Revisiting Irrational Numbers Questions in English

(N/A) Let us assume,to the contrary,that $\sqrt{3}$ is rational.
That is,we can find integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{3} = \frac{a}{b}$.
Suppose $a$ and $b$ have a common factor other than $1$. Then we can divide by the common factor and assume that $a$ and $b$ are coprime.
So,$b\sqrt{3} = a$.
Squaring on both sides,we get $3b^2 = a^2$.
This implies that $a^2$ is divisible by $3$,and by the fundamental theorem of arithmetic,$a$ is also divisible by $3$.
So,we can write $a = 3c$ for some integer $c$.
Substituting $a = 3c$ in $3b^2 = a^2$,we get $3b^2 = (3c)^2 = 9c^2$,which simplifies to $b^2 = 3c^2$.
This means that $b^2$ is divisible by $3$,and consequently,$b$ is also divisible by $3$.
Therefore,$a$ and $b$ have at least $3$ as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence,we conclude that $\sqrt{3}$ is irrational.

(N/A) Let us assume,to the contrary,that $5-\sqrt{3}$ is rational.
That is,we can find 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 $\sqrt{3} = \frac{5b - a}{b}$.
Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.
Therefore,we conclude that $5-\sqrt{3}$ is irrational.

(N/A) Let us assume,to the contrary,that $3 \sqrt{2}$ is rational.
That is,we can find coprime integers $a$ and $b$ $(b \neq 0)$ such that $3 \sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{3b}$.
Since $3$,$a$,and $b$ are integers,$\frac{a}{3b}$ must be a rational number,which implies that $\sqrt{2}$ is rational.
However,this contradicts the established fact that $\sqrt{2}$ is irrational.
Therefore,our initial assumption is false,and we conclude that $3 \sqrt{2}$ is irrational.

(N/A) Assume that $\sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{5} = \frac{a}{b}$,where $a$ and $b$ are co-prime (i.e.,their only common factor is $1$).
Squaring both sides,we get $a^2 = 5b^2$.
This implies that $a^2$ is divisible by $5$,and by the Fundamental Theorem of Arithmetic,$a$ is also divisible by $5$.
Let $a = 5k$ for some integer $k$.
Substituting this into the equation $a^2 = 5b^2$,we get $(5k)^2 = 5b^2$,which simplifies to $25k^2 = 5b^2$,or $b^2 = 5k^2$.
This implies that $b^2$ is divisible by $5$,and consequently,$b$ is also divisible by $5$.
Since both $a$ and $b$ are divisible by $5$,they have a common factor of $5$,which contradicts our initial assumption that $a$ and $b$ are co-prime.
Therefore,our assumption that $\sqrt{5}$ is rational is false,and we conclude that $\sqrt{5}$ is irrational.

(N/A) Assume that $3+2 \sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $3+2 \sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2 \sqrt{5} = \frac{a}{b} - 3$.
Simplifying the right side,$2 \sqrt{5} = \frac{a-3b}{b}$.
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 rational.
However,this contradicts the established fact that $\sqrt{5}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption that $3+2 \sqrt{5}$ is rational.
Therefore,we conclude that $3+2 \sqrt{5}$ is irrational.

(B) Assume that $\frac{1}{\sqrt{2}}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $\frac{1}{\sqrt{2}} = \frac{a}{b}$.
By rearranging the terms,we get $\sqrt{2} = \frac{b}{a}$.
Since $a$ and $b$ are integers,$\frac{b}{a}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the established fact that $\sqrt{2}$ is an irrational number.
Therefore,our initial assumption is false,and $\frac{1}{\sqrt{2}}$ must be an irrational number.

(N/A) Assume that $7 \sqrt{5}$ is a rational number.
Therefore,we can find two 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 $\sqrt{5}$ must also be a rational number.
However,this contradicts the established fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption that $7 \sqrt{5}$ is rational is incorrect.
Hence,$7 \sqrt{5}$ is an irrational number.

(N/A) Assume that $6+\sqrt{2}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $6+\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{b} - 6$.
Since $a$ and $b$ are integers,$\frac{a}{b} - 6 = \frac{a-6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the established fact that $\sqrt{2}$ is an irrational number.
This contradiction has arisen due to our incorrect assumption that $6+\sqrt{2}$ is rational.
Therefore,we conclude that $6+\sqrt{2}$ is an irrational number.

