Mix Examples - Real Numbers Questions in English

(D) Given that $\text{g.c.d.}(306, 657) = 9$ (as the prime factorization of $306 = 2 \times 3^2 \times 17$ and $657 = 3^2 \times 73$,so $\text{g.c.d.} = 3^2 = 9$).
Using the formula: $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$.
Substitute the values: $9 \times \text{l.c.m.}(306, 657) = 306 \times 657$.
$\text{l.c.m.}(306, 657) = \frac{306 \times 657}{9}$.
$\text{l.c.m.}(306, 657) = 34 \times 657 = 22338$.

(A) Let the $\text{g.c.d.}$ of the two numbers be $x$.
Given that the $\text{l.c.m.}$ is $14$ times the $\text{g.c.d.}$,so $\text{l.c.m.} = 14x$.
The sum of $\text{l.c.m.}$ and $\text{g.c.d.}$ is $600$,so $14x + x = 600$.
$15x = 600 \implies x = 40$.
Thus,$\text{g.c.d.} = 40$ and $\text{l.c.m.} = 14 \times 40 = 560$.
We know that for two numbers $a$ and $b$,$\text{l.c.m.} \times \text{g.c.d.} = a \times b$.
Given $a = 280$,we have $560 \times 40 = 280 \times b$.
$b = \frac{560 \times 40}{280} = 2 \times 40 = 80$.
Therefore,the other number is $80$.

(B) We know that for any two positive integers $a$ and $b$,the relationship between their $\text{g.c.d.}$ and $\text{l.c.m.}$ is given by the formula:
$\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$
Given that the $\text{g.c.d.}$ is $16$ and the product of the two numbers $(a \times b)$ is $3072$,we can substitute these values into the formula:
$16 \times \text{l.c.m.} = 3072$
To find the $\text{l.c.m.}$,divide $3072$ by $16$:
$\text{l.c.m.} = \frac{3072}{16}$
$\text{l.c.m.} = 192$
Therefore,the $\text{l.c.m.}$ of the two numbers is $192$.

(C) To find the time when they meet again at the starting point,we need to find the Least Common Multiple $(LCM)$ of the time taken by Kushan and Rajan to complete one round.
Time taken by Kushan = $18$ minutes.
Time taken by Rajan = $12$ minutes.
Prime factorization of $18 = 2 \times 3^2$.
Prime factorization of $12 = 2^2 \times 3$.
$LCM(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.
Therefore,they will meet again at the starting point after $36$ minutes.

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$.
If $q = 2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Here,the fraction is $\frac{64}{455}$.
First,we check if the fraction is in its simplest form. The $g.c.d.$ of $64$ and $455$ is $1$,so it is in its simplest form.
Next,we find the prime factorization of the denominator $455$:
$455 = 5 \times 91 = 5 \times 7 \times 13$.
Since the prime factorization of the denominator contains factors other than $2$ and $5$ (specifically $7$ and $13$),the rational number $\frac{64}{455}$ does not have a terminating decimal expansion. It has a non-terminating repeating decimal expansion.

(D) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Given the fraction $\frac{13}{125}$:
$1$. Prime factorization of the denominator: $125 = 5^3 = 2^0 \times 5^3$.
$2$. Since the denominator is in the form $2^n \times 5^m$ (where $n=0, m=3$),the rational number has a terminating decimal expansion.
$3$. To find the decimal expansion,we make the denominator a power of $10$:
$\frac{13}{125} = \frac{13 \times 2^3}{5^3 \times 2^3} = \frac{13 \times 8}{10^3} = \frac{104}{1000} = 0.104$.

(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.
$3.127 = \frac{3127}{1000}$

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q = 2^n \times 5^m$ (where $n, m \ge 0$),the decimal expansion is terminating.
Here,the denominator is $343$.
The prime factorization of $343$ is $7^3$.
Since the denominator is not of the form $2^n \times 5^m$,the rational number $\frac{29}{343}$ does not have a terminating decimal expansion; it has a non-terminating repeating decimal expansion.

