Prove that $\displaystyle \sqrt{2}$, $\displaystyle \sqrt{3}$ and $\displaystyle \sqrt{5}$ can't be terms 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.
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 $\displaystyle \mathbb{Q}$.
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.
My solution was a little more humble...
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 $\displaystyle \sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}$
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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 $\displaystyle \frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}$
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.