Question 1

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\frac{7}{20}, \frac{4}{15}$ and $\frac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Accepted Answer

To determine whether the decimal expansion of a rational number $\frac{p}{q}$ is terminating or repeating, factorize the denominator $q$ into primes. If the denominator (in simplest form) has only 2 and/or 5 as prime factors, the decimal is terminating; otherwise, it is repeating.

1) $\frac{7}{20}$: Denominator 20 = 2^2 * 5. Only 2 and 5 as prime factors, so decimal terminates.
Long division: 7 ÷ 20 = 0.35 (terminating)

2) $\frac{4}{15}$: Denominator 15 = 3 * 5. Since 3 is a prime factor other than 2 or 5, decimal repeats.
Long division: 4 ÷ 15 = 0.2666... (0.2\overline{6}) repeating

3) $\frac{13}{250}$: Denominator 250 = 2 * 5^3. Only 2 and 5 as prime factors, so decimal terminates.
Long division: 13 ÷ 250 = 0.052 (terminating)

Hence, $\frac{7}{20}$ and $\frac{13}{250}$ have terminating decimals, while $\frac{4}{15}$ has a repeating decimal. — Step 1: Factorize denominators.
Step 2: Check prime factors.
Step 3: Perform long division to verify.

This confirms the rule that denominators with only 2 and/or 5 as prime factors yield terminating decimals; otherwise, decimals repeat.

Question 2

Perform the long division for  $\frac{1}{13}$ . Identify the repeating block of digits. Does it show cyclic properties if you evaluate  $\frac{2}{13}$ ? Now compute  $\frac{3}{13}, \frac{4}{13}$ , etc. What do you notice?

Accepted Answer

Performing long division for $\frac{1}{13}$:

1 ÷ 13 = 0.076923076923... The repeating block is '076923' (6 digits).

For $\frac{2}{13}$: 2 ÷ 13 = 0.153846153846... Repeating block '153846'.

Similarly:
$\frac{3}{13} = 0.230769230769...$ repeating '230769'
$\frac{4}{13} = 0.307692307692...$ repeating '307692'
$\frac{5}{13} = 0.384615384615...$ repeating '384615'
$\frac{6}{13} = 0.461538461538...$ repeating '461538'

Notice that the repeating blocks are cyclic permutations of the digits '076923'. This is a property of 1/13 and its multiples, showing cyclic behavior in their decimal expansions. — Step 1: Perform long division for 1/13 to find repeating block.
Step 2: Perform for multiples 2/13, 3/13, etc.
Step 3: Observe that the repeating blocks are rotations (cyclic permutations) of the same 6-digit sequence.

This cyclic property is a special characteristic of the denominator 13.

Question 3

Classify the following numbers as rational or irrational:

(i)  $\sqrt{81}$
(ii)  $\sqrt{12}$
(iii) 0.33333 ...
(iv) 0.123451234512345 ...
(v) 1.01001000100001 ... (Notice the pattern: Is it repeating a single block?)
(vi) 23.560185612239874790120

Find the explicit fractions in case they are rational.

Accepted Answer

(i) $\sqrt{81} = 9$, which is a rational number (integer).

(ii) $\sqrt{12} = 2\sqrt{3}$, irrational because $\sqrt{3}$ is irrational.

(iii) 0.33333 ... is a repeating decimal (0.\overline{3}), rational.
Explicit fraction: $\frac{1}{3}$.

(iv) 0.123451234512345 ... is a repeating decimal with block '12345', rational.
Explicit fraction: Let x = 0.1234512345..., multiply by 10^5 = 100000:
100000x = 12345.1234512345...
Subtracting x:
100000x - x = 12345.1234512345... - 0.1234512345... = 12345
99999x = 12345
x = \frac{12345}{99999} = \frac{1371}{11111} (after simplification).

(v) 1.01001000100001 ... The pattern is not a repeating block but increasing number of zeros between 1's, so non-repeating, irrational.

(vi) 23.560185612239874790120 is a non-terminating, non-repeating decimal (given), so irrational.

