Representation of irrational numbers on number line - class-XI
Description: representation of irrational numbers on number line | |
Number of Questions: 18 | |
Created by: Mira Shah | |
Tags: real numbers basic algebra number system maths number systems real numbers (rational and irrational numbers) |
Which of the following numbers lie between $1$ and $3$?
Between any $2$ real numbers, __________ can always be represented on a number line.
Following are the steps to represent $\sqrt5$ on number line.
Arrange them in order.
1) Draw OC on line with $l(OC)=l(OB)$,
2) Draw $AB \perp OA\ and\ l(AB) =1$
3) Take $l(OA)=2$
4) $l(OC)=\sqrt5$, C is required point on real line.
Which of the following irrational numbers lie between $6$ and $8$?
The number $\sqrt{10}$ lies between $2$ integers $a$ and $b$ such that $b-a = 1$. Then $b+a = \, ?$
Which one of the following is not true?
The greater number between $\sqrt{17}-\sqrt{12}$ and $\sqrt{11}-\sqrt{6}$ is ____.
The value of $0.\overline{2}$ in the form $\frac{p}{q}$ , where p and q are integers and $q\ne 0$ is :
Which of the following is/are correct?
To represent a rational number $\sqrt{2}$ on number line, take sides of right triangle as:
Use ______________ to represent an irrational number on number line.
$D$ is a real number with non terminating digits $a _1$ and $a _2$ after the decimal point. Let $D = 0, a _1 a _2 a _1 a _2 ........ $ with $a _1 & a _2$ both not zero which of the following when multiplied by $D$ will necessarily give an integer ?
The number $5\sqrt{34}$ lies between
Can $\sqrt { 3 } -3$ be represented on the number line.
Give an example of two irrational numbers whose difference is an irrational number.
Which is the wrong step that shows $\displaystyle 5-\sqrt{3}$ is irrational?
(I) Contradiction : Assume that $\displaystyle 5-\sqrt{3}$ is rational
(II) Find coprime a & b $\displaystyle \left ( b\neq 0 \right )$ such that $\displaystyle 5-\sqrt{3}=\frac{a}{b},\therefore 5-\frac{a}{b}=\sqrt{3}$
Rearranging above equation $\displaystyle \sqrt{3}=5-\frac{a}{b}=\frac{5b-a}{b}$
(III) Since a & b are integers we get $\displaystyle 5-\frac{a}{b}$ is irrational and so $\displaystyle \sqrt{3}$ is irrational
(IV) But this contradicts the fact that $\displaystyle \sqrt{3}$ is irrational Hence $\displaystyle 5-\sqrt{3}$ is irrational
Which of the following irrational numbers lie between $4$ and $7$?
The ascending order of the surds $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$ is