Tag: introduction to irrational numbers

Questions Related to introduction to irrational numbers

State whether the following statement is true or not:
$7-\sqrt { 2 } $ is irrational.

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:
Let us suppose $7-\sqrt 2$ is rational.

$=>7-\sqrt 2$ is in the form of $\dfrac pq$ where $p$ and $q$ are integers and $q\neq0$

$=>\sqrt2=-\dfrac pq+7$

​$=>\sqrt2=\dfrac{-p+7q}{q}$

as $p, q$ and $7$ are integers $\dfrac{-p+7q}{q}$ is a rational number.
$=>\sqrt 2$ is a rational number.

But we know that $\sqrt 2$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that $7-\sqrt 2$ is a rational number.

Hence, $7- \sqrt2$  is an irrational 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. $22/7$

  2. $3.14$

  3. $3.1401140014000$

  4. $3.\overline {14}$

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Correct Option: A

$7+\sqrt7$ is irrational

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

Assuming  that x,y,z  are positive real numbers,simplify the following :


$ (\sqrt{x})^{-2/3}\sqrt{y^{4}}\div \sqrt{xy^{-1/2}} $

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

  1. $ \dfrac{y^{9/4}}{x^{5}} $

  2. $ \dfrac{y^{9/4}}{x^{5/6}} $

  3. $ \dfrac{y^{9/4}}{x^{-5/6}} $

  4. $ \dfrac{y^{-9/4}}{x^{5/6}} $

Show answer
Correct Option: B
Explanation:
Given,

$(\sqrt{x})^{-\frac{2}{3}}\sqrt{y^4}\div \sqrt{xy^{-\frac{1}{2}}}$

$=\dfrac{x^{-\frac{1}{3}}y^2}{x^{\frac{1}{2}}y^{-\frac{1}{4}}}$

$=x^{-\frac{1}{3}-\frac{1}{2}}y^{2+\frac{1}{4}}$

$=x^{-\frac{5}{6}}y^{\frac{9}{4}}$

$=\dfrac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$

Which of the following is an irrational number?

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

  1. $\sqrt{41616}$

  2. $23.232323...$

  3. $\displaystyle\frac{(1+\sqrt3)^3 - (1-\sqrt3)^3}{\sqrt3}$

  4. $23.10100100010000...$

Show answer
Correct Option: D
Explanation:

$23.10100100010000...$ will be irrational number because it's non terminating and non repeating.

Correct option is (D)

The multiplicative inverse of $-1 + \sqrt{2}$ is

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

  1. $-1-\sqrt{2}$

  2. $1-\sqrt{2}$

  3. $1+\sqrt{2}$

  4. $\sqrt{2}$

  5. $2-\sqrt{2}$

Show answer
Correct Option: C
Explanation:

Multiplicative inverse of $\sqrt2-1$ is $\dfrac{1}{\sqrt2-1}$
Now multiply and divide with $\sqrt2+1$
We get $\dfrac{\sqrt2+1}{(\sqrt2+1)(\sqrt2-1)} $

$=\dfrac {\sqrt 2+1}{(\sqrt 2)^2-1^2}$
$= \sqrt2+1$

If a = 0.1039, then the value of $\sqrt{4a^2-4a+1}+3a$ is :

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

  1. 0.1039

  2. 0.2078

  3. 1.1039

  4. 2.1039

Show answer
Correct Option: C
Explanation:
$\sqrt{4a^{2}+4a+1}+3a=\sqrt{(2a-1)^{2}}+3a$ 
(as $2a< 1$ take $(1-2a)$) 
$\Rightarrow (1-2a)+3a=a+1=\boxed{1.1039}$

Which one of the following is not true?

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

  1. $\sqrt{2}$ is an irrational number

  2. If a is a rational number and $\sqrt{b}$ is an irrational number then $a\sqrt{b}$ is irrational number

  3. Every surd is an irrational number

  4. The square root of every positive integer is always irrational

Show answer
Correct Option: D
Explanation:

(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:


$π = 3.141592…$
$\sqrt {2} = 1.414213…$

Therefore, $\sqrt {2}$ is an irrational number.

(b) Let us take a rational number $a=\dfrac {2}{1}$ and an irrational number $b=\sqrt {2}$, then their product can be determined as:

$a\times b=2\times \sqrt { 2 } =2\sqrt { 2 }$ which is also an irrational number.

Therefore, if $a$ is a rational number and $\sqrt {b}$ is an irrational number than $a\sqrt {b}$ is an irrational number.

(c) By definition, a surd is a irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt {2}$ is a surd since $2$ is rational and $\sqrt {2}$ is irrational.

Surds are numbers left in root form $\sqrt {}$ to express its exact value. It has an infinite number of non-recurring decimals. 

Therefore, every surd is an irrational number.

(d) Let us take a positive integer $4$, now square root of $4$ will be:

$\sqrt {4}=2$ which is not an irrational number 

Hence, the square root of every positive integer is not always irrational.

Which one of the following is not true?

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

  1. When x is not a perfect square, $\sqrt{x}$ is an irrational number

  2. The index form of $\sqrt[m]{x^n}$ is $x^{\frac{n}{m}}$

  3. The radical form of $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}$ is $\sqrt[m]{x^n}$

  4. Every real number is an irrational number

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Correct Option: D
Explanation:

$(D)\,\, Real = Rational + Irrational$

If $a$ is an irrational number then which of the following describe the additive inverse of $a$.

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

  1. $a+a=2a$

  2. $a+0=a$

  3. $a\times=0$

  4. $a+(-a)=0$

Show answer
Correct Option: D
Explanation:

If the sum of two  irrational numbers is 0 then the two numbers are called additive inverse.
So option $D$ is correct.