Question

Arrange the three values

a = \sqrt{27} - \sqrt{26}, b = \sqrt{28} - \sqrt{27}, c = \sqrt{29} - \sqrt{28}

in ascending order.

Accepted Answer

We have $a = \sqrt{27} - \sqrt{26}, b = \sqrt{28} - \sqrt{27}$ and $c = \sqrt{29} - \sqrt{28}$ .

On rationalising all the three numbers, we get

$a = \frac{(\sqrt{27} - \sqrt{26}) (\sqrt{27} + \sqrt{26})}{(\sqrt{27} + \sqrt{26})}, b = \frac{(\sqrt{28} - \sqrt{27}) (\sqrt{28} + \sqrt{27})}{(\sqrt{28} + \sqrt{27})}$ and $c = \frac{(\sqrt{29} - \sqrt{28}) (\sqrt{29} + \sqrt{28})}{(\sqrt{29} + \sqrt{28})}$

Now, using $(x + y) (x - y) = x^{2} - y^{2},$ we get

$\Rightarrow a = \frac{{(\sqrt{27})}^{2} - {(\sqrt{26})}^{2}}{\sqrt{27} + \sqrt{26}}, b = \frac{{(\sqrt{28})}^{2} - {(\sqrt{27})}^{2}}{\sqrt{28} + \sqrt{27}}$ and $c = \frac{{(\sqrt{29})}^{2} - {(\sqrt{28})}^{2}}{\sqrt{29} + \sqrt{28}}$

$\Rightarrow a = \frac{27 - 26}{\sqrt{27} + \sqrt{26}}, b = \frac{28 - 27}{\sqrt{28} + \sqrt{27}}$ and $c = \frac{29 - 28}{\sqrt{29} + \sqrt{28}}$

$\Rightarrow a = \frac{1}{\sqrt{27} + \sqrt{26}}, b = \frac{1}{\sqrt{28} + \sqrt{27}}$ and $c = \frac{1}{\sqrt{29} + \sqrt{28}}$

As, we know that $\sqrt{29} > \sqrt{28} > \sqrt{27} > \sqrt{26}$

$\Rightarrow \sqrt{29} > \sqrt{27}$ and $\sqrt{28} > \sqrt{26}$

$\Rightarrow \sqrt{29} + \sqrt{28} > \sqrt{27} + \sqrt{28}$ and $\sqrt{28} + \sqrt{27} > \sqrt{26} + \sqrt{27}$

$\Rightarrow \sqrt{29} + \sqrt{28} > \sqrt{28} + \sqrt{27} > \sqrt{27} + \sqrt{26}$

$\Rightarrow \frac{1}{\sqrt{29} + \sqrt{28}} < \frac{1}{\sqrt{28} + \sqrt{27}} < \frac{1}{\sqrt{27} + \sqrt{26}}$

We know that, if $x, y, z$ are positive real numbers such that $x > y > z,$ then $\frac{1}{x} < \frac{1}{y} < \frac{1}{z} .$

$∴ c < b < a .$

Embibe Experts Solutions for Chapter: Number System, Exercise 4: EXERCISE - 2.4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Number System, Exercise 4: EXERCISE - 2.4

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