Mix Examples - Real Numbers Questions in English

(C) The greatest common divisor (g.c.d.) of two integers $a$ and $b$,denoted as $d = \text{gcd}(a, b)$,is the largest positive integer that divides both $a$ and $b$.
$1$. By definition,$d$ must divide both $a$ and $b$,so $d \mid a$ and $d \mid b$ is true.
$2$. Any common divisor $c$ of $a$ and $b$ must be less than or equal to the greatest common divisor $d$. Thus,$c \leqslant d$ is true.
$3$. By the property of the greatest common divisor,any common divisor $c$ of $a$ and $b$ must also divide their greatest common divisor $d$. Thus,$c \mid d$ is true.
$4$. The statement $c > d$ is false because if $c$ is a common divisor,it cannot be strictly greater than the greatest common divisor $d$.

(A) According to the Fundamental Theorem of Arithmetic and the properties of rational numbers,a rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime) has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^{m} 5^{n}$,where $m$ and $n$ are non-negative integers.
Therefore,the decimal form of $\frac{p}{q}$ is a terminating decimal expansion.

(B) For the fraction $\frac{12}{35}$,we first check the denominator $q = 35$.
The prime factorization of the denominator is $q = 5 \times 7$.
$A$ 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.
Since the denominator $35$ contains a prime factor $7$ (which is not $2$ or $5$),the decimal expansion of $\frac{12}{35}$ is non-terminating and recurring.
Therefore,the correct option is $B$.

(C) First,simplify the fraction $\frac{42}{35}$ by dividing both the numerator and the denominator by their greatest common divisor,which is $7$.
$\frac{42 \div 7}{35 \div 7} = \frac{6}{5}$.
Now,convert the fraction to a decimal by making the denominator a power of $10$:
$\frac{6}{5} = \frac{6 \times 2}{5 \times 2} = \frac{12}{10} = 1.2$.
Since the decimal expansion $1.2$ ends after a finite number of digits,it is a terminating decimal expansion.

(D) To find the number of digits after the decimal point,we convert the fraction to a decimal by making the denominator a power of $10$.
$\frac{47}{500} = \frac{47 \times 2}{500 \times 2} = \frac{94}{1000} = 0.094$.
In the decimal expansion $0.094$,there are $3$ digits after the decimal point.

(A) To find the number of decimal places after which the expansion of $\frac{9}{1600}$ terminates,we first express the denominator in prime factors.
$1600 = 16 \times 100 = 2^4 \times 10^2 = 2^4 \times (2 \times 5)^2 = 2^4 \times 2^2 \times 5^2 = 2^6 \times 5^2$.
Now,the fraction is $\frac{9}{2^6 \times 5^2}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^4$:
$\frac{9 \times 5^4}{2^6 \times 5^2 \times 5^4} = \frac{9 \times 625}{2^6 \times 5^6} = \frac{5625}{(2 \times 5)^6} = \frac{5625}{10^6}$.
Since the denominator is $10^6$,the decimal expansion will terminate after $6$ digits.

(C) To find the decimal expansion of $\frac{337}{125}$,we can convert the denominator into a power of $10$ by multiplying both the numerator and the denominator by $8$.
$\frac{337}{125} = \frac{337 \times 8}{125 \times 8}$
$= \frac{2696}{1000}$
$= 2.696$

(C) Two surds are called like surds if their radicands are the same after simplifying them to their simplest form.
First,simplify $\sqrt{12}$:
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.
Now,check the options:
$A) \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$
$B) \sqrt{36} = 6$ (This is a rational number,not a surd)
$C) \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$D) \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$
Since $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{48} = 4\sqrt{3}$,they share the same irrational factor $\sqrt{3}$.
Therefore,$\sqrt{12}$ and $\sqrt{48}$ are like surds.