(B) Let $r$ be a non-zero rational number and $i$ be an irrational number.
Assume the product $p = r \times i$ is rational.
Then $p = \frac{a}{b}$ for some integers $a, b$ where $b \neq 0$.
Since $r$ is a non-zero rational,$r = \frac{p}{q}$ for some integers $p, q$ where $p, q \neq 0$.
Then $i = \frac{p}{r} = \frac{p}{q} \times \frac{b}{a} = \frac{pb}{qa}$.
Since $p, b, q, a$ are integers,$\frac{pb}{qa}$ is a rational number.
This contradicts the fact that $i$ is irrational.
Therefore,the product of a non-zero rational and an irrational number is always irrational.
For example,$\frac{3}{4} \times \sqrt{2} = \frac{3\sqrt{2}}{4}$,which is an irrational number.

Let us assume that $\sqrt{2}+\sqrt{3}$ is rational. Let $\sqrt{2}+\sqrt{3} = a$,where $a$ is a rational number.
Then,$\sqrt{2} = a - \sqrt{3}$.
Squaring both sides,we get:
$(\sqrt{2})^2 = (a - \sqrt{3})^2$
$2 = a^2 + 3 - 2a\sqrt{3}$
Rearranging the terms to isolate $\sqrt{3}$:
$2a\sqrt{3} = a^2 + 3 - 2$
$2a\sqrt{3} = a^2 + 1$
$\sqrt{3} = \frac{a^2 + 1}{2a}$
Since $a$ is a rational number,$\frac{a^2 + 1}{2a}$ must also be a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the known fact that $\sqrt{3}$ is an irrational number. This contradiction has arisen because of our initial assumption that $\sqrt{2}+\sqrt{3}$ is rational.
Therefore,we conclude that $\sqrt{2}+\sqrt{3}$ is irrational.

(N/A) Let us assume that $\sqrt{3}+\sqrt{5}$ is a rational number.
Let $\sqrt{3}+\sqrt{5} = a$,where $a$ is a rational number.
Then,$\sqrt{3} = a - \sqrt{5}$.
Squaring both sides,we get:
$(\sqrt{3})^2 = (a - \sqrt{5})^2$
$3 = a^2 + 5 - 2a\sqrt{5}$ (using the identity $(x - y)^2 = x^2 + y^2 - 2xy$).
Rearranging the terms to isolate $\sqrt{5}$:
$2a\sqrt{5} = a^2 + 5 - 3$
$2a\sqrt{5} = a^2 + 2$
$\sqrt{5} = \frac{a^2 + 2}{2a}$.
Since $a$ is a rational number,$\frac{a^2 + 2}{2a}$ must also be a rational number.
This implies that $\sqrt{5}$ is a rational number,which contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption is incorrect,and $\sqrt{3}+\sqrt{5}$ is an irrational number.

(D) To determine if the rational number $\frac{29}{343}$ has a terminating decimal expansion,we examine the prime factorization of the denominator.
$1$. The denominator is $343 = 7^3$.
$2$. $A$ rational number $\frac{p}{q}$ has a terminating decimal expansion if and only if the prime factorization of $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
$3$. Since the prime factorization of $343$ is $7^3$,which does not contain factors of $2$ or $5$,the decimal expansion is non-terminating and repeating.
$4$. Therefore,$\frac{29}{343}$ does not have a terminating decimal expansion.

(C) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n 5^m$,where $n$ and $m$ are non-negative integers.
In the given rational number $\frac{129}{2^{2} 5^{7} 7^{5}}$,the denominator contains a factor of $7^5$ in addition to $2$ and $5$.
Since the prime factorization of the denominator includes a prime factor other than $2$ or $5$ (which is $7$),the decimal expansion of this rational number is non-terminating and repeating.
Therefore,it does not have a terminating decimal expansion.

