse you'll get 0, a rational number! Am I reading this correctly?

15. Jan 5, 2008

### Gib Z

TYPO!! TYPO!!! *tears* damn I meant neither of the numbers :(

16. Jan 5, 2008

### morphism

Ah I figured you did; unfortunately, it's necessary but not sufficient, e.g. (1-sqrt(2)) + sqrt(2) = 1.

17. Jan 5, 2008

### Gib Z

Sigh ok, thanks morphism =]

18. Jan 5, 2008

### Fizzicist

This is fairly simple if you use induction. Okay, here we go

Step 1: Establish your base case

We have to establish that this is irrational for some number n. It's always good to try n=1 at first because if you establish this as your base case you won't have to use strong or "backwards" induction (i.e. proving the statement is true for all n smaller than the base case as well). Using n=1, we find that the expression evaluates to sqrt(2) + 1, which is irrational (I'm not going to bother proving this. If your professor doesn't accept this as axiomatic and wants you to prove it, do it yourself).

Step 2: Prove that the expression is true for all n

We have already established the truth of the statement for n=1. Since 1 is the lowest possible value of n, all we need to do is establish the truth of the statement for n+1 (which proves that it's true for all values of n greater than 1 as well). So, let's substitute n+1 into the expression. This gives us sqrt(n+1) + sqrt(n+2). Now let's use proof by contradiction. Assume this expression is equal to a/b, where a and b are integers (and hence a/b is rational). From here, I'll let you take over, but I'll give you a hint: first use a conjugate to get rid of the square roots.

Last edited: Jan 5, 2008
19. Jan 5, 2008

### Gib Z

That seems to be the original question...

20. Jan 5, 2008

### Fizzicist

How is that the original question? The original problem was to prove that sqrt(n) + sqrt(n+1) is irrational for all n. Establishing a base case and proving that sqrt(n+1) + sqrt (n+2) is irrational will prove this.