# sqrt 2, sqrt 3, sqrt 5 can't be in the arithmetic/geometric progression

• January 5th 2008, 10:30 AM
james_bond
sqrt 2, sqrt 3, sqrt 5 can't be in the same arithmetic/geometric progression
Prove that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ can't be terms in the same
• arithmetic
• geometric

progression.
• January 5th 2008, 10:55 AM
janvdl
Quote:

Originally Posted by james_bond
Prove that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ can't be terms in the same
• arithmetic
• geometric

progression.

Calculate $r$ and $d$. You'll see they're not the same between $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{3}$, $\sqrt{5}$
• January 5th 2008, 12:33 PM
mr fantastic
Quote:

Originally Posted by janvdl
Calculate $r$ and $d$. You'll see they're not the same between $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{3}$, $\sqrt{5}$

Trivial with a calculator of course. But proving $\sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}$ is a bit more interesting (for a student at this level) without one .....
• January 5th 2008, 07:11 PM
ThePerfectHacker
Quote:

Originally Posted by mr fantastic
Trivial with a calculator of course. But proving $\sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}$ is a bit more interesting (for a student at this level) without one .....

It is not hard. Assume that is is then $2\sqrt{3} = \sqrt{5} + \sqrt{2}$. Square and there is a condatriction.

In general given, $p_1,...,p_n$ distinct primes and $p_{n+1}$ is also distinct. Then $\sqrt{p_{n+1}}$ cannot be expressed as a linear combination of $\sqrt{p_1},...,\sqrt{p_n}$ over $\mathbb{Q}$.
• January 5th 2008, 10:30 PM
james_bond
But they don't have to be neighbors!
• January 5th 2008, 10:37 PM
mr fantastic
Quote:

Originally Posted by ThePerfectHacker
It is not hard. Assume that is is then $2\sqrt{3} = \sqrt{5} + \sqrt{2}$. Square and there is a condatriction.

In general given, $p_1,...,p_n$ distinct primes and $p_{n+1}$ is also distinct. Then $\sqrt{p_{n+1}}$ cannot be expressed as a linear combination of $\sqrt{p_1},...,\sqrt{p_n}$ over $\mathbb{Q}$.

The proof of this theorem (at least for the concrete case arising in this question) is the interesting bit (for a student at this level).
• January 6th 2008, 12:49 AM
janvdl
Quote:

Originally Posted by ThePerfectHacker
It is not hard. Assume that is is then $2\sqrt{3} = \sqrt{5} + \sqrt{2}$. Square and there is a condatriction.

I just went through the trouble of doing so, and then I saw this post... :mad:

EDIT: But i'll show what i did anyway :D
• January 6th 2008, 12:50 AM
red_dog
Suppose $\sqrt{2},\sqrt{3},\sqrt{5}$ are not consecutive terms of an arithmetic or geometric progression.

For the arithmetic progression:

Let $\sqrt{2}=a_m=a_1+(m-1)r$ (1)
$\sqrt{3}=a_n=a_1+(n-1)r$ (2)
$\sqrt{5}=a_p=a_1+(p-1)r$ (3)
Substracting (2) from (1) and (3) from (2) we get
$\sqrt{3}-\sqrt{2}=(n-m)r$
$\sqrt{5}-\sqrt{3}=(p-n)r$
Then $\displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{n-m}{p-n}$
But, the left side member is irational and the right side member is rational.

For the geometric progression:

Let $\sqrt{2}=b_1q^{m-1}$ (1)
$\sqrt{3}=b_1q^{n-1}$ (2)
$\sqrt{5}=b_1q^{p-1}$ (3)
Then $\displaystyle q=\left(\frac{2}{3}\right)^{\frac{m-n}{2}}=\left(\frac{3}{5}\right)^{\frac{n-p}{2}}$
$\displaystyle \Rightarrow 2^{\frac{m-n}{2}}5^{\frac{n-p}{2}}=3^{\frac{m-p}{2}}\Rightarrow 2^{m-n}5^{n-p}=3^{m-p}$
The last equality is not true.
• January 6th 2008, 01:04 AM
janvdl
Quote:

Originally Posted by red_dog
Suppose $\sqrt{2},\sqrt{3},\sqrt{5}$ are not consecutive terms of an arithmetic or geometric progression.

