Question
Prove that the following number is irrational:$ \sqrt{3} + \sqrt{2}$
✓
Answer
$\sqrt{3}+\sqrt{2}$
Let $\sqrt{3}+\sqrt{2}$ be a rational number.
$\Rightarrow \sqrt{3}+\sqrt{2}=x$
Squaring on both the sides, we get
$(\sqrt{3}+\sqrt{2})^2=x^2$
$ \Rightarrow 3+2+2 \times \sqrt{3} \times \sqrt{2}=x^2$
$ \Rightarrow \mathrm{x}^2-5=2 \sqrt{6 } $
$ \Rightarrow \sqrt{6}=\frac{x^2-5}{2}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow x^2-5$ is a rational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a rational number.
But $\sqrt{6 } $ is an irrational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a irrational number.
$\Rightarrow x^2-5$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $x$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{3}+\sqrt{2}$ is a rational number is wrong.
$\therefore \sqrt{3}+\sqrt{2}$ is an irrational number.
Let $\sqrt{3}+\sqrt{2}$ be a rational number.
$\Rightarrow \sqrt{3}+\sqrt{2}=x$
Squaring on both the sides, we get
$(\sqrt{3}+\sqrt{2})^2=x^2$
$ \Rightarrow 3+2+2 \times \sqrt{3} \times \sqrt{2}=x^2$
$ \Rightarrow \mathrm{x}^2-5=2 \sqrt{6 } $
$ \Rightarrow \sqrt{6}=\frac{x^2-5}{2}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow x^2-5$ is a rational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a rational number.
But $\sqrt{6 } $ is an irrational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a irrational number.
$\Rightarrow x^2-5$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $x$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{3}+\sqrt{2}$ is a rational number is wrong.
$\therefore \sqrt{3}+\sqrt{2}$ is an irrational number.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the following figure;$ P$ and $Q$ are the points of intersection of two circles with centers $O$ and $O'$. If straight lines $\text{APB}$ and $\text{CQD}$ are parallel to $OO';$prove that: $(i) OO' = \frac{1}{2} AB ; (ii) AB = CD$
→If $\sin(A +B) = 1(A -B) = 1,$ find $A$ and $B.$
→At what rate of interest per annum will a sum of $Rs. 62,500$ earn a compound interest of $Rs. 5,100$ in one year? The interest is to be compounded half yearly.
→Solve the following simultaneous equations :$6x + 3y = 7xy,3x + 9y = 11xy$
→If (a + b + c) = 14 and (a2 + b2 +c2) = 74, find the value of (ab + bc + ca).
If $\left(x-\frac{1}{x}\right)=8$, find the values of $\left(x^2-\frac{1}{x^2}\right)$.
→In $\triangle ABC, D$ and $E$ are the mid$-$point on $AB$ and $AC$ such that $DE \| BC.$If $AD = 4, AE = 8, DB = x - 4$ and $EC = 3x - 19$, find $x.$
→The weekly wages (in rupees) of 30 workers in a factory are given below :
630, 635, 690, 610, 635, 636, 639, 645, 698, 690, 620, 660, 632, 633, 655, 645, 604, 608, 612, 640, 685, 635, 636, 678, 640, 668, 690, 606, 640, 690
Represent the data in the form of a frequency distribution with class size 10.
630, 635, 690, 610, 635, 636, 639, 645, 698, 690, 620, 660, 632, 633, 655, 645, 604, 608, 612, 640, 685, 635, 636, 678, 640, 668, 690, 606, 640, 690
Represent the data in the form of a frequency distribution with class size 10.
Solve the following pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$:1.5x + 0.1y = 6.2,3x - 0.4y = 11.2$
→Simplify the following and express with positive index :$\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{-1}$
→