(N/A) The given decimal expansion is $0.120120012000120000 \ldots$
$1$. $A$ number is rational if its decimal expansion is either terminating or non-terminating recurring.
$2$. $A$ number is irrational if its decimal expansion is non-terminating and non-recurring.
$3$. In the given number $0.120120012000120000 \ldots$,the pattern of digits does not repeat in a fixed block,and it does not terminate.
$4$. Since the decimal expansion is non-terminating and non-recurring,the given number is an irrational number.
$5$. Because it is irrational,it cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

(N/A) The given number is $43. \overline{123456789}$.
Since the decimal expansion is non-terminating and repeating (recurring),the given number is a rational number.
$A$ rational number has a terminating decimal expansion if and only if the prime factorization of its denominator $q$ is of the form $2^m \times 5^n$,where $m$ and $n$ are non-negative integers.
Since this decimal expansion is non-terminating repeating,the denominator $q$ must have prime factors other than $2$ or $5$.

(N/A) Let us assume that $\sqrt{p}+\sqrt{q}$ is a rational number.
Let $\sqrt{p}+\sqrt{q} = a$,where $a$ is a non-zero rational number.
Then,$\sqrt{q} = a - \sqrt{p}$.
Squaring both sides,we get:
$(\sqrt{q})^2 = (a - \sqrt{p})^2$
$q = a^2 + p - 2a\sqrt{p}$.
Rearranging the terms to isolate $\sqrt{p}$:
$2a\sqrt{p} = a^2 + p - q$
$\sqrt{p} = \frac{a^2 + p - q}{2a}$.
Since $a$ is a rational number,the expression $\frac{a^2 + p - q}{2a}$ must also be a rational number.
However,we know that $\sqrt{p}$ is an irrational number because $p$ is a prime number.
This leads to a contradiction because an irrational number cannot be equal to a rational number.
Therefore,our initial assumption is false,and $\sqrt{p} + \sqrt{q}$ must be an irrational number.

(N/A) Assume,to the contrary,that $\sqrt{3}$ is a rational number.
Then,there exist coprime positive integers $a$ and $b$ such that $\sqrt{3} = \frac{a}{b}$,where $gcd(a, b) = 1$.
Squaring both sides,we get $3 = \frac{a^2}{b^2}$,which implies $a^2 = 3b^2$ .......... $(1)$.
Since $3$ divides $a^2$,it follows that $3$ must also divide $a$ (by the Fundamental Theorem of Arithmetic).
Let $a = 3k$ for some integer $k$.
Substituting this into equation $(1)$,we get $(3k)^2 = 3b^2$,which simplifies to $9k^2 = 3b^2$,or $b^2 = 3k^2$.
This implies that $3$ divides $b^2$,and consequently,$3$ must divide $b$.
Thus,$3$ is a common factor of both $a$ and $b$,which contradicts our initial assumption that $gcd(a, b) = 1$.
Therefore,our assumption that $\sqrt{3}$ is rational is false.
Hence,$\sqrt{3}$ is an irrational number.

(N/A) Assume,for the sake of contradiction,that $\sqrt[3]{6}$ is a rational number.
Let $\sqrt[3]{6} = \frac{a}{b}$,where $a, b \in N$ and $g.c.d.(a, b) = 1$.
Since $1 < 6 < 8$,we have $\sqrt[3]{1} < \sqrt[3]{6} < \sqrt[3]{8}$,which implies $1 < \frac{a}{b} < 2$.
This means $b > 1$,because if $b = 1$,then $\frac{a}{b}$ would be an integer,but there is no integer between $1$ and $2$.
Cubing both sides,we get $6 = \frac{a^3}{b^3}$,which implies $6b^3 = a^3$.
Since $g.c.d.(a, b) = 1$,it follows that $g.c.d.(a^3, b^3) = 1$.
From $6b^3 = a^3$,we see that $b^3$ must divide $a^3$. Since $g.c.d.(a^3, b^3) = 1$,this is only possible if $b^3 = 1$,which means $b = 1$.
However,we already established that $b > 1$. This is a contradiction.
Therefore,our assumption that $\sqrt[3]{6}$ is rational is false.
Hence,$\sqrt[3]{6}$ is an irrational number.

