Question
Insert two irrational numbers between $3$ and $4.$
✓
Answer
Since $3$ and $4$ are rational numbers and $3 \times 4=12$ is not a perfect square.
$\therefore$ One irrational number between $3$ and $4=\sqrt{3 \times 4}=\sqrt{12}$
And , an irrational number between $3$ and $\sqrt{12}=\sqrt{3 \times \sqrt{12}}=\sqrt{3 \sqrt{12}}$
$\therefore$ Required irrational numbers between $3$ and $4$ are : $\sqrt{12}$ and $\sqrt{3 \sqrt{12}}$
$\therefore$ One irrational number between $3$ and $4=\sqrt{3 \times 4}=\sqrt{12}$
And , an irrational number between $3$ and $\sqrt{12}=\sqrt{3 \times \sqrt{12}}=\sqrt{3 \sqrt{12}}$
$\therefore$ Required irrational numbers between $3$ and $4$ are : $\sqrt{12}$ and $\sqrt{3 \sqrt{12}}$
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
If $\sin \theta=\frac{1}{\sqrt{2}}$, find the values of other trigonometrical ratios for $\theta$.
→If $\left(x+\frac{1}{x}\right)=3$, show that $\left(x^3+\frac{1}{x^3}\right)=0$.
→Draw the graph for the each linear equation given below:$y=\frac{3 x}{2}+\frac{2}{3}$
→Express each of the following as the difference of two squares:$(x^2 - 2x + 3)(x^2 + 2x + 3)$
→In the given figure, $\triangle A B C$ is right angled at $B$.
If $AC =20 cm$ and $\tan A =\frac{3}{4}$, find the lengths of AB and BC .
→If $AC =20 cm$ and $\tan A =\frac{3}{4}$, find the lengths of AB and BC .
Express the following as a single logarithm:$2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}$
→Factorise : $(a+4)^2-5 a b-20 b-6 b^2$
→Solve the following equations for the unknown$: 2x - (3x - 4) = 3x - 4$
→There are two examination halls A and B. If 12 pupils are sent from A to B, the number of pupils in each room becomes the same. If 11 pupils are sent from room B to room A, then the number of pupils in A is double their number in B. Find the number of pupils in each room.
Without using trigonometric tables, prove that:
$\cos ^2 23^{\circ}+\cos ^2 67^{\circ}=1$
→$\cos ^2 23^{\circ}+\cos ^2 67^{\circ}=1$