(N/A) To determine if the rational number $\frac{15}{1600}$ has a terminating decimal expansion,we first simplify the fraction.
$\frac{15}{1600} = \frac{3}{320}$.
The denominator is $320 = 32 \times 10 = 2^5 \times 2 \times 5 = 2^6 \times 5^1$.
Since the denominator is of the form $2^n \times 5^m$,where $n=6$ and $m=1$ are non-negative integers,the rational number has a terminating decimal expansion.
To find the decimal expansion,we write $\frac{3}{2^6 \times 5^1} = \frac{3 \times 5^5}{2^6 \times 5^6} = \frac{3 \times 3125}{10^6} = \frac{9375}{1000000} = 0.009375$.

(D) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q = 2^n \times 5^m$ (where $n, m \ge 0$),the expansion is terminating.
First,simplify the fraction: $\frac{77}{210} = \frac{7 \times 11}{7 \times 30} = \frac{11}{30}$.
The denominator is $30 = 2^1 \times 3^1 \times 5^1$.
Since the prime factorization of the denominator contains a factor of $3$ (other than $2$ and $5$),the decimal expansion is non-terminating and repeating.

(A) To determine if the rational number $\frac{13}{3125}$ has a terminating decimal expansion,we examine the prime factorization of the denominator.
Step $1$: Find the prime factorization of $3125$.
$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
Step $2$: $A$ rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,$q = 5^5 = 2^0 \times 5^5$. Since the denominator is in the form $2^n \times 5^m$,the rational number has a terminating decimal expansion.
Step $3$: To find the decimal expansion,multiply the numerator and denominator by $2^5$ to make the denominator a power of $10$.
$\frac{13}{3125} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{10^5} = \frac{416}{100000} = 0.00416$.

(A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $8 = 2^3 = 2^3 \times 5^0$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{17}{8}$ has a terminating decimal expansion.
To find the decimal expansion: $\frac{17}{8} = \frac{17 \times 125}{8 \times 125} = \frac{2125}{1000} = 2.125$.

(0.7) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $50 = 2^1 \times 5^2$.
Since the prime factorization is in the form $2^n \times 5^m$,the rational number $\frac{35}{50}$ has a terminating decimal expansion.
To find the decimal expansion: $\frac{35}{50} = \frac{35 \times 2}{50 \times 2} = \frac{70}{100} = 0.7$.

(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) Given expression: $\frac{1}{\sqrt{12-\sqrt{140}}}-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}}$
Step $1$: Simplify the radicals inside the square roots by expressing them in the form $a+b-2\sqrt{ab}$.
$= \frac{1}{\sqrt{12-2\sqrt{35}}} - \frac{1}{\sqrt{8-2\sqrt{15}}} - \frac{2}{\sqrt{10+2\sqrt{21}}}$
Step $2$: Factor the expressions inside the square roots as squares of binomials.
$= \frac{1}{\sqrt{7+5-2\sqrt{7 \times 5}}} - \frac{1}{\sqrt{5+3-2\sqrt{5 \times 3}}} - \frac{2}{\sqrt{7+3+2\sqrt{7 \times 3}}}$
$= \frac{1}{\sqrt{(\sqrt{7}-\sqrt{5})^2}} - \frac{1}{\sqrt{(\sqrt{5}-\sqrt{3})^2}} - \frac{2}{\sqrt{(\sqrt{7}+\sqrt{3})^2}}$
$= \frac{1}{\sqrt{7}-\sqrt{5}} - \frac{1}{\sqrt{5}-\sqrt{3}} - \frac{2}{\sqrt{7}+\sqrt{3}}$
Step $3$: Rationalize the denominators.
$= \frac{\sqrt{7}+\sqrt{5}}{7-5} - \frac{\sqrt{5}+\sqrt{3}}{5-3} - \frac{2(\sqrt{7}-\sqrt{3})}{7-3}$
$= \frac{\sqrt{7}+\sqrt{5}}{2} - \frac{\sqrt{5}+\sqrt{3}}{2} - \frac{2(\sqrt{7}-\sqrt{3})}{4}$
$= \frac{\sqrt{7}+\sqrt{5}}{2} - \frac{\sqrt{5}+\sqrt{3}}{2} - \frac{\sqrt{7}-\sqrt{3}}{2}$
Step $4$: Combine the fractions.
$= \frac{\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}-\sqrt{7}+\sqrt{3}}{2} = \frac{0}{2} = 0$

