Tag: representation of irrational numbers on number line

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Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

Use ______________ to represent an irrational number on number line.

  1. Isosceles-angle theorem

  2. Scalene angle theorem

  3. Right-angled theorem

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Using the Pythagoras Theorem, we can represent some irrational numbers, which are surds, on a number line.
Since it involves Pythagoras Theorem, we get to use Right Angle Theorem.

Hence, to represent an irrational number, we generally use right angled theorem.

Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

$D$ 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 ?

  1. $99$
  2. $18$
  3. $125$
  4. $75$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

its straight question
give $D=0.abababab$ $(say - 1)$ 
Multiply both sides by $100$ $i.e.$ 
$100D = ab.abababab$ $(say - 2)$
now subtract $1$ from $2 .$ That gives
$99D = ab => D = ab/99$ hence it should be multiplied by $99k$ to get an integer ab$.$

Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

Give an example of two irrational numbers whose difference is an irrational number.

  1. $\sqrt{3},-\sqrt{3}$
  2. $\sqrt{5,}-\sqrt{5}$
  3. $4\sqrt{3},-2\sqrt{3}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$4\sqrt{3},2\sqrt{3}$ are the irrational numbers and thier difference,


$4\sqrt{3}-2\sqrt{3}=2\sqrt 3$ is also an irrational number.

Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

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

  1. Both I and II

  2. Only III

  3. Only II

  4. Both II and III
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Step III is incorrect because it claims that since a and b are integers, 5 - a/b is irrational. In reality, 5 - a/b is rational if a and b are integers, which is the basis of the contradiction proof.

Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

Which of the following irrational numbers lie between $4$ and $7$?

  1. $\sqrt{25}$
  2. $\sqrt{19}$
  3. $\sqrt{47}$
  4. $\sqrt{50}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$4^{2} = 16$

$5^{2} = 25$
$6^{2} = 36$
$7^{2} = 49$


$\Rightarrow \sqrt19$ and $\sqrt47$ are irrational numbers which lie between $4$ and $7$


$\sqrt25 = 5$ which is a rational number

Multiple choice maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

The ascending order of the surds $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$ is 

  1. $\sqrt[9]{4}, \sqrt[6]{3}, \sqrt[3]{2}$
  2. $\sqrt[9]{4}, \sqrt[3]{2}, \sqrt[6]{3}$
  3. $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$
  4. $\sqrt[6]{3}, \sqrt[9]{4}, \sqrt[3]{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lets keep in mind the following common log values:

$log(2)=0.30$
$log(3)=0.47$
Lets compare these three surds by their log values.
Reason: Since $log(x) $ is an increasing function  when $x>1$ so we can compare these values by comparing their log values.

$log(\sqrt[3]{2})=log(2^{\frac{1}3})=\left(\dfrac{1}3\right)log(2)=\left(\dfrac{1}3\right) \times 0.3=0.100$

$log(\sqrt[6]{3})=log(3^{\frac{1}6})=\left(\dfrac{1}6\right)log(3)=\left(\dfrac{1}6\right) \times 0.47=0.078$

$log(\sqrt[9]{4})=log(\sqrt[9]{2^2})=log(2^{\frac{2}9})=\left(\dfrac{2}9\right)log(2)=\left(\dfrac{2}9\right) \times 0.3=0.060$

By looking at the log values,it is clear that,

$\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$