(N/A) First,rationalize the denominator: $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Assume,for the sake of contradiction,that $\frac{\sqrt{2}}{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\frac{\sqrt{2}}{2} = \frac{a}{b}$.
This implies $\sqrt{2} = \frac{2a}{b}$.
Since $a$ and $b$ are integers,$\frac{2a}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the well-known fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $\frac{1}{\sqrt{2}}$ is rational must be false.
Hence,$\frac{1}{\sqrt{2}}$ is an irrational number.

(N/A) Assume that $3+2 \sqrt{5}$ is a rational number.
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 $2 \sqrt{5} = \frac{a}{b} - 3 = \frac{a-3b}{b}$.
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 an irrational number.
Therefore,our assumption that $3+2 \sqrt{5}$ is rational is incorrect.
Hence,$3+2 \sqrt{5}$ is an irrational number.

(B) Assume,to the contrary,that $\sqrt{7}$ is a rational number.
Then,there exist coprime positive integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{7} = \frac{a}{b}$.
Squaring both sides,we get $7 = \frac{a^2}{b^2}$,which implies $a^2 = 7b^2$.
This means that $a^2$ is divisible by $7$,and by the Fundamental Theorem of Arithmetic,$a$ must also be divisible by $7$.
So,we can write $a = 7k$ for some integer $k$.
Substituting this into the equation $a^2 = 7b^2$,we get $(7k)^2 = 7b^2$,which simplifies to $49k^2 = 7b^2$,or $b^2 = 7k^2$.
This implies that $b^2$ is divisible by $7$,and therefore $b$ must also be divisible by $7$.
Since both $a$ and $b$ are divisible by $7$,they have a common factor of $7$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption that $\sqrt{7}$ is rational is false,and $\sqrt{7}$ is an irrational number.

(N/A) Assume,to the contrary,that $\sqrt{21}$ is a rational number.
Then,there exist two coprime positive integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{21} = \frac{a}{b}$.
Squaring both sides,we get $21 = \frac{a^2}{b^2}$,which implies $a^2 = 21b^2$.
This means $a^2$ is divisible by $21$,so $a$ must also be divisible by $21$ (since $21$ is a product of two distinct primes $3$ and $7$).
Let $a = 21k$ for some integer $k$.
Substituting this into the equation,we get $(21k)^2 = 21b^2$,which simplifies to $441k^2 = 21b^2$,or $b^2 = 21k^2$.
This implies $b^2$ is divisible by $21$,so $b$ must also be divisible by $21$.
Thus,$a$ and $b$ have a common factor of $21$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{21}$ is an irrational number.

(B) Assume that $\sqrt{5}+1$ is a rational number.
Then,we can write $\sqrt{5}+1 = \frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Rearranging the equation,we get $\sqrt{5} = \frac{p}{q} - 1$.
This simplifies to $\sqrt{5} = \frac{p-q}{q}$.
Since $p$ and $q$ are integers,$\frac{p-q}{q}$ must be a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the known fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption is false,and $\sqrt{5}+1$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{3}+\sqrt{7}$ is a rational number. Let $\sqrt{3}+\sqrt{7} = r$,where $r$ is a rational number.
Squaring both sides,we get: $(\sqrt{3}+\sqrt{7})^2 = r^2$.
$3 + 7 + 2\sqrt{21} = r^2$.
$10 + 2\sqrt{21} = r^2$.
$2\sqrt{21} = r^2 - 10$.
$\sqrt{21} = \frac{r^2 - 10}{2}$.
Since $r$ is a rational number,$r^2$ is also a rational number. Therefore,$\frac{r^2 - 10}{2}$ is a rational number.
This implies that $\sqrt{21}$ is a rational number.
However,we know that $\sqrt{21}$ is an irrational number because $21$ is not a perfect square.
This is a contradiction to our assumption that $\sqrt{3}+\sqrt{7}$ is rational.
Therefore,our assumption is false,and $\sqrt{3}+\sqrt{7}$ must be an irrational number.

(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 $3\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{3b}$.
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.
Our assumption that $3\sqrt{2}$ is rational is incorrect.
Therefore,$3\sqrt{2}$ is an irrational number.