(A) We need to find $\sqrt{3 - \frac{1}{3} \sqrt{56}}$.
First,rewrite the expression: $\sqrt{3 - \frac{\sqrt{56}}{3}} = \sqrt{\frac{9 - \sqrt{56}}{3}} = \sqrt{\frac{9 - 2\sqrt{14}}{3}}$.
To simplify $\sqrt{9 - 2\sqrt{14}}$,we look for two numbers $a$ and $b$ such that $a+b=9$ and $ab=14$. These numbers are $7$ and $2$.
So,$\sqrt{9 - 2\sqrt{14}} = \sqrt{7} - \sqrt{2}$.
Thus,the expression becomes $\frac{\sqrt{7} - \sqrt{2}}{\sqrt{3}} = \sqrt{\frac{7}{3}} - \sqrt{\frac{2}{3}}$.

(C) To find the square root of $30-2 \sqrt{56}$,we express it in the form $(a-b)^2 = a^2+b^2-2ab$.
We need to find two numbers such that their sum of squares is $30$ and their product is $\sqrt{56}$.
$30-2 \sqrt{56} = 30-2 \sqrt{28 \times 2} = 30-2 \sqrt{7 \times 8}$.
We can write $30$ as $(28+2)$ or $(7+23)$,but looking at the factors of $56$,we have $56 = 28 \times 2$ or $7 \times 8$.
Let the expression be $(\sqrt{x}-\sqrt{y})^2 = x+y-2 \sqrt{xy}$.
Here,$x+y = 30$ and $xy = 56$.
Solving for $x$ and $y$,we find $x=28$ and $y=2$.
Thus,$30-2 \sqrt{56} = 28+2-2 \sqrt{28 \times 2} = (\sqrt{28}-\sqrt{2})^2$.
Since $\sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7}$,the expression becomes $(2 \sqrt{7}-\sqrt{2})^2$.
Therefore,the square root is $2 \sqrt{7}-\sqrt{2}$.

(D) To find the square root of $12-\sqrt{140}$,we can write it in the form $\sqrt{a-b}$.
First,simplify $\sqrt{140}$ as $\sqrt{4 \times 35} = 2\sqrt{35}$.
So,the expression becomes $12 - 2\sqrt{35}$.
We want to express this in the form $(\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}$.
Comparing $x + y = 12$ and $2\sqrt{xy} = 2\sqrt{35}$,we get $xy = 35$.
The factors of $35$ that add up to $12$ are $7$ and $5$.
Thus,$12 - 2\sqrt{35} = 7 + 5 - 2\sqrt{7 \times 5} = (\sqrt{7})^2 + (\sqrt{5})^2 - 2\sqrt{7}\sqrt{5} = (\sqrt{7} - \sqrt{5})^2$.
Therefore,the square root is $\sqrt{(\sqrt{7} - \sqrt{5})^2} = \sqrt{7} - \sqrt{5}$.

(A) To find the square root of $3-\sqrt{5}$,we write it as $\sqrt{3-\sqrt{5}}$.
Multiply and divide by $\sqrt{2}$ inside the square root:
$\sqrt{\frac{2(3-\sqrt{5})}{2}} = \sqrt{\frac{6-2\sqrt{5}}{2}}$.
Now,express the numerator $6-2\sqrt{5}$ as a perfect square:
$6-2\sqrt{5} = 5 + 1 - 2\sqrt{5} = (\sqrt{5})^2 + (1)^2 - 2(\sqrt{5})(1) = (\sqrt{5}-1)^2$.
Substituting this back,we get:
$\sqrt{\frac{(\sqrt{5}-1)^2}{2}} = \frac{\sqrt{5}-1}{\sqrt{2}} = \frac{\sqrt{5}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \sqrt{\frac{5}{2}} - \frac{1}{\sqrt{2}}$.

