Tag: proofs of irrationality

Questions Related to proofs of irrationality

State whether true or false: 

 $3+\sqrt{6}$ 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:

AS we know 3 is an irrational number and $\sqrt{6}$ is also a rational number 

Addition of two rational numbers are always a rational number 
So,  3 + $\sqrt{6}$ is also a rational number

Which of the following is an irrational number? 

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

  1. $0.14$

  2. $0.14 \overline{16}$

  3. $1.1 {416}$

  4. $0.4014001400014....$

Show answer
Correct Option: D
Explanation:
The decimal expansion of a rational number is either terminating or non-terminating repeating.

(A) $0.14$ is terminating, so it is a rational number

(B) $0.14\bar{16}=0.141616....$ 

is also rational ( non-terminating repeating ), where digits $16$ are repeating.

(C) $0.1416$ is terminating, so it is a rational number

(D) $0.4014001400014.....$

is an irrational number because it is neither terminating nor repeating.

Each of the following numbers is irrational
i) $(5 + 3\sqrt{2})$
ii) $3 \sqrt{7}$
iii) $\dfrac{3}{\sqrt{5}}$
iv) $(2 - 3\sqrt{5})$
v) $(\sqrt{3} + \sqrt{5})$

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

State whether the following statement is true or false.

The following number is irrational
$7\sqrt {5}$

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:
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $7\sqrt{5}$ is rational.
Hence, $7\sqrt{5}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $7\sqrt{5}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{5}=\dfrac{1}{7}\times\dfrac{a}{b}$

Here, $\dfrac{a}{7b}$ is a rational number, but $\sqrt{5}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $7\sqrt{5}$ is irrational number. 

 $2-\sqrt {3}$ 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:

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

Since, the difference of a rational and irrational number is irrational, hence, $2-\sqrt{3}$ is an irrational number.

State whether the following statement is true or false.

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

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:
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $6+\sqrt{2}$ is rational.
Hence, $6+\sqrt{2}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $6+\sqrt{2}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{2}=\dfrac{a}{b}-6$

$\Rightarrow$  $\sqrt{2}=\dfrac{a-6b}{b}$

Here, $\dfrac{a-6b}{b}$ is a rational number, but $\sqrt{2}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $6+\sqrt{2}$ is irrational number. 

Which of the following is always true 

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

  1. $irrational + irrational =irrational $

  2. $\dfrac{rational }{rational }=rational $

  3. $\dfrac{integer }{integer}=integer$

  4. None of these

Show answer
Correct Option: B
Explanation:

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C: $\dfrac{1}{2}=0.5$

Proof for B:

Let $q _1, q _2$ be two rational numbers such that $q _2\neq0$.


As they are rational, they can be written as $a/b, c/d$ respectively for some integers $a, b, c, d$. $(b,c,d\neq0)$

$\dfrac{q _1}{q _2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since, $a,b,c,d$ were integers, even $ad$ and $bc(\neq0)$ are integers and therefore, the above expression is rational.

If the product of two irrational numbers is rational, then which of the following can be concluded?

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

  1. The ratio of the greater and the smaller numbers is an integer.

  2. The sum of the numbers must be rational.

  3. The excess of the greater irrational number over the irrational number must be rational.

  4. None of the above

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Correct Option: A

 $\frac { 2 } { 2 + \sqrt { 3 } }$ 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

State whether the statement is true/false.
$\sqrt{72}$ is irrational

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:

72 is not an Irrational number because it can be expressed as the quotient of two integers: 72 ÷ 1

So, given statement is true.