(N/A) The decimal expansion $0.02003000400005 \ldots$ is non-terminating and non-recurring.
Since a number is rational if and only if its decimal expansion is either terminating or non-terminating repeating,this number does not satisfy the condition.
Therefore,$0.02003000400005 \ldots$ is an irrational number.

(N/A) The decimal form $5 . \overline{123456}$ is non-terminating and recurring. Therefore,it is a rational number.
Let $x = 5 . \overline{123456}$.
$\therefore x = 5.123456123456 \ldots$ (Equation $1$)
Since there are $6$ repeating digits,multiply both sides by $10^6 = 1000000$:
$\therefore 1000000x = 5123456.123456123456 \ldots$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$1000000x - x = 5123456.123456 \ldots - 5.123456 \ldots$
$999999x = 5123451$
$\therefore x = \frac{5123451}{999999}$
Simplifying the fraction by dividing both numerator and denominator by $9$:
$x = \frac{569272.333...}{111111}$ (Note: The fraction is $\frac{5123451}{999999}$ which is the standard $\frac{p}{q}$ form).

(A) The decimal form $3.127$ is a terminating decimal.
Since every terminating decimal can be expressed as a fraction,it is a rational number.
To express $3.127$ in the form of $\frac{p}{q}$,we remove the decimal point and divide by $1000$ because there are three digits after the decimal point.
$3.127 = \frac{3127}{1000}$

(A) Let $x = 0.2\overline{35}$.
This means $x = 0.2353535...$ (Equation $1$).
Multiply both sides by $10$ to shift the decimal point: $10x = 2.353535...$ (Equation $2$).
Multiply Equation $1$ by $1000$: $1000x = 235.353535...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$:
$1000x - 10x = 235.353535... - 2.353535...$
$990x = 233$.
Therefore,$x = \frac{233}{990}$.
Since the number can be expressed in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(A) Let $x = 18.484848 \ldots$ (Equation $1$).
Since the repeating part is $48$,multiply both sides by $100$:
$100x = 1848.484848 \ldots$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$100x - x = 1848.484848 \ldots - 18.484848 \ldots$
$99x = 1830$.
$x = \frac{1830}{99}$.
Dividing both numerator and denominator by $3$,we get $x = \frac{610}{33}$.
Since the number can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(A) Let $x = 0.\overline{001}$.
This means $x = 0.001001001...$ (Equation $1$).
Since there are $3$ repeating digits,multiply both sides by $10^3 = 1000$:
$1000x = 1.001001001...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1000x - x = 1.001001001... - 0.001001001...$
$999x = 1$.
Therefore,$x = \frac{1}{999}$.
Since the number can be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.

(N/A) number is rational if its decimal expansion is either terminating or non-terminating repeating.
In the given decimal expansion $0.05005000500005 \ldots$,the pattern of digits does not repeat,and it does not terminate.
Since the decimal expansion is non-terminating and non-repeating,the number is irrational.
Therefore,it cannot be expressed in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

(D) Let $\sqrt{7+2 \sqrt{10}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $7 + 2 \sqrt{10} = x + y + 2 \sqrt{xy}$.
Comparing the rational and irrational parts,we have $x + y = 7$ and $2 \sqrt{xy} = 2 \sqrt{10}$,which implies $xy = 10$.
We need two numbers whose sum is $7$ and product is $10$.
The factors of $10$ are $(5, 2)$ such that $5 + 2 = 7$ and $5 \times 2 = 10$.
Thus,$x = 5$ and $y = 2$.
Therefore,$\sqrt{7+2 \sqrt{10}} = \sqrt{5} + \sqrt{2}$.