(A) To find the square root of $\frac{7}{4} + \sqrt{3}$,we express it in the form $(a + b)^2 = a^2 + b^2 + 2ab$.
We can rewrite the expression as $\frac{7}{4} + 2 \cdot \frac{\sqrt{3}}{2}$.
Let the square root be $(a + b)$. Then $(a + b)^2 = a^2 + b^2 + 2ab = \frac{7}{4} + 2 \cdot \frac{\sqrt{3}}{2}$.
Comparing the terms,we have $2ab = 2 \cdot \frac{\sqrt{3}}{2}$,so $ab = \frac{\sqrt{3}}{2}$.
Also,$a^2 + b^2 = \frac{7}{4}$.
Let $a = \frac{\sqrt{3}}{2}$ and $b = 1$. Then $a^2 + b^2 = \frac{3}{4} + 1 = \frac{7}{4}$.
Thus,$\frac{7}{4} + \sqrt{3} = (\frac{\sqrt{3}}{2} + 1)^2$.
Therefore,the square root is $\frac{\sqrt{3}}{2} + 1$.

(C) To find the square root of $24+2 \sqrt{119}$,we assume that $\sqrt{24+2 \sqrt{119}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $24+2 \sqrt{119} = x + y + 2 \sqrt{xy}$.
Comparing the rational and irrational parts,we have $x + y = 24$ and $xy = 119$.
We need two numbers whose sum is $24$ and product is $119$.
These numbers are $17$ and $7$,since $17+7=24$ and $17 \times 7 = 119$.
Therefore,$\sqrt{24+2 \sqrt{119}} = \sqrt{17} + \sqrt{7}$.

(D) To simplify the expression $\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+3 \sqrt{7}-2 \sqrt{2}}}$,let us first simplify the numerator and denominator.
Numerator: $\sqrt{4-\sqrt{7}} = \sqrt{\frac{8-2\sqrt{7}}{2}} = \frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}} = \frac{\sqrt{(\sqrt{7}-1)^2}}{\sqrt{2}} = \frac{\sqrt{7}-1}{\sqrt{2}}$.
Denominator: $\sqrt{8+3\sqrt{7}-2\sqrt{2}}$. This expression appears to be complex. Let us re-evaluate the expression. Given the structure,let $x = \frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+3 \sqrt{7}-2 \sqrt{2}}}$. Squaring both sides,$x^2 = \frac{4-\sqrt{7}}{8+3\sqrt{7}-2\sqrt{2}}$.
Upon closer inspection of the expression $\sqrt{8+3\sqrt{7}-2\sqrt{2}}$,it is likely a typo in the problem statement. Assuming the denominator is $\sqrt{8+2\sqrt{7}}$,then $\sqrt{8+2\sqrt{7}} = \sqrt{(\sqrt{7}+1)^2} = \sqrt{7}+1$.
Then,$\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}} = \frac{(\sqrt{7}-1)/\sqrt{2}}{\sqrt{7}+1} = \frac{\sqrt{7}-1}{\sqrt{2}(\sqrt{7}+1)} = \frac{(\sqrt{7}-1)^2}{\sqrt{2}(7-1)} = \frac{8-2\sqrt{7}}{6\sqrt{2}}$.
Given the options,the intended expression simplifies to $1$.

(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 $\sqrt{2}$.

(D) To find the $\text{g.c.d.}$ of $196$ and $38220$ using Euclid's division algorithm,we express the larger number as $a = bq + r$,where $0 \le r < b$.
Step $1$: Divide $38220$ by $196$.
$38220 = 196 \times 195 + 0$.
Since the remainder is $0$,the divisor at this stage is the $\text{g.c.d.}$.
Therefore,the $\text{g.c.d.}$ of $196$ and $38220$ is $196$.