Summary:
(i) Rational
(ii) Irrational
(iii) Rational, $\frac{1}{3}$
(iv) Rational, $\frac{1371}{11111}$
(v) Irrational
(vi) Irrational — Step 1: Identify if the number is a perfect square or not.
Step 2: Check decimal expansions for repeating patterns.
Step 3: Use algebraic method to find fraction for repeating decimals.
Step 4: Recognize non-repeating, non-terminating decimals as irrational.

Question 4

The number  $0.9$  (which means  $0.99999 \ldots$ ) is a rational number. Using algebra (let  $x = 0.9$ , multiply by 10, and subtract), explain why  $0.9$  is exactly equal to 1.

Accepted Answer

Let $x = 0.99999...$

Multiply both sides by 10:
$10x = 9.99999...$

Subtract the original equation from this:
$10x - x = 9.99999... - 0.99999...$
$9x = 9$

Divide both sides by 9:
$x = 1$

Since $x = 0.99999...$, it follows that $0.99999... = 1$.

Hence, the decimal 0.9 (repeating) is exactly equal to 1. — Step 1: Assign the repeating decimal to a variable.
Step 2: Multiply by power of 10 to shift decimal.
Step 3: Subtract to eliminate repeating part.
Step 4: Solve for variable to find equality.

Question 5

*5. We have seen that the repeating block of  $\frac{1}{7}$  is a cyclic number.

Try to find more numbers  $(n)$  whose reciprocals  $\left(\frac{1}{n}\right)$  produce decimals with repeating blocks that are cyclic.

Non-uniqueness of decimal representations. Just as  $1 = \frac{10}{10} = \frac{100}{100}$ , rational numbers can have two decimal forms. Any terminating decimal has an alternative with repeating 9s:  $1.000 \ldots = 0.999 \ldots$ ,  $2.47000 \ldots = 2.46999 \ldots$ .

Is it not surprising that  $0.999 \ldots = 1$ ? Many would have guessed that it is slightly less than 1.

Accepted Answer

The question asks to explore reciprocals of numbers $n$ whose decimal expansions have cyclic repeating blocks, similar to $\frac{1}{7}$.

Known cyclic numbers occur for denominators which are prime and for which 10 is a primitive root modulo $n$. Examples include:
- $\frac{1}{7} = 0.142857...$ (6-digit cyclic block)
- $\frac{1}{13} = 0.076923...$ (6-digit cyclic block)
- $\frac{1}{17} = 0.0588235294117647...$ (16-digit cyclic block)
- $\frac{1}{19} = 0.052631578947368421...$ (18-digit cyclic block)

These reciprocals produce repeating decimals where the repeating block cycles through permutations when multiplied by integers less than $n$.

Regarding non-uniqueness of decimal representations:
Any terminating decimal can be represented alternatively with repeating 9s. For example:
$1.000... = 0.999...$
$2.47000... = 2.46999...$

This shows that decimals are not unique representations of rational numbers.

It is indeed surprising that $0.999... = 1$, but as shown in Q4, they are exactly equal. — Step 1: Identify primes where 10 is a primitive root modulo $n$.
Step 2: Recognize cyclic behavior in decimal expansions of their reciprocals.
Step 3: Understand that decimal representations can have two forms, terminating or repeating 9s.
Step 4: Accept the equality $0.999... = 1$ as a mathematical fact.

Question 6

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) $\frac{3}{50}$

(ii) $\frac{2}{9}$

Accepted Answer

Solution:
(i) $\frac{3}{50} = 0.06$ (Terminating decimal)
Divide 3 by 50 using long division:
3.000 ÷ 50 = 0.06 exactly.

(ii) $\frac{2}{9} = 0.2222...$ (Non-terminating repeating decimal)
Divide 2 by 9 using long division:
2 ÷ 9 = 0.2222... with 2 repeating infinitely.

Hence, $\frac{3}{50}$ is a terminating decimal and $\frac{2}{9}$ is a non-terminating repeating decimal. — By performing long division:
(i) 3 ÷ 50 = 0.06 exactly, so terminating decimal.
(ii) 2 ÷ 9 = 0.2222... repeating 2 infinitely, so non-terminating repeating decimal.