(A) To simplify the expression,we rationalize each term individually:
$1$. $\frac{1}{\sqrt{6}-\sqrt{5}} = \frac{1(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})} = \frac{\sqrt{6}+\sqrt{5}}{6-5} = \sqrt{6}+\sqrt{5}$
$2$. $\frac{3}{\sqrt{5}-\sqrt{2}} = \frac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})} = \frac{3(\sqrt{5}+\sqrt{2})}{5-2} = \frac{3(\sqrt{5}+\sqrt{2})}{3} = \sqrt{5}+\sqrt{2}$
$3$. $\frac{4}{\sqrt{6}+\sqrt{2}} = \frac{4(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{4(\sqrt{6}-\sqrt{2})}{6-2} = \frac{4(\sqrt{6}-\sqrt{2})}{4} = \sqrt{6}-\sqrt{2}$
Now,substitute these back into the original expression:
$(\sqrt{6}+\sqrt{5}) - (\sqrt{5}+\sqrt{2}) - (\sqrt{6}-\sqrt{2})$
$= \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2}$
$= (\sqrt{6} - \sqrt{6}) + (\sqrt{5} - \sqrt{5}) + (\sqrt{2} - \sqrt{2})$
$= 0 + 0 + 0 = 0$

(B) Step $1$: Rationalize the first term $\frac{2}{\sqrt{5}-\sqrt{3}}$.
Multiply numerator and denominator by $(\sqrt{5} + \sqrt{3})$:
$\frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} = \frac{2(\sqrt{5} + \sqrt{3})}{2} = \sqrt{5} + \sqrt{3}$.
Step $2$: Simplify the second term $\frac{4}{\sqrt{10+\sqrt{84}}}$.
Note that $\sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$.
So,$\sqrt{10 + 2\sqrt{21}} = \sqrt{(\sqrt{7} + \sqrt{3})^2} = \sqrt{7} + \sqrt{3}$.
Thus,the term becomes $\frac{4}{\sqrt{7} + \sqrt{3}}$.
Rationalize it: $\frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$.
Step $3$: Rationalize the third term $\frac{2}{\sqrt{7}-\sqrt{5}}$.
Multiply numerator and denominator by $(\sqrt{7} + \sqrt{5})$:
$\frac{2(\sqrt{7} + \sqrt{5})}{7 - 5} = \frac{2(\sqrt{7} + \sqrt{5})}{2} = \sqrt{7} + \sqrt{5}$.
Step $4$: Combine all terms:
$(\sqrt{5} + \sqrt{3}) + (\sqrt{7} - \sqrt{3}) - (\sqrt{7} + \sqrt{5})$
$= \sqrt{5} + \sqrt{3} + \sqrt{7} - \sqrt{3} - \sqrt{7} - \sqrt{5} = 0$.

(N/A) To simplify the expression,we first multiply the numerator and denominator of each term by $\sqrt{2}$ to eliminate the nested radicals.
Let the expression be $E = \frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2-\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}$.
Multiply the numerator and denominator of each term by $\sqrt{2}$:
$E = \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{2}(\sqrt{2}+\sqrt{2-\sqrt{3}})} + \frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{2}(\sqrt{2}-\sqrt{2+\sqrt{3}})}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4-2\sqrt{3}}} + \frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4+2\sqrt{3}}}$
Note that $4-2\sqrt{3} = (\sqrt{3}-1)^2$ and $4+2\sqrt{3} = (\sqrt{3}+1)^2$.
So,$\sqrt{4-2\sqrt{3}} = \sqrt{3}-1$ and $\sqrt{4+2\sqrt{3}} = \sqrt{3}+1$.
Substituting these values:
$E = \frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3}-1} + \frac{2\sqrt{2}-\sqrt{6}}{2-(\sqrt{3}+1)}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{\sqrt{3}+1} + \frac{2\sqrt{2}-\sqrt{6}}{1-\sqrt{3}}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{\sqrt{3}+1} - \frac{2\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}$
Taking the common denominator $(\sqrt{3}+1)(\sqrt{3}-1) = 3-1 = 2$:
$E = \frac{(2\sqrt{2}+\sqrt{6})(\sqrt{3}-1) - (2\sqrt{2}-\sqrt{6})(\sqrt{3}+1)}{2}$
$E = \frac{(2\sqrt{6}-2\sqrt{2}+3\sqrt{2}-\sqrt{6}) - (2\sqrt{6}+2\sqrt{2}-3\sqrt{2}-\sqrt{6})}{2}$
$E = \frac{(\sqrt{6}+\sqrt{2}) - (\sqrt{6}-\sqrt{2})}{2}$
$E = \frac{\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$.
Hence,the expression is proved to be $\sqrt{2}$.

