Illustrate Euclid's Lemma for each of the following pairs of integers: $(a)$ $42 \div 5$; $(b)$ $75 \div 4$

Solution:

$(a)$ When we divide $42$ by $5$, we get quotient $8$ and remainder $2$. So, Euclid's Lemma shows that $42 = 5 \times 8 + 2$. Note that the remainder $r =2 \lt 5$.
$(b)$ When we divide $75$ by $4$, we get remainder $18$ and remainder $3$. The Lemma shows that $75 = 4 \times 18 + 3$. Here again, note that $r = 3 \lt 4$.

Find the HCF of each of the following pairs of positive integers: $(a)$ $560, 38$; $(b)$ $4052, 420$

Solution:

Divide the bigger number by the smaller number and find the quotient. If the remainder is not $0$, then divide the divisor by the remainder and continue this process until you get a remainder $0$. The divisor of the last division is the required HCD. The processis shown below:
$(a)$ \begin{eqnarray} 560 &=& 38 \times 14 + 28\\ 38 &=& 28 \times 1 + 10\\ 28 &=& 10 \times 2 + 8\\ 10 &=& 8 \times 1 + 2\\ 8 &=& 2 \times 4 + 0\\ \end{eqnarray}
The divisor of the last step is $2$. So, HCF of $560$ and $38$ is $2$.
$(b)$ \begin{eqnarray} 4052 &=& 420 \times 9 + 272\\ 420 &=& 272 \times 1 + 148\\ 272 &=& 148 \times 1 + 124\\ 148 &=& 124 \times 1 + 24\\ 124 &=& 24 \times 5 + 4\\ 24 &=& 4 \times 6 + 0\\ \end{eqnarray}
The HCF of $4052$ and $420$ is $4$, as that is the divisor of the last step.

Show that any positive integer of the form $2m-1$, where $m$ is any integer, is odd.

Solution:

Let $a$ be any positive integer of the form $2m-1$ where $m$ is any integer. If $a$ is not odd, then it should be even. This means that $a = 2q$ for some positive integer $q$. This results in $2m-1 = 2q$. Rearranging the terms, we get $2m-2q = 1$. Factoring $2$ and rewriting, we get $2(m-q) = 1$. Now, as $m$ and $q$ both are integers, $m-q$ is also an integer. Let $m-q=b$ for some integer $b$. This gives us $2b = 1$, so that $b = \frac{1}{2}$. This contradicts the truth that $b$ is some integer. So, our assumption that $a$ is even is false.Thus, we show thatany positive integer of the form $2m-1$ is always odd.

A book seller has $384$ fiction books and $216$ non-fiction books. He wants to stack these books in such a way that each stack has the same number, the take up the least area in his shelf. What is the least number of books that can be placed in each stack?

Solution:

Here, the HCF of $384$ and $216$ gives the maximum number of books that can be stacked in order to consume least space. \begin{eqnarray} 384 &=& 216 \times 1 + 168\\ 216 &=& 168 \times 1 + 48\\ 168 &=& 48 \times 3 + 5\\ 48 &=& 24 \times 2 + 0\\ \end{eqnarray} So, stacking each type of books such that 24 books are in a row would consume least space.

Write the prime factorization of each of the following numbers: $(a)$ $600$; $(b)$ $4851$; $(c)$ $7875$

Solution:

$(a)$ From the tree diagram shown below, we get $600 = 2^3 \times 3 \times 5^2$. Though these factors can be written in any order, they will have three $2$'s, one $3$, and two $5$'s.

$(b)$ From the tree diagram, we get $4851 = 3^2 \times 7^2 \times 11$, which is unique except for the order in which these factors can be written.

$(c)$ From the tree diagram, we get, $7875 = 3^2 \times 5^2 \times 7$.

Show that the numbers of the form $3^n$ where $n$ is a natural number cannot end with the digit $2$

Solution:

If possible, let a number $3^n$ for a natural number $n$ end with $2$. This would mean that $2$ is a factor of $3^n$ as it becomes an even number. But, we know that $3^n = 3 \times 3 \times 3......$, where $3$ occurs $n$ times. So, our presumption that $3^n$ ends with $2$ for some $n$ is false. This means that any number of the form $3^n$ for a natural number $n$ cannot end with $2$.

Find the HCF and LCM of $24$ and $30$ by prime factorization method.