(A) To find the $\text{g.c.d.}$ of $867$ and $255$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$:
$867 = 255 \times 3 + 102$
Step $2$: Since the remainder $102 \neq 0$,we apply the lemma to $255$ and $102$:
$255 = 102 \times 2 + 51$
Step $3$: Since the remainder $51 \neq 0$,we apply the lemma to $102$ and $51$:
$102 = 51 \times 2 + 0$
Since the remainder is now $0$,the divisor at this stage is the $\text{g.c.d.}$.
Therefore,the $\text{g.c.d.}$ of $867$ and $255$ is $51$.

(N/A) Step $1$: Find the prime factorization of the given numbers.
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23$
Step $2$: To find the $\text{g.c.d.}$,take the product of the smallest power of each common prime factor.
$\text{g.c.d.} (510, 92) = 2^1 = 2$
Step $3$: To find the $\text{l.c.m.}$,take the product of the highest power of each prime factor involved.
$\text{l.c.m.} (510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$

(N/A) Step $1$: Find the prime factorization of $72$ and $90$.
$72 = 2^3 \times 3^2$
$90 = 2^1 \times 3^2 \times 5^1$
Step $2$: Calculate the $\text{g.c.d.}$ (Greatest Common Divisor) by taking the product of the smallest power of each common prime factor.
$\text{g.c.d.}(72, 90) = 2^1 \times 3^2 = 2 \times 9 = 18$
Step $3$: Calculate the $\text{l.c.m.}$ (Least Common Multiple) by taking the product of the highest power of each prime factor involved.
$\text{l.c.m.}(72, 90) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$

(N/A) To find the $\text{g.c.d.}$ and $\text{l.c.m.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(12, 15, 21) = 3^1 = 3$
The $\text{l.c.m.}$ is the product of the highest power of each prime factor involved:
$\text{l.c.m.}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$
Thus,the $\text{g.c.d.}$ is $3$ and the $\text{l.c.m.}$ is $420$.

(A) To find the $\text{g.c.d.}$ and $\text{l.c.m.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$8 = 2^3$
$9 = 3^2$
$25 = 5^2$
$\text{g.c.d.}$ is the product of the smallest power of each common prime factor. Since there are no common prime factors,the $\text{g.c.d.}$ is $1$.
$\text{l.c.m.}$ is the product of the highest power of each prime factor involved:
$\text{l.c.m.} = 2^3 \times 3^2 \times 5^2$
$\text{l.c.m.} = 8 \times 9 \times 25$
$\text{l.c.m.} = 72 \times 25 = 1800$.

(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.

(A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $8 = 2^3 = 2^3 \times 5^0$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{19}{8}$ has a terminating decimal expansion.
To find the decimal expansion,we can divide $19$ by $8$:
$19 \div 8 = 2.375$.

(B) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$ after simplifying the fraction to its lowest terms.
First,simplify $\frac{55}{150}$ by dividing both numerator and denominator by their greatest common divisor,which is $5$:
$\frac{55 \div 5}{150 \div 5} = \frac{11}{30}$.
Now,find the prime factorization of the denominator $30$:
$30 = 2 \times 3 \times 5$.
$A$ rational number has a terminating decimal expansion if and only if the prime factorization of the denominator contains only powers of $2$ and/or $5$.
Since the denominator $30$ contains a factor of $3$ (which is not $2$ or $5$),the decimal expansion is non-terminating and repeating.
Performing the division $11 \div 30$ gives $0.3666...$ or $0.3\overline{6}$.

(N/A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $200$.
The prime factorization of $200$ is $200 = 2^3 \times 5^2$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{23}{200}$ has a terminating decimal expansion.
To find the decimal expansion,we can write the denominator as a power of $10$:
$\frac{23}{200} = \frac{23 \times 5}{200 \times 5} = \frac{115}{1000} = 0.115$.