(N/A) 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.
This contradiction has arisen because of our incorrect assumption that $7 \sqrt{5}$ is rational.
Therefore,we conclude that $7 \sqrt{5}$ is an irrational number.

(N/A) Let us assume,to the contrary,that $6+\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $6+\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{b} - 6$.
This simplifies to $\sqrt{2} = \frac{a - 6b}{b}$.
Since $a$ and $b$ are integers,$\frac{a - 6b}{b}$ 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 $6+\sqrt{2}$ is rational is false.
Hence,$6+\sqrt{2}$ is an irrational number.

(N/A) Assume,to the contrary,that $2+\sqrt{3}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $2+\sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{3} = \frac{a}{b} - 2$.
This simplifies to $\sqrt{3} = \frac{a - 2b}{b}$.
Since $a$ and $b$ are integers,$\frac{a - 2b}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
Therefore,our initial assumption is false,and $2+\sqrt{3}$ must be an irrational number.

(N/A) Assume,to the contrary,that $\sqrt{11}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{11} = \frac{a}{b}$.
Squaring both sides,we get $11 = \frac{a^2}{b^2}$,which implies $a^2 = 11b^2$.
This means $a^2$ is divisible by $11$. Since $11$ is a prime number,$a$ must also be divisible by $11$.
Let $a = 11k$ for some integer $k$.
Substituting this into the equation,we get $(11k)^2 = 11b^2$,which simplifies to $121k^2 = 11b^2$,or $b^2 = 11k^2$.
This implies $b^2$ is divisible by $11$,and consequently,$b$ is divisible by $11$.
Since both $a$ and $b$ are divisible by $11$,they have a common factor $11$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{11}$ is an irrational number.

(N/A) 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 $\sqrt{3} = \frac{a}{b} - 5$.
This simplifies to $\sqrt{3} = \frac{a - 5b}{b}$.
Since $a$ and $b$ are integers,$\frac{a - 5b}{b}$ 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 that $5+\sqrt{3}$ is rational is incorrect.
Hence,$5+\sqrt{3}$ is an irrational number.

(N/A) 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 $2\sqrt{5} = \frac{a}{b} - 3$.
Simplifying the right side,$2\sqrt{5} = \frac{a-3b}{b}$.
Dividing by $2$,we get $\sqrt{5} = \frac{a-3b}{2b}$.
Since $a$ and $b$ are integers,$\frac{a-3b}{2b}$ must be 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 that $3+2\sqrt{5}$ is rational is false.
Hence,$3+2\sqrt{5}$ is irrational.

(N/A) Assume,to the contrary,that $3 \sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3 \sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ 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 $3 \sqrt{5}$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{5}-\sqrt{3}$ is a rational number.
Let $\sqrt{5}-\sqrt{3} = r$,where $r$ is a rational number.
Then,$\sqrt{5} = r + \sqrt{3}$.
Squaring both sides,we get: $(\sqrt{5})^2 = (r + \sqrt{3})^2$.
$5 = r^2 + 3 + 2r\sqrt{3}$.
$5 - 3 - r^2 = 2r\sqrt{3}$.
$2 - r^2 = 2r\sqrt{3}$.
$\sqrt{3} = \frac{2 - r^2}{2r}$.
Since $r$ is a rational number,$\frac{2 - r^2}{2r}$ must also be 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 that $\sqrt{5}-\sqrt{3}$ is rational is incorrect.
Hence,$\sqrt{5}-\sqrt{3}$ is an irrational number.

(N/A) Assume,to the contrary,that $\frac{1}{\sqrt{3}}$ is a rational number.
Then,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers,$q \neq 0$,and $p, q$ are coprime.
$\frac{1}{\sqrt{3}} = \frac{p}{q}$
Rearranging the equation,we get $\sqrt{3} = \frac{q}{p}$.
Since $p$ and $q$ are integers,$\frac{q}{p}$ must be a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
Therefore,our initial assumption is false.
Hence,$\frac{1}{\sqrt{3}}$ is an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{7}-\sqrt{2}$ is a rational number.
Let $\sqrt{7}-\sqrt{2} = r$,where $r$ is a non-zero rational number.
Then,$\sqrt{7} = r + \sqrt{2}$.
Squaring both sides,we get: $(\sqrt{7})^2 = (r + \sqrt{2})^2$.
$7 = r^2 + 2 + 2r\sqrt{2}$.
$7 - r^2 - 2 = 2r\sqrt{2}$.
$5 - r^2 = 2r\sqrt{2}$.
$\sqrt{2} = \frac{5 - r^2}{2r}$.
Since $r$ is a rational number,$\frac{5 - r^2}{2r}$ must also be 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 is incorrect,and $\sqrt{7}-\sqrt{2}$ is an irrational number.