For the arithmetic progression:

Let $\sqrt{2}=a_m=a_1+(m-1)r$ (1)
$\sqrt{3}=a_n=a_1+(n-1)r$ (2)
$\sqrt{5}=a_p=a_1+(p-1)r$ (3)
Substracting (2) from (1) and (3) from (2) we get
$\sqrt{3}-\sqrt{2}=(n-m)r$
$\sqrt{5}-\sqrt{3}=(p-n)r$
Then $\displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{n-m}{p-n}$
But, the left side member is irational and the right side member is rational.

For the geometric progression:

Let $\sqrt{2}=b_1q^{m-1}$ (1)
$\sqrt{3}=b_1q^{n-1}$ (2)
$\sqrt{5}=b_1q^{p-1}$ (3)
Then $\displaystyle q=\left(\frac{2}{3}\right)^{\frac{m-n}{2}}=\left(\frac{3}{5}\right)^{\frac{n-p}{2}}$
$\displaystyle \Rightarrow 2^{\frac{m-n}{2}}5^{\frac{n-p}{2}}=3^{\frac{m-p}{2}}\Rightarrow 2^{m-n}5^{n-p}=3^{m-p}$
The last equality is not true.

My solution was a little more humble... :D

Arithmetic:

Assume they are equal.

$\sqrt{3} - \sqrt{2} = \sqrt{5} - \sqrt{3}$

Set $\sqrt{3} = x$

$x - \sqrt{2} = \sqrt{5} - x$

$2x = \sqrt{5} + \sqrt{2}$

Square both sides

$4x^2 = 25 + 2( \sqrt{5} \cdot \sqrt{2} ) + 4$

$4(\sqrt{3})^2 = 29 + 2 \sqrt{10}$

$12 - 29 = 2 \sqrt{10}$

$-17 \neq 2 \sqrt{10}$

Therefore $\sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}$

----

Geometric

Once again, to make it easier, set $\sqrt{3} = x$

Assume equality

$\frac{x}{ \sqrt{2} } = \frac{ \sqrt{5} }{x}$

$x^2 = \sqrt{10}$

$x^4 = 10$

$(\sqrt{3})^4 \neq 10$

Therefore $\frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}$
• January 6th 2008, 02:06 AM
Isomorphism
Quote:

Originally Posted by janvdl
My solution was a little more humble... :D

Arithmetic:

Assume they are equal.

$\sqrt{3} - \sqrt{2} = \sqrt{5} - \sqrt{3}$

Set $\sqrt{3} = x$

$x - \sqrt{2} = \sqrt{5} - x$

$2x = \sqrt{5} + \sqrt{2}$

Square both sides

$4x^2 = 25 + 2( \sqrt{5} \cdot \sqrt{2} ) + 4$

$4(\sqrt{3})^2 = 29 + 2 \sqrt{10}$

$12 - 29 = 2 \sqrt{10}$

$-17 \neq 2 \sqrt{10}$

Therefore $\sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}$

----

Geometric

Once again, to make it easier, set $\sqrt{3} = x$

Assume equality

$\frac{x}{ \sqrt{2} } = \frac{ \sqrt{5} }{x}$

$x^2 = \sqrt{10}$

$x^4 = 10$

$(\sqrt{3})^4 \neq 10$

Therefore $\frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}$

You are again assuming they are consecutive terms of the progressions. See red_dog's solution, he does not assume it
• January 6th 2008, 06:33 AM
janvdl
Quote:

Originally Posted by Isomorphism
You are again assuming they are consecutive terms of the progressions. See red_dog's solution, he does not assume it

TPH said it was possible to do it this way. And if it's not wrong, what's your problem with me to do it this way?
• January 6th 2008, 09:33 AM
ThePerfectHacker
Quote:

Originally Posted by janvdl
TPH said it was possible to do it this way. And if it's not wrong, what's your problem with me to do it this way?

It is just that you were doing a different problem than Red_dog. You were saying that sqrt(2),sqrt(3),sqrt(5) cannot be right next to each other, and what you did is correct. But Red_dog did a stronger problem he showed that you cannot have these three numbers in any arithmetic progession. Meaning you cannot have sqrt(2) as a 3rd term, sqrt(3) as a 10th term, and sqrt(5) as a 29th term.