Question
Prove that $5\sqrt2$ is an irrational number.
✓
Answer
Let $5\sqrt2$ is a rational number.
$\therefore5\sqrt2=\frac{\text{p}}{\text{q}},$ where p and q are some integers and HCF (p, q) = 1 ...(1)
$\Rightarrow5\sqrt2\text{q}=\text{p}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{p}^2$
$\Rightarrow2(25\text{q}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by 2
⇒ p is divisible by 2 ...(2)
Let p = 2m, where m is some integer.
$\Rightarrow5\sqrt2\text{q}=\text{2m}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{2m}^2$
$\Rightarrow2(25\text{q}^2)=\text{4m}^2$
$\Rightarrow\text{25q}^2=\text{2m}^2$
$\Rightarrow q^2$ is divisible by 2
⇒ q is divisible by 2 ...(3)
From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, $5\sqrt2$ is irrational.
$\therefore5\sqrt2=\frac{\text{p}}{\text{q}},$ where p and q are some integers and HCF (p, q) = 1 ...(1)
$\Rightarrow5\sqrt2\text{q}=\text{p}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{p}^2$
$\Rightarrow2(25\text{q}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by 2
⇒ p is divisible by 2 ...(2)
Let p = 2m, where m is some integer.
$\Rightarrow5\sqrt2\text{q}=\text{2m}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{2m}^2$
$\Rightarrow2(25\text{q}^2)=\text{4m}^2$
$\Rightarrow\text{25q}^2=\text{2m}^2$
$\Rightarrow q^2$ is divisible by 2
⇒ q is divisible by 2 ...(3)
From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, $5\sqrt2$ 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
Find the number of metallic circular disc with 1.5cm base diameter and of height 0.2cm to be melted to form a right circular cylinder of height 10cm and diameter 4.5cm.
A carpet is laid on the floor of a room 8m by 5m. There is a border of constant width all around the carpet. If the area of the border is $12m^2$, find its width.
→In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. if OA = 20cm, find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$
→The perimeter of a rectangular field is 82m and its area is $400m^2$. Find the breadth of the rectangle.
→Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:
$\text{f}(\text{x})=2\text{x}^2+\text{x}^2-5\text{x}+2;\frac{1}{2},1,-2$
→$\text{f}(\text{x})=2\text{x}^2+\text{x}^2-5\text{x}+2;\frac{1}{2},1,-2$
Find the value of k for which the root are real and equal in the following equations:
$(k + 1)x^2 - 2(3k + 1)x + 8k + 1 = 0$
→$(k + 1)x^2 - 2(3k + 1)x + 8k + 1 = 0$
Prove the following identities:
$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
→$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
In the given figure, OABC is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region.
Solve the following systems of equations:
$\frac{\text{x}}{3}+\frac{\text{y}}{4}=11,$
$\frac{5\text{x}}{6}-\frac{\text{y}}{3}=-7.$
→$\frac{\text{x}}{3}+\frac{\text{y}}{4}=11,$
$\frac{5\text{x}}{6}-\frac{\text{y}}{3}=-7.$
The angle of elevation of the top of a vertical tower from a point on the ground is 60º. From another point 10m vertically above the first, its angle of elevation is 45º. Find the height of the tower.