Revisiting Irrational Numbers Questions in English

(B) The given number $0.123123123...$ can be written as $0.\overline{123}$.
Since the decimal expansion is non-terminating and repeating,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Let $x = 0.123123...$ (Equation $1$).
Multiplying both sides by $1000$,we get $1000x = 123.123123...$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$1000x - x = 123.123123... - 0.123123...$
$999x = 123$
$x = \frac{123}{999} = \frac{41}{333}$.
Since the number can be expressed as a ratio of two integers,it is a rational number.

(D) rationalising factor is a number that,when multiplied by a given irrational number,results in a rational number.
For the expression $5 \sqrt{2}$,the irrational part is $\sqrt{2}$.
To make $\sqrt{2}$ rational,we multiply it by $\sqrt{2}$ because $\sqrt{2} \times \sqrt{2} = 2$,which is a rational number.
Therefore,the rationalising factor of $5 \sqrt{2}$ is $\sqrt{2}$.

(A) surd is defined as an irrational root of a rational number,such as $\sqrt[n]{x}$,where $x$ is a rational number and the result is irrational.
$\pi$ is a transcendental number. It is irrational,but it cannot be expressed as the $n$-th root of any rational number. Therefore,$\pi$ is an irrational number that is not a surd.
$\sqrt{16} = 4$,which is a rational number.
$\sqrt{8} = 2\sqrt{2}$,which is a surd.
$3\sqrt{27} = 3 \times 3\sqrt{3} = 9\sqrt{3}$,which is a surd.
Thus,the correct option is $A$.

(B) The square root of $3$ is an irrational number. Its decimal expansion is $1.7320508\ldots$
Rounding this value to $2$ decimal places,we look at the third decimal digit,which is $2$. Since $2 < 5$,we keep the second decimal digit as it is.
Therefore,the approximate value of $\sqrt{3}$ up to $2$ decimal places is $1.73$.

(C) We know that the value of $\sqrt{3} \approx 1.732$.
Therefore,$\frac{1}{\sqrt{3}} = \frac{1}{1.732}$.
By performing the division,we get $\frac{1}{1.732} \approx 0.5773$.
Rounding this value to $2$ decimal places,we get $0.58$.

(D) The square root of $2$ is an irrational number,which is approximately $1.41421356...$
Rounding this value to $2$ decimal places,we look at the third decimal digit,which is $4$. Since $4 < 5$,we keep the second decimal digit as it is.
Therefore,the approximate value of $\sqrt{2}$ up to $2$ decimal places is $1.41$.

(C) Let the two conjugate binomial surds be $(a + \sqrt{b})$ and $(a - \sqrt{b})$.
Their product is $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b$.
Since $a$ and $b$ are rational numbers,$a^2 - b$ is also a rational number.
Therefore,the product of two conjugate binomial surds is always a rational number.

(C) conjugate surd is formed by changing the sign between the two terms of a binomial surd.
For a binomial surd of the form $a+\sqrt{b}$,its conjugate surd is $a-\sqrt{b}$.
Given the expression $3+\sqrt{2}$,we change the sign of the radical part.
Therefore,the conjugate surd of $3+\sqrt{2}$ is $3-\sqrt{2}$.

(D) To simplify the expression $\sqrt{5+2 \sqrt{6}}$,we look for two numbers whose sum is $5$ and whose product is $6$.
These two numbers are $3$ and $2$.
We can rewrite the expression as:
$\sqrt{5+2 \sqrt{6}} = \sqrt{(3+2)+2 \sqrt{3 \times 2}}$
Using the algebraic identity $(a+b)^2 = a^2+b^2+2ab$,we have:
$\sqrt{3+2+2 \sqrt{3} \sqrt{2}} = \sqrt{(\sqrt{3})^2+(\sqrt{2})^2+2 \sqrt{3} \sqrt{2}}$
$= \sqrt{(\sqrt{3}+\sqrt{2})^2}$
$= \sqrt{3}+\sqrt{2}$

(A) To rationalise the denominator of $\frac{1}{3+\sqrt{8}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3-\sqrt{8})$.
$\frac{1}{3+\sqrt{8}} = \frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{3-\sqrt{8}}{(3)^2 - (\sqrt{8})^2}$
$= \frac{3-\sqrt{8}}{9-8}$
$= \frac{3-\sqrt{8}}{1}$
$= 3-\sqrt{8}$

(C) To simplify the expression $\sqrt{12-\sqrt{140}}$,we first rewrite $\sqrt{140}$ as $2\sqrt{35}$.
Thus,the expression becomes $\sqrt{12-2\sqrt{35}}$.
We look for two numbers whose sum is $12$ and whose product is $35$. These numbers are $7$ and $5$.
We can write $12$ as $(7+5)$ and $35$ as $(7 \times 5)$.
So,$\sqrt{12-2\sqrt{35}} = \sqrt{(7+5)-2\sqrt{7 \times 5}}$.
Using the identity $(a-b)^2 = a^2+b^2-2ab$,we have $\sqrt{(\sqrt{7}-\sqrt{5})^2}$.
Therefore,the result is $\sqrt{7}-\sqrt{5}$.

