Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether True or False :


All the following numbers are irrationals.
(i) $\dfrac { 2 }{ \sqrt { 7 }  } $ (ii) $\dfrac { 3 }{ 2\sqrt { 5 }  }$ (iii) $4+\sqrt { 2 } $ (iv) $5\sqrt { 2 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In all of the above questions $\sqrt7,\sqrt5,\sqrt2$ are a $irrational$ numbers


And $\text{the addition, subtraction, division and product between  rational and irrational gives  an irrational number}$

So that all of the above are irrational numbers.

hence option A is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$2-3\sqrt { 5 }$ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here 2 is a rational number

and $3\sqrt5$ is an irrational number  and $\text{difference of rational and irrational is always an irrational number}$


So that $2 - 3\sqrt5 $  is an $irrational$ number

hence option A is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$\sqrt { 3 } +\sqrt { 4 } $ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt4= 2 \text{ is a  rational number and }$$\sqrt3$ is a $irrational$ number


And $\text{the addition of rational and irrational is always an irrational number}$

So that $(\sqrt3+\sqrt4)$ is an irrational number.

hence option A is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

The number $6+\sqrt { 2 } $ is irrational.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Here \  6 \text{ is a  rational number and }$$\sqrt2$ is a $irrational$ number


And $\text{the addition of rational and irrational is always an irrational number}$

So that $(6+\sqrt2)$ is an irrational number.

hence option A is correct

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :
If $p,  q $ are prime positive integers, then $\sqrt { p } +\sqrt { q } $ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\sqrt{p}+\sqrt{q}$ is rational ........ assumption
$\sqrt{p}+\sqrt{q}=\dfrac{a}{b}$
squaring, we get
$p+q+2\sqrt{pq}=\left(\dfrac{a}{b}\right)^{2}$

$\sqrt{pq}=\dfrac{1}{2}\left[\left(\dfrac{a}{b}\right)^{2}-p-q\right] - (i)$

Now, $p$ & $q$ are prime positive numbers so, $\sqrt{p}$ and $\sqrt{q}$ is irrational also $\sqrt{pq}$ 

so in (i)
Irrational $=$ rational $\Rightarrow$ which is a contradiction

$\Rightarrow\ \sqrt p+\sqrt q$ is a irrational number if $p,q$ are prime positive numbers