(B) Let the square root be $\sqrt{a} + \sqrt{b}$.
By squaring both sides,we get $a + b + 2\sqrt{ab} = 8 + \sqrt{63}$.
Comparing the rational and irrational parts,we have $a + b = 8$ and $2\sqrt{ab} = \sqrt{63}$.
Squaring the second equation,$4ab = 63$,so $ab = \frac{63}{4}$.
Now,consider a quadratic equation with roots $a$ and $b$: $t^2 - (a+b)t + ab = 0$,which is $t^2 - 8t + \frac{63}{4} = 0$.
Multiplying by $4$,we get $4t^2 - 32t + 63 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{32 \pm \sqrt{1024 - 1008}}{8} = \frac{32 \pm 4}{8}$.
Thus,$t_1 = \frac{36}{8} = \frac{9}{2}$ and $t_2 = \frac{28}{8} = \frac{7}{2}$.
The square root is $\sqrt{\frac{9}{2}} + \sqrt{\frac{7}{2}} = \frac{3}{\sqrt{2}} + \frac{\sqrt{7}}{\sqrt{2}} = \frac{3 + \sqrt{7}}{\sqrt{2}}$.
Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{2}$,we get $\frac{(3 + \sqrt{7})\sqrt{2}}{2} = \frac{3\sqrt{2} + \sqrt{14}}{2}$.

(B) To simplify the expression,we rationalize each term individually.
First term: $\frac{2 \sqrt{2}}{\sqrt{5}+\sqrt{3}} = \frac{2 \sqrt{2}(\sqrt{5}-\sqrt{3})}{5-3} = \frac{2(\sqrt{10}-\sqrt{6})}{2} = \sqrt{10}-\sqrt{6}$.
Second term: $\frac{3 \sqrt{3}}{\sqrt{5}+\sqrt{2}} = \frac{3 \sqrt{3}(\sqrt{5}-\sqrt{2})}{5-2} = \frac{3(\sqrt{15}-\sqrt{6})}{3} = \sqrt{15}-\sqrt{6}$.
Third term: $\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{5}(\sqrt{3}-\sqrt{2})}{3-2} = \sqrt{15}-\sqrt{10}$.
Now,substitute these back into the original expression:
$(\sqrt{10}-\sqrt{6}) - (\sqrt{15}-\sqrt{6}) + (\sqrt{15}-\sqrt{10})$
$= \sqrt{10} - \sqrt{6} - \sqrt{15} + \sqrt{6} + \sqrt{15} - \sqrt{10}$
$= 0$.

(B) number is rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Any terminating decimal or non-terminating repeating decimal is a rational number.
The given number $2.031\overline{2}$ is a non-terminating repeating decimal because the digit $2$ repeats infinitely.
Therefore,$2.031\overline{2}$ is a rational number.

(A) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ such that the product is a rational number.
Given the expression $-2 - \sqrt{5}$.
Let the rationalising factor be $x$.
We know that $(-2 - \sqrt{5}) \times (-2 + \sqrt{5}) = (-2)^2 - (\sqrt{5})^2$.
$= 4 - 5 = -1$.
Since $-1$ is a rational number,the expression $-2 + \sqrt{5}$ is the rationalising factor.
Alternatively,we can also multiply by $2 + \sqrt{5}$:
$(-2 - \sqrt{5}) \times (2 + \sqrt{5}) = -(2 + \sqrt{5}) \times (2 + \sqrt{5}) = -(4 + 5 + 4\sqrt{5}) = -(9 + 4\sqrt{5})$,which is not rational.
Therefore,the correct rationalising factor is $-2 + \sqrt{5}$.