(C) An irrational number is a number that cannot be expressed as a fraction $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Its decimal expansion is non-terminating and non-recurring.
$1$. $\sqrt{4} = 2$,which is a rational number.
$2$. $0.010101...$ is a recurring decimal,which can be expressed as $1/99$,so it is a rational number.
$3$. $\pi$ (pi) is a well-known irrational number because its decimal expansion is non-terminating and non-recurring.
$4$. $5$ can be written as $5/1$,which is a rational number.
Therefore,the correct option is $C$.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter.
It is a non-terminating and non-repeating decimal, which means it cannot be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore, $\pi$ is an irrational number.

(D) rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
$1$. $\sqrt{7}$ is an irrational number because $7$ is not a perfect square.
$2$. $3\pi$ is an irrational number because $\pi$ is irrational.
$3$. $0.101000...$ is a non-terminating and non-repeating decimal,which makes it an irrational number.
$4$. $0.25$ can be written as $\frac{25}{100} = \frac{1}{4}$,which is in the form $\frac{p}{q}$.
Therefore,$0.25$ is a rational number.

(A) The given number is $0.01111...$,which can be written as $0.0\bar{1}$.
Since the decimal expansion is non-terminating and repeating,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Let $x = 0.0111...$ (Equation $1$).
Multiply by $10$: $10x = 0.1111...$ (Equation $2$).
Multiply by $100$: $100x = 1.1111...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$: $100x - 10x = 1.1111... - 0.1111...$.
$90x = 1$.
$x = \frac{1}{90}$.
Since the number can be expressed as a fraction of two integers,it is a rational number.

(B) number is called rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
$\pi$ is a mathematical constant defined as the ratio of a circle's circumference to its diameter.
Its decimal representation is non-terminating and non-repeating.
Therefore, $\pi$ is an irrational number.

(B) number is called rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
Since $\sqrt{2}$ cannot be expressed as a ratio of two integers, it is an irrational number.
Therefore, $\sqrt{2}$ is an irrational number.

(B) The given expression is $\sqrt{1+1}$.
Calculating the sum inside the square root: $1+1 = 2$.
Therefore,the expression becomes $\sqrt{2}$.
Since $2$ is not a perfect square,$\sqrt{2}$ is an irrational number.
Thus,the correct option is $B$.

(A) The given number is $0.01111...$,which can be written as $0.0\overline{1}$.
Since the decimal expansion is non-terminating and repeating,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Let $x = 0.0111...$ (Equation $1$).
Multiply by $10$: $10x = 0.1111...$ (Equation $2$).
Multiply by $100$: $100x = 1.1111...$ (Equation $3$).
Subtracting Equation $2$ from Equation $3$: $100x - 10x = 1.1111... - 0.1111...$.
$90x = 1$.
$x = \frac{1}{90}$.
Since the number can be expressed as a fraction $\frac{1}{90}$,it is a rational number.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter.
It is a non-terminating and non-repeating decimal,which means it cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore,$\pi$ is an irrational number.

(B) number is called rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Since $\sqrt{2}$ cannot be expressed as a ratio of two integers,it is an irrational number.
Therefore,$\sqrt{2}$ is an irrational number.

(B) Given expression is $\sqrt{1+1}$.
First,calculate the sum inside the square root: $1+1 = 2$.
So,the expression becomes $\sqrt{2}$.
Since $2$ is not a perfect square,$\sqrt{2}$ is an irrational number.
Therefore,the correct option is $B$.

(B) The decimal expansion $0.9999...$ can be written as $0.\bar{9}$.
Since the digit $9$ repeats infinitely after the decimal point,it is a non-terminating repeating decimal expansion.
Furthermore,$0.9999... = 1$,which is a rational number,and all rational numbers have either terminating or non-terminating repeating decimal expansions.

(D) The number of decimal places in the decimal expansion of a rational number $\frac{p}{q}$ (where $q$ is of the form $2^n \cdot 5^m$) is given by $\max(n, m)$.
However,for a fraction $\frac{p}{q}$ where $q$ is not of the form $2^n \cdot 5^m$,the decimal expansion is non-terminating and repeating.
For $\frac{3}{113}$,the denominator $113$ is a prime number and is not of the form $2^n \cdot 5^m$.
Therefore,the decimal expansion is non-terminating and repeating.
If the question implies the length of the period of the repeating decimal,for a prime $p$,the period length divides $p-1$.
For $p = 113$,the period length is $112$.

(B) The decimal expansion $0.9999...$ can be written as $0.\bar{9}$.
Since the digit $9$ repeats infinitely,it is a non-terminating repeating decimal expansion.
Furthermore,mathematically,$0.9999... = 1$,which is a rational number,and all rational numbers have either terminating or non-terminating repeating decimal expansions.

