Prove that $\displaystyle \sqrt{2}$, $\displaystyle \sqrt{3}$ and $\displaystyle \sqrt{5}$ can't be terms in the same

- arithmetic
- geometric

progression.

- Jan 5th 2008, 10:30 AMjames_bondsqrt 2, sqrt 3, sqrt 5 can't be in the same arithmetic/geometric progression
Prove that $\displaystyle \sqrt{2}$, $\displaystyle \sqrt{3}$ and $\displaystyle \sqrt{5}$ can't be terms in the same

- arithmetic
- geometric

progression. - Jan 5th 2008, 10:55 AMjanvdl
- Jan 5th 2008, 12:33 PMmr fantastic
- Jan 5th 2008, 07:11 PMThePerfectHacker
It is not hard. Assume that is is then $\displaystyle 2\sqrt{3} = \sqrt{5} + \sqrt{2}$. Square and there is a condatriction.

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

- Jan 5th 2008, 10:37 PMmr fantastic
- Jan 6th 2008, 12:49 AMjanvdl
- Jan 6th 2008, 12:50 AMred_dog
Suppose $\displaystyle \sqrt{2},\sqrt{3},\sqrt{5}$ are not consecutive terms of an arithmetic or geometric progression.

For the arithmetic progression:

Let $\displaystyle \sqrt{2}=a_m=a_1+(m-1)r$ (1)

$\displaystyle \sqrt{3}=a_n=a_1+(n-1)r$ (2)

$\displaystyle \sqrt{5}=a_p=a_1+(p-1)r$ (3)

Substracting (2) from (1) and (3) from (2) we get

$\displaystyle \sqrt{3}-\sqrt{2}=(n-m)r$

$\displaystyle \sqrt{5}-\sqrt{3}=(p-n)r$

Then $\displaystyle \displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{n-m}{p-n}$

But, the left side member is irational and the right side member is rational.

For the geometric progression:

Let $\displaystyle \sqrt{2}=b_1q^{m-1}$ (1)

$\displaystyle \sqrt{3}=b_1q^{n-1}$ (2)

$\displaystyle \sqrt{5}=b_1q^{p-1}$ (3)

Then $\displaystyle \displaystyle q=\left(\frac{2}{3}\right)^{\frac{m-n}{2}}=\left(\frac{3}{5}\right)^{\frac{n-p}{2}}$

$\displaystyle \displaystyle \Rightarrow 2^{\frac{m-n}{2}}5^{\frac{n-p}{2}}=3^{\frac{m-p}{2}}\Rightarrow 2^{m-n}5^{n-p}=3^{m-p}$

The last equality is not true. - Jan 6th 2008, 01:04 AMjanvdl
My solution was a little more humble... :D

**Arithmetic:**

Assume they are equal.

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

Set $\displaystyle \sqrt{3} = x$

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

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

Square both sides

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

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

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

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

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

----

**Geometric**

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

Assume equality

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

$\displaystyle x^2 = \sqrt{10}$

$\displaystyle x^4 = 10$

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

Therefore $\displaystyle \frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}$ - Jan 6th 2008, 02:06 AMIsomorphism
- Jan 6th 2008, 06:33 AMjanvdl
- Jan 6th 2008, 09:33 AMThePerfectHacker
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.