Comparing the difference of irrational number

Question

i know this is a very basic question but i am not able to do this

The question is -: Compare these two value

square root(13) - square root(12) , square root(14) - square root(13)

Can this question be solved without putting the value

Thank you for helping me . I am very grateful to you

Answer 1

Do you know Calculus? The derivative of the function $f(x)=\sqrt x$ is a positive decreasing function, meaning $f(x)$ looks like this:

Since the slope is getting shallower, the difference $f(x+1)-f(x)$ is getting smaller as $x$ increases. You can prove this using either the Fundamental Theorem of Calculus, or using the Mean value theorem. You could also just calculate the derivative of $g(x)=f(x+1)-f(x)$ and show that it's negative.

Therefore $\sqrt{14}-\sqrt{13}\leq\sqrt{13}-\sqrt{12}$.

Answer 2

You want to compare $\sqrt{13}-\sqrt{12}$ to $\sqrt{14}-\sqrt{13}$.

So, you want to compare $2\sqrt{13}$ to $\sqrt{14}+\sqrt{12}$.

Squaring, you want to compare $52$ to $26+2\sqrt{168}$, or $26$ to $2\sqrt{168}$, or $13$ to $\sqrt{168}$.

Squaring again, you want to compare $169$ to $168$. Which is easy.

Answer 3

Note that

$$\begin{align} \sqrt{14}-\sqrt{13}\lt\sqrt{13}-\sqrt{12} &\iff\sqrt{14}+\sqrt{12}\lt2\sqrt{13}\\ &\iff\sqrt{1+{1\over13}}+\sqrt{1-{1\over13}}\lt2 \end{align}$$

But

$$\begin{align} \sqrt{1+{1\over13}}+\sqrt{1-{1\over13}} &\lt\sqrt{1+{1\over13}+\left(1\over26\right)^2}+\sqrt{1-{1\over13}+\left(1\over26\right)^2}\\ &=\sqrt{\left(1+{1\over26}\right)^2}+\sqrt{\left(1-{1\over26}\right)^2}\\ &=\left(1+{1\over26}\right)+\left(1-{1\over26}\right)\\ &=2 \end{align}$$

Remark: The numbers $12$, $13$, and $14$ can be replaced by $x-1$, $x$, and $x+1$, as long as $x\ge1$.