(D) rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
$1$. $\sqrt{7}$ is an irrational number because $7$ is not a perfect square.
$2$. $3\pi$ is an irrational number because $\pi$ is irrational.
$3$. $0.101000...$ is a non-terminating and non-repeating decimal,which makes it an irrational number.
$4$. $0$ can be written as $\frac{0}{1}$,which satisfies the condition of a rational number.
Therefore,$0$ is the correct rational number.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter.
It is a non-terminating and non-repeating decimal,which means it cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore,$\pi$ is an irrational number.

(A) Let $x = 1. \overline{23} = 1.232323...$
Multiplying by $100$,we get $100x = 123.232323...$
Subtracting $x$ from $100x$,we get $99x = 122$,so $x = \frac{122}{99}$.
Now,$3.2 = \frac{32}{10} = \frac{16}{5}$.
The sum is $\frac{122}{99} + \frac{16}{5} = \frac{122 \times 5 + 16 \times 99}{99 \times 5} = \frac{610 + 1584}{495} = \frac{2194}{495}$.
Since the sum can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(B) Given expression: $(\sqrt{2} - \sqrt{3})(\sqrt{3} + \sqrt{2})$
This is in the form of $(a - b)(a + b) = a^2 - b^2$.
Here,$a = \sqrt{2}$ and $b = \sqrt{3}$ is not quite right,let us rewrite it as $(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})$.
Using the identity $(a - b)(a + b) = a^2 - b^2$:
$(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$.
Since $-1$ is an integer,it is a rational number.
Therefore,the expression is not an irrational number.

(B) rational number is defined as a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
$\sqrt{2}$ is a non-terminating and non-repeating decimal $(1.41421356...)$.
Since it cannot be expressed as a ratio of two integers,it is an irrational number.
Therefore,$\sqrt{2}$ is not a rational number.

(B) Let us assume,to the contrary,that $7\sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $7\sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{7b}$.
Since $a$ and $b$ are integers,$\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our assumption is incorrect,and $7\sqrt{5}$ must be an irrational number.

(B) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is even,so $a$ must also be even.
Let $a = 2k$ for some integer $k$.
Substituting this into the equation,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This shows that $b^2$ is even,so $b$ must also be even.
Since both $a$ and $b$ are even,they have a common factor $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is irrational.

(B) Let us assume,to the contrary,that $3 + 2\sqrt{5}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3 + 2\sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2\sqrt{5} = \frac{a}{b} - 3$.
$2\sqrt{5} = \frac{a - 3b}{b}$.
Dividing by $2$,we get $\sqrt{5} = \frac{a - 3b}{2b}$.
Since $a$ and $b$ are integers,$\frac{a - 3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is irrational.
Therefore,our assumption is incorrect,and $3 + 2\sqrt{5}$ is irrational.

(B) Assume,to the contrary,that $\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is even,and therefore $a$ must also be even.
Let $a = 2k$ for some integer $k$.
Substituting this into the equation,we get $(2k)^2 = 2b^2$,so $4k^2 = 2b^2$,which simplifies to $b^2 = 2k^2$.
This shows that $b^2$ is even,and therefore $b$ must also be even.
Since both $a$ and $b$ are even,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Thus,our assumption is false,and $\sqrt{2}$ is irrational.

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is divisible by $2$,and consequently,$a$ is also divisible by $2$ (by the Fundamental Theorem of Arithmetic).
So,we can write $a = 2k$ for some integer $k$.
Substituting this into the equation $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This shows that $b^2$ is divisible by $2$,and consequently,$b$ is also divisible by $2$.
Thus,both $a$ and $b$ have at least $2$ as a common factor,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption that $\sqrt{2}$ is rational is false,and $\sqrt{2}$ must be irrational.

(A) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This means $a^2$ is divisible by $2$,and consequently,$a$ is also divisible by $2$ (by the Fundamental Theorem of Arithmetic).
So,we can write $a = 2k$ for some integer $k$.
Substituting this into $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This implies $b^2$ is divisible by $2$,and consequently,$b$ is also divisible by $2$.
Thus,both $a$ and $b$ have at least $2$ as a common factor,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is irrational.

(N/A) Let us assume,to the contrary,that $3\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3\sqrt{2} = \frac{a}{b}$.
This can be rearranged as $\sqrt{2} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $3\sqrt{2}$ is a rational number is incorrect.
Hence,$3\sqrt{2}$ is an irrational number.

(B) Let us assume,to the contrary,that $5 - \sqrt{3}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $5 - \sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $5 - \frac{a}{b} = \sqrt{3}$,which simplifies to $\frac{5b - a}{b} = \sqrt{3}$.
Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the fact that $\sqrt{3}$ is an irrational number.
Therefore,our assumption is incorrect,and $5 - \sqrt{3}$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This means $a^2$ is divisible by $2$,so $a$ must also be divisible by $2$ (by the Fundamental Theorem of Arithmetic).
Let $a = 2k$ for some integer $k$.
Substituting this into $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This means $b^2$ is divisible by $2$,so $b$ must also be divisible by $2$.
Since both $a$ and $b$ are divisible by $2$,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is an irrational number.