Question
Prove that $7 \sqrt{5}$ is irrational.
✓
Answer
We can prove $7 \sqrt{5}$ irrational by contradiction.
Let us suppose that $7 \sqrt{5}$ is rational.
It means we have some co$-$prime integers $a$ and $b(b \neq 0)$ such that
$7 \sqrt{5}=\frac{a}{b}$
$\Rightarrow \sqrt{5}=\frac{a}{7 b}$
$\ce{R.H.S}$ of $(1)$ is rational but we know that $\sqrt{5}$ is irrational.
It is not possible which means our assumption is wrong.
Therefore, $7 \sqrt{5}$ cannot be rational.
Hence, it is irrational.
Let us suppose that $7 \sqrt{5}$ is rational.
It means we have some co$-$prime integers $a$ and $b(b \neq 0)$ such that
$7 \sqrt{5}=\frac{a}{b}$
$\Rightarrow \sqrt{5}=\frac{a}{7 b}$
$\ce{R.H.S}$ of $(1)$ is rational but we know that $\sqrt{5}$ is irrational.
It is not possible which means our assumption is wrong.
Therefore, $7 \sqrt{5}$ cannot be rational.
Hence, it is irrational.
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
Prove that a line draw through the mid point of one side of a triangle parallel to another side bisects the third side.
Find the ratio in which the y-axis divides the line segment joining the points (4, −5) and (−1,2). Also find the point of intersection.
Express the following as a fraction in simplest form:
$0.\overline{24}$
→$0.\overline{24}$
Solve the following quadratic equation:$\text{x}^2-3\sqrt5\text{x}+10=0$
→Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$(2a - 1)x + 3y = 5,$
$3x + (b - 1)y = 2$
→$(2a - 1)x + 3y = 5,$
$3x + (b - 1)y = 2$
In the given figure $, AB$ and $CD$ are two diameters of a circle with centre $O,$ which are perpendicular to each other. $OB$ is the diameter of the smaller circle. If $OA =7 \ cm,$ find the area of the shaded region. $[$Use $\pi=\frac{22}{7} ]$
→ For which value of $k$ will the following pair of linear equations have no solution? $3 x+y=1,(2 k-1) x+(k-1) y=2 k+1$
→In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides AB and AC respectively such that DE || BC.
If $\frac{\text{AD}}{\text{DB}}=\frac{3}{4}$ and AC = 15cm, find AE.
→If $\frac{\text{AD}}{\text{DB}}=\frac{3}{4}$ and AC = 15cm, find AE.
If $\alpha,\ \beta$ are the zeros of the polynomial $f(x) = x^2 - 5x + k$ such that $\alpha-\beta=1,$ find the value of k.
→Find the area of an isoscale triangle each of whose equal sides is 13cm and whose base is 24cm.