1. ## Prove irrationality!

Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.

I've planned my approach in 2 ways:

1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.

But how to go about it?

2. ## Trying to help

It appears to me that since the first term doesn't cancel the remaining terms, then all you have to do is prove that the first term is irrational.

Any whole number that's squared which you add one to has to be irrational when you take its square root. I think you can work through the rest.

3. Originally Posted by wonderboy1953
It appears to me that since the first term doesn't cancel the remaining terms, then all you have to do is prove that the first term is irrational.

Any whole number that's squared which you add one to has to be irrational when you take its square root. I think you can work through the rest.
Are you really sure thats enough? What about $\sqrt{2}+2,1-\sqrt{2}$ then they don't cancel out but it is nonetheless rational. You need to say a little more if you want to use that argument.

4. Originally Posted by usagi_killer
Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.
This is a theorem: If N is a non-square positive integer then $\sqrt{N}$ is irrational.
What do you know about the sum of irrational numbers?

Does that help? HOW?

5. Originally Posted by usagi_killer
Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.

I've planned my approach in 2 ways:

1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.

But how to go about it?
This is equivalent to $\sum_{x= 1001}^{2000} \sqrt{1+x^2}$. And $\sqrt{x^2+1}$ is not a square number which implies it is irrational.

Since it is monotonic increasing, the sum of the irrationals can't be rational (e.g. we can't have expressions of the form $(2-\sqrt{2}) +(4+\sqrt{2})$).

6. Originally Posted by Sampras
This is equivalent to $\sum_{x= 1001}^{2000} \sqrt{1+x^2}$. And $\sqrt{x^2+1}$ is not a square number which implies it is irrational.

Since it is monotonic increasing, the sum of the irrationals can't be rational (e.g. we can't have expressions of the form $(2-\sqrt{2}) +(4+\sqrt{2})$).

Perhaps I'm missing something here, but your very example is an increasing sequence of irrational numbers whose sum is rational! There are others, of course: $\sqrt{2}-1\,,\,\sqrt{2}-\frac{1}{2}\,,\,\sqrt{2}\,,\,7-3\sqrt{2}$, etc.

Tonio

7. Originally Posted by tonio
Perhaps I'm missing something here, but your very example is an increasing sequence of irrational numbers whose sum is rational! There are others, of course: $\sqrt{2}-1\,,\,\sqrt{2}-\frac{1}{2}\,,\,\sqrt{2}\,,\,7-3\sqrt{2}$, etc.

Tonio
It's not an alternating sequence, so the sum should be irrational.

8. Okay so had another look at this.

So to prove the sum is not an integer.

$n < \sqrt{n^2+1} < n + \frac{1}{n}$

So, $1001 < \sqrt{1001^2+1} < 1001 + \frac{1}{1001}$

Let $\frac{1}{n} = q$

An overestimate for the sum would be $1001 + q_1 + 1002 + q_2 + 1003 + q_3 + ... + 2000 + q_{1000}$

An underestimate for the sum would be $1001 + 1002 + 1003 + ... + 2000$

But since $0 < q_1 + q_2 + q_3 + ... + q_{1000} < \frac{1}{1001}$

Thus the sum is not an integer.

Use induction to prove it is a zero of a monic polynomial.

Base case:

Let the sum be written as $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_n}$

Let $n = 1$.

Thus $\sqrt{a_1} = y$

$y^2 - a_1 = 0$

Thus $a_1$ is a zero.

Now for the inductive hypothesis, assume $n = k$ is true.

Thus $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k}$ is true.

Now we wish to prove that $n = k+1$ is also true.

We will denote this as $x$.

$x = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k} + \sqrt{a_{k+1}}$

$\implies x = y + \sqrt{a_{k+1}}$

$\implies y = x - \sqrt{a_{k+1}}$

Let this be a zero of a monic polynomial $P(x) = x^r + c_{r-1}x^{r-1} + c_{r-2}x^{r-2} + ... + c_0$

So using the inductive hypothesis, $y$ is a zero $P(x)$, namely $P(y) = 0$

so $P(y) = P(x - \sqrt{a_{k+1}}) = (x - \sqrt{a_{k+1}})^r + c_{r-1}(x - \sqrt{a_{k+1}})^{r-1} + c_{r-2}(x - \sqrt{a_{k+1}})^{r-2} + ... + c_0$

Now what...?

9. Originally Posted by usagi_killer
Okay so had another look at this.

So to prove the sum is not an integer.

$n < \sqrt{n^2+1} < n + \frac{1}{n}$

So, $1001 < \sqrt{1001^2+1} < 1001 + \frac{1}{1001}$

Let $\frac{1}{n} = q$

An overestimate for the sum would be $1001 + q_1 + 1002 + q_2 + 1003 + q_3 + ... + 2000 + q_{1000}$

An underestimate for the sum would be $1001 + 1002 + 1003 + ... + 2000$

But since $0 < q_1 + q_2 + q_3 + ... + q_{1000} < \frac{1}{1001}$

Thus the sum is not an integer.

Use induction to prove it is a zero of a monic polynomial.

Base case:

Let the sum be written as $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_n}$

Let $n = 1$.

Thus $\sqrt{a_1} = y$

$y^2 - a_1 = 0$

Thus $a_1$ is a zero.

Now for the inductive hypothesis, assume $n = k$ is true.

Thus $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k}$ is true.

Now we wish to prove that $n = k+1$ is also true.

We will denote this as $x$.

$x = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k} + \sqrt{a_{k+1}}$

$\implies x = y + \sqrt{a_{k+1}}$

$\implies y = x - \sqrt{a_{k+1}}$

Let this be a zero of a monic polynomial $P(x) = x^r + c_{r-1}x^{r-1} + c_{r-2}x^{r-2} + ... + c_0$

So using the inductive hypothesis, $y$ is a zero $P(x)$, namely $P(y) = 0$

so $P(y) = P(x - \sqrt{a_{k+1}}) = (x - \sqrt{a_{k+1}})^r + c_{r-1}(x - \sqrt{a_{k+1}})^{r-1} + c_{r-2}(x - \sqrt{a_{k+1}})^{r-2} + ... + c_0$

Now what...?
if you are familliar with basic concepts of algebraic number theory, you should know this much that the set of integral elements (an elements is called integral if it's a root of a monic polynomial
with integer coefficient) is a ring and hence it's closed under addition. now each $\sqrt{a_i}$ is integral and thus $A=\sum_{i=1}^n \sqrt{a_i}$ is integral too. it's also a basic fact that $\mathbb{Z}$ is integrally closed, that is an
integral element which is also a rational number has to be an integer. but, as you almost showed, $A$ is not an integer and so it cannot be rational. Q.E.D.

10. Originally Posted by Sampras
It's not an alternating sequence, so the sum should be irrational.

Uuh? But it is not!
$\sqrt{2}-1+\sqrt{2}-\frac{1}{2}+\sqrt{2}+7-3\sqrt{2}=7-\frac{3}{2}=\frac{11}{2}\,\in\mathbb{Q}$

Tonio