Tag: maths

Questions Related to maths

State the following statement is True or False
 35.251252253...is an irrational number

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

  1. True

  2. False

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Correct Option: A
Explanation:

$35.251522253...\Rightarrow$ It is a non-terminating decimal. This number cannot be written as a simple fraction.

Therefore, it's an irrational number.

For three irrational numbers $p,q$ and $r$ then $p.(q+r)$ can be 

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

  1. A rational number

  2. An irrational number

  3. An integer

  4. All of the above

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Correct Option: D
Explanation:

$p,q$ and $r$ are all irrational 
Let $p=q=r=\sqrt2$
$p(q+r)=p.q+p.r$
$\Rightarrow \sqrt { 2 } (\sqrt { 2 } +\sqrt { 2 } )=\sqrt { 2 } .\sqrt { 2 } +\sqrt { 2 } .\sqrt { 2 } =2+2=4$
which is rational as well as integer
Let us take another case in which $p=\sqrt2$ and $q=r=\sqrt3$
$\Rightarrow \sqrt { 2 } (\sqrt { 3 } +\sqrt { 3 } )=\sqrt { 2 } .\sqrt { 3 } +\sqrt { 2 } .\sqrt { 3 } =\sqrt { 6 } +\sqrt { 6 } =2\sqrt { 6 } $
which is an irrational number.
So on applying distributive property on three irrational numbers we can get an integer,a rational as well as an irrational number .
So option $D$ is correct.

Which of the following irrational number lies between $\dfrac{3}{5}$ and $\dfrac{9}{10}$

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

  1. $\dfrac{\sqrt80}{10}$

  2. $\dfrac{\sqrt85}{10}$

  3. $\dfrac{\sqrt82}{10}$

  4. $\dfrac{\sqrt83}{10}$

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Correct Option: A
Explanation:
$\sqrt36<\sqrt{80}<\sqrt{81}$
On dividing with 10, we get
$\dfrac{6}{10}<\dfrac{\sqrt{80}}{10}<\dfrac{9}{10}$

Which one of the following statements is not correct?

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

  1. If $a$ is a rational number and $b$ is irrational, then $a+b$ is irrational.

  2. The product of non-zero rational number with an irrational number is always irrational.

  3. The addition of any two rational numbers can be an integer.

  4. The division of any two integers is an integer.

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Correct Option: A
Explanation:

Sum of rational and an irrational number is rational (i.e., need not to be irrational)

State whether the given statement is True or False :

$2\sqrt { 3 }-1 $ is an irrational number.
maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

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Correct Option: A
Explanation:

$\text{Here 1 is a rational number and }$$2\sqrt3$ is a $irrational$ number


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

So that $(2\sqrt3- 1)$ is an irrational number.

hence option A is correct.

State whether the given statement is true/false:

$\sqrt{p} + \sqrt{q}$, is irrational, where p,q are primes.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Assume that $\sqrt p  + \sqrt q $ is rational. So,

$\sqrt p  + \sqrt q  = \frac{a}{b}$

$p + q + 2\sqrt {pq}  = \frac{{{a^2}}}{{{b^2}}}$

$\sqrt {pq}  = \frac{1}{2}\left( {\frac{{{a^2}}}{{{b^2}}} – p - q} \right)$

Since the RHS of the above equation is rational but $\sqrt {pq} $ is an irrational number, so the assumption is wrong.

Therefore, it is true that $\sqrt p  + \sqrt q $ is irrational.

State true or false:
$\sqrt{2}$ is not a rational number.
maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:


Let us assume that $\sqrt{2}$ is a rational number
$\Rightarrow \sqrt{2}=\dfrac{p}{q}\left [ \dfrac{p}{q}\      is\ in\    simplest\      form\  \right ]$

$\Rightarrow 2q^{2}=p^{2}$
$p^{2}$ is even $\Rightarrow p$ is even $\Rightarrow p=2k$
$2q^{2}=4k^{2}\Rightarrow q^{2}=2k^{2}\Rightarrow q^{2}$ is even
$\Rightarrow q$ is even.
$\Rightarrow p,q$ have 2 as a common factor which is contradiction to assumption.
$\therefore \sqrt{2}=\dfrac{p}{q}$ is a false
$\Rightarrow \sqrt{2}$ is not a rational.

Is the following are irrational numbers
$\sqrt{6}+\sqrt{2}$

State True or False

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

  1. True

  2. False

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Correct Option: A

Given that $\sqrt {3}$; rational. Then  " $2 + \sqrt {3}$ is irrational. "is true/false 

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

  1. True

  2. False

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Correct Option: A

If a, b and c are real numbers and $\dfrac{a+1}{ b}=\dfrac{7}{3}, \ \  \dfrac{b+1}{ c}=4 , \ \ \dfrac{c+1}{ a}=1$, then what is the value of $abc$

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

  1. 3

  2. 1

  3. 4

  4. 2

Show answer
Correct Option: B