Tag: proof of irrationality of numbers

Questions Related to proof of irrationality of numbers

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.

Show answer
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

Show answer
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

 $\sqrt{3}-\sqrt{5}$ is an 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: B

State true or false. 
$\sqrt { 3 } + \sqrt { 4 }$ is an 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: B
Explanation:
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
$\Rightarrow$  All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

$\sqrt{3}=1.732$ is an irrational.
$\sqrt{4}=2$ is rational.
Now,
$\Rightarrow$  $ \sqrt{3}+\sqrt{4}=1.732+2$
                        $=3.732$
$3.732$ cannot be converted into fraction.
$\therefore$  $\sqrt{3}+\sqrt{4}$ is an irrational number.
$\therefore$  Given statement is false.

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}$ up to $\infty$ is?

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

  1. $2$

  2. $3$

  3. $30$

  4. $5$

Show answer
Correct Option: B
Explanation:

We have,

$y=\sqrt { 6+\sqrt { 6+....\infty  }  } =\sqrt { 6+y } $

On squaring both sides, we get
$\Rightarrow { y }^{ 2 }-y-6=0$

$\Rightarrow y=3,-2$


But $y$ can not be $-ve$.

Hence, $y=3$ is the answer.