Solution:

Prime factorization of $24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$. Prime factorization of $30 = 2 \times 3 \times \ 5$. HCF = $2 \times 3 = 6$. LCM = $2^3 \times 3 \times 5 = 120$.

Find the HCF of $120$ and$176$ by prime factorization method. Hence find their LCM.

Solution:

Prime factorization of $120 = 2^3 \times 3 \times 5$. Prime factorization of $176 = 2^4 \times 11$. HCS (120, 176) is the product of smallest powers of the prime factors: $2^3 = 8$. Using the property discussed above, we have LCM = $\frac{120 \times 176}{8} = 2640$.

HCF and LCM of two positive integers is $9$ and $6930$ respectively. If one of the numbers is $198$, find the other number.

Solution:

Let the other number be $x$. Then, we have \begin{eqnarray} 198x &=& 9 \times 6930\\ x &=& \frac{9 \times 6930}{198}\\ x &=& 315\\ \end{eqnarray}.

Find the HCF and LCM of $18$, $21$, and $33$ by prime factorization method.

Solution:

\begin{eqnarray} 18 &=&2 \times 3^2\\ 21 &=& 3 \times 7 \\ 33 &=& 3 \times 11\\ HCF &=& 3\\ LCM &=& 2 \times 3^2 \times 7 \times 11\\ &=& 1386\\ \end{eqnarray}

Find the HCF and LCM of $12$, $45$ and $72$ by prime factorization method.

Solution:

\begin{eqnarray} 12 &=& 2^2 \times 3\\ 45 &=& 3^2 \times 5\\ 72 &=& 2^3 \times 3^2\\ HCF &=& 3\\ LCM &=& 2^3 \times 3^2 \times 5\\ &=& 360\\ \end{eqnarray}

Prove that $\sqrt{5}$ is irrational.

Solution:

Let us assume that $\sqrt{5}$ is rational. Then, we should be able to express $\sqrt{5}$ as a fraction $\frac{p}{q}$, where $p,q$ are coprime integers with $q\neq 0$. That is $\sqrt{5} = \frac{p}{q}$. Squaring both sides, we get $5 = \frac{p^2}{q^2}$. Rearranging, we get $p^2 = 5q^2$. This means that $5$ is a factor of $p^2$. Then, by the above theorem, $5$ should also be a factor of $p$. Let $p = 5k$ for some integer $k$. So, $p^2 = (5k)^2 = 25k^2$. Subtituting for $p$, we have $25k^2 = 5q^2$. Diving both sides by $5$ gives $q^2 = 5k^2$. This means that $5$ is a factor of $q$. That is, $5$ is a factor of both $p$ and $q$. This contradicts the fact the $p$ and $q$ are coprime. So, our assumption is false. Hence, $\sqrt{5}$ is an irrational number.

Prove that $2 - \sqrt{5}$ is irrational.

Solution:

Let us assume that $2 - \sqrt{5}$ is rational. Then, it can be expressed as $\frac{p}{q}$, where $p,q$ are coprime integers, and $q \neq 0$. That is, $2 - \sqrt{5} = \frac{p}{q}$. Rearranging, we get $\sqrt{5} = 2 - \frac{p}{q} = \frac{2q - p}{q}$ As $p$ and $q$ are integers, this would mean that $\sqrt{5}$ is expressed as a fraction $\frac{2q-p}{q}$, and so $\sqrt{5}$ is rational. But we know that $\sqrt{5}$ is irrational, which is a contradiction. So, our assumption is false. Hence, $2 - \sqrt{5}$ is irrational.

Prove that $3 + 4 \sqrt{2}$ is irrational.

Solution:

Let us assume that $3 + 4\sqrt{2}$ is rational. Then $3 + 4\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are coprime integers and $q \neq 0$. Rearranging this equation, we get $4\sqrt{2} = \frac{p}{q} - 3 = \frac{p - 3q}{q}$. This gives $\sqrt{2} = \frac{p-3q}{4q}$. Since $p$ and $q$ are integers, $p-3q$ and $4q$ are also integers.This means that $\sqrt{2}$ can be expressed as a fraction, which makes it a rational number. This contradicts the fact that $\sqrt{2}$ is irrational. So, our assumption is false. Thus, $3 + 4\sqrt{2}$ is irrational.

Prove that $\frac{\sqrt{7}}{2\sqrt{3}}$ is irrational.

