Prove irrationality!

**usagi_killer** · November 16th 2009, 01:23 AM

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?

**wonderboy1953** · November 16th 2009, 09:07 AM

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.

**Drexel28** · November 16th 2009, 10:42 AM

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.

**Plato** · November 16th 2009, 12:25 PM

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?

**Sampras** · November 16th 2009, 12:35 PM

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 $(2-\sqrt{2}) +(4+\sqrt{2})$ ).

**tonio** · November 16th 2009, 01:27 PM

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

**Sampras** · November 16th 2009, 01:33 PM

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.

**usagi_killer** · November 16th 2009, 05:42 PM

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...?

**NonCommAlg** · November 16th 2009, 07:21 PM

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.

**tonio** · November 16th 2009, 07:23 PM

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

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