(B) Two surds are called 'like surds' if they have the same radicand after simplification.
$A) \sqrt{8} = 2\sqrt{2}$ and $\sqrt{16} = 4$. These are not like surds.
$B) \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ and $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. Both have the same surd part $\sqrt{2}$,so they are like surds.
$C) \sqrt{72} = 6\sqrt{2}$ and $\sqrt{6}$. These are not like surds.
$D) \sqrt{24} = 2\sqrt{6}$ and $\sqrt{48} = 4\sqrt{3}$. These are not like surds.
Therefore,the correct pair is $\sqrt{18}$ and $\sqrt{50}$.

(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 $3-\sqrt{2}$.

(B) To solve $(\sqrt{3}+2)^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.
Here,$a = \sqrt{3}$ and $b = 2$.
Substituting these values into the identity:
$(\sqrt{3}+2)^{2} = (\sqrt{3})^{2} + 2(\sqrt{3})(2) + (2)^{2}$
$= 3 + 4\sqrt{3} + 4$
$= 7 + 4\sqrt{3}$.

(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 $\sqrt{7}-\sqrt{5}$.

(D) Any odd integer $n$ can be expressed in the form $2k+1$ where $k$ is an integer.
Alternatively,any odd integer $n$ can be represented as $n = 4k \pm 1$ or simply by testing values.
Let $n = 1$,then $n^{2}-1 = 1^{2}-1 = 0$ (divisible by $8$).
Let $n = 3$,then $n^{2}-1 = 3^{2}-1 = 9-1 = 8$ (divisible by $8$).
Let $n = 5$,then $n^{2}-1 = 5^{2}-1 = 25-1 = 24$ (divisible by $8$).
General proof: Since $n$ is odd,$n = 2k+1$.
$n^{2}-1 = (2k+1)^{2}-1 = 4k^{2}+4k+1-1 = 4k(k+1)$.
Since $k(k+1)$ is the product of two consecutive integers,it is always even,i.e.,$k(k+1) = 2m$.
Therefore,$n^{2}-1 = 4(2m) = 8m$.
Thus,$n^{2}-1$ is always divisible by $8$.

(A) Euclid's division lemma states that for any two positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that $a = bq + r,$ where the remainder $r$ must satisfy the condition $0 \leq r < b.$ This means the remainder $r$ is greater than or equal to $0$ and strictly less than the divisor $b.$ Therefore,the correct condition is $0 \leq r < b.$

(B) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leqslant r < b$.
In this problem,we are given $b = 7$.
Substituting $b = 7$ into the condition $0 \leqslant r < b$,we get $0 \leqslant r < 7$.
Therefore,the correct condition for the remainder $r$ is $0 \leqslant r < 7$.

(D) According to Euclid's division lemma,for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \le r < b$.
In this question,$b = 5$.
Therefore,the possible values for $r$ are $0, 1, 2, 3, 4$.
Since $r$ must be less than $5$,the value $r = 6$ is not possible.

(B) To solve this,we calculate the Least Common Multiple ($L$.$C$.$M$.) and Greatest Common Divisor ($G$.$C$.$D$.) for each pair:
$1$. $\text{l.c.m.}(8, 16, 24)$:
Prime factors: $8 = 2^3$,$16 = 2^4$,$24 = 2^3 \times 3^1$.
$L$.$C$.$M$. = $2^4 \times 3^1 = 16 \times 3 = 48$. (Matches $b$)
$2$. $\text{g.c.d.}(8, 16, 24)$:
Common factors: $8 = 2^3$,$16 = 2^4$,$24 = 2^3 \times 3^1$.
$G$.$C$.$D$. = $2^3 = 8$. (Matches $c$)
$3$. $\text{l.c.m.}(6, 12, 18)$:
Prime factors: $6 = 2 \times 3$,$12 = 2^2 \times 3$,$18 = 2 \times 3^2$.
$L$.$C$.$M$. = $2^2 \times 3^2 = 4 \times 9 = 36$. (Matches $a$)
$4$. $\text{g.c.d.}(6, 12, 18)$:
Common factors: $6 = 2 \times 3$,$12 = 2^2 \times 3$,$18 = 2 \times 3^2$.
$G$.$C$.$D$. = $2^1 \times 3^1 = 6$. (Matches $e$)
Thus,the correct matching is $(1-b), (2-c), (3-a), (4-e)$.