Solution:

Let us assume that $\frac{\sqrt{7}}{2\sqrt{3}}$ is rational. Then $\frac{\sqrt{7}}{2\sqrt{3}} = \frac{p}{q}$, where $p$ and $q$ are coprime integers and $q \neq 0$. Rearranging this equation, we get $q\sqrt{7} = 2p\sqrt{3}$. Squaring both sides gives $7q^2 = 12p^2$. So, $q^2 = \left(\frac{12}{7}\right)p^2$. As $p$ and $q$ are coprime integers, by the theorem, $\frac{12}{7}$ is a factor of $q$ also. That is, $q = \left(\frac{12}{7}\right)r$ for some interger $r$. So, we end up with $q^2 = \left(\frac{12}{7}\right)^2r^2 = \left(\frac{12}{7}\right)p^2$. Thus we have $p = \left(\frac{12}{7}\right)r$. This means that $p$ and $q$ are not coprime integers. This is a contradiction. Hence our assumption is falso. So, $\frac{\sqrt{7}}{2\sqrt{3}}$ is irrational.

Find decimal expansions of $(i) \frac{3}{8}; \quad (ii) \frac{12}{5} \quad (iii) \frac{9}{20} \quad (iv) \frac{67}{40}$

Solution:

We know that place values of decimal expansions are tenths, hundredths thousandths, etc. Therefore, if we can write an equivalent fraction for each of the fractions such that the denominators are powers of ten, then we can write their decimal expansions without actual division.
$(i)$ The number $8$ can be made power of ten ($10^3$) by multiplying it by $125$ \begin{eqnarray} \frac{3}{8} &=& \frac{3}{8} \times \frac{125}{125}\\ &=& \frac{3\times 125}{8 \times 125}\\ &=& \frac{375}{1000}\\ &=& 0.375\\ \end{eqnarray}
$(ii)$ The number $5 $ multiplied by $2$ gives $10$: \begin{eqnarray} \frac{12}{5} &=& \frac{12 \times 2}{5 \times 2}\\ &=& \frac{24}{10}\\ &=& 2.4\\ \end{eqnarray}
$(iii)$ The number $20$ multiplied by $5$ gives $100$: \begin{eqnarray} \frac{9}{20} &=& \frac{9 \times 5}{20 \times 5}\\ &=& \frac{45}{100}\\ &=& 0.45\\ \end{eqnarray}
$(iv)$ The number $40$ multiplied by $25$ gives $1000$: \begin{eqnarray} \frac{67}{40} &=& \frac{67 \times 25}{40 \times 25}\\ &=& \frac{1675}{1000}\\ &=& 1.675\\ \end{eqnarray}

Express each of the following decimals as fractions: $(i) 0.325; \quad ii) 2.25; \quad (iii) 3.728$.

Solution:

$(i)$ \begin{eqnarray} 0.325 &=& \frac{325}{1000} \\ &=& \frac{25 \times 13}{25 \times 40}\\ &=& \frac{13}{40}\\ &=& \frac{13}{2^3 \times 5}\\ \end{eqnarray}
$(ii)$ \begin{eqnarray} 2.25 &=& 2\frac{25}{100}\\ &=& 2\frac{1}{4}\\ &=&2 \frac{1}{2^2}\\ \end{eqnarray}
$(iii)$ \begin{eqnarray} 3.726 &=& 3\frac{728}{1000}\\ &=& 3\frac{182}{225}\\ &=& 3 \frac{182}{3^2 \times 5^2}\\ \end{eqnarray}

Express each of the following fractions as decimals: $(i) \frac{2}{3} \quad (ii) \frac{11}{6} \quad (iii) \frac{5}{9} \quad (iv) \frac{2}{11}$

Solution:

Since the denominators of these fractions have factors other then $2$s and $5$s, we need to use actual divison process in order to get their decimal expansion.
$(i)$ We note from the actual division shown below that the remainder $2$ repeats after every step, making the expansion = $0. \overline{6}$

$(ii)$ In this problem, the remainder $2$ repeats after the second step, making the expansion $1.8\overline{3}$
$(iii)$ In this problem, $5$ is the repeating remainder. Thus we have $\frac{5}{9} = 0.\overline{5}$
$(iv)$ Here, we have remainders $9$ and $2$ repeating alternately. So, we have $\frac{11}{20} = 0.\overline{18}$

Following examples illustrate process of converting recurring decimal expansions into fractions:

Example 4:

In conclusion, we infer that the decimal expansion of a rational number is either terminating or repeating (recurring).