Question 7

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

Accepted Answer

Proof by contradiction:
Assume $\sqrt{5}$ is rational, so $\sqrt{5} = \frac{p}{q}$ where $p, q$ are integers with no common factors and $q \neq 0$.
Then,
$$ \sqrt{5} = \frac{p}{q} \implies 5 = \frac{p^2}{q^2} \implies p^2 = 5q^2 $$
This means $p^2$ is divisible by 5, so $p$ is divisible by 5.
Let $p = 5k$ for some integer $k$.
Substitute back:
$$ (5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies 5k^2 = q^2 $$
Thus, $q^2$ is divisible by 5, so $q$ is divisible by 5.
Therefore, both $p$ and $q$ are divisible by 5, contradicting the assumption that $p/q$ is in lowest terms.
Hence, $\sqrt{5}$ is irrational. — Assuming $\sqrt{5}$ rational leads to both numerator and denominator divisible by 5, contradicting lowest terms assumption. Hence irrational.

Question 8

Convert the following decimal numbers in the form of $\frac{p}{q}$.

(i) 12.6

(ii) 0.0120

(iii) $3.052$

(iv) 1.235

(v) 0.23

(vi) 2.05

(vii) 2.125

(viii) 3.125

(ix) 2.1625

Accepted Answer

Solutions:
(i) 12.6 = $\frac{126}{10} = \frac{63}{5}$

(ii) 0.0120 = $\frac{120}{10000} = \frac{3}{250}$

(iii) 3.052 = $\frac{3052}{1000} = \frac{763}{250}$

(iv) 1.235 = $\frac{1235}{1000} = \frac{247}{200}$

(v) 0.23 = $\frac{23}{100}$

(vi) 2.05 = $\frac{205}{100} = \frac{41}{20}$

(vii) 2.125 = $\frac{2125}{1000} = \frac{17}{8}$

(viii) 3.125 = $\frac{3125}{1000} = \frac{25}{8}$

(ix) 2.1625 = $\frac{21625}{10000} = \frac{865}{400}$ — Convert decimals to fractions by writing over appropriate power of 10 and simplifying.

Question 9

Locate the following rational numbers on the number line.

(i) 0.532

(ii) 1.15

Accepted Answer

Solution:
(i) 0.532 lies between 0.5 and 0.6 on the number line, closer to 0.53.
(ii) 1.15 lies between 1.1 and 1.2 on the number line, closer to 1.15.
Mark these points accordingly on the number line. — Identify approximate positions between two integers and mark on number line.

Question 10

Find 6 rational numbers between 3 and 4.

Accepted Answer

Solution:
Between 3 and 4, rational numbers can be found by taking fractions with denominator greater than 1.
For example: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 are six rational numbers between 3 and 4.
Alternatively, $\frac{31}{10}, \frac{32}{10}, \frac{33}{10}, \frac{34}{10}, \frac{35}{10}, \frac{36}{10}$. — Rational numbers are dense; infinitely many exist between any two numbers. Choose convenient fractions.

Question 11

Find 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$.

Accepted Answer

Solution:
Convert to decimals: $\frac{2}{5} = 0.4$, $\frac{3}{5} = 0.6$.
Five rational numbers between 0.4 and 0.6 can be:
0.42, 0.44, 0.46, 0.48, 0.5
Or as fractions:
$\frac{21}{50}, \frac{22}{50}, \frac{23}{50}, \frac{24}{50}, \frac{25}{50}$. — Choose fractions with denominator 50 between 20/50 and 30/50.

Question 12

Find 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$.

Accepted Answer

Solution:
Convert to decimals:
$\frac{1}{6} \approx 0.1667$, $\frac{2}{5} = 0.4$.
Choose denominator 30:
$\frac{5}{30} = 0.1667$, $\frac{6}{30} = 0.2$, $\frac{7}{30} = 0.2333$, $\frac{8}{30} = 0.2667$, $\frac{9}{30} = 0.3$, $\frac{10}{30} = 0.3333$, $\frac{11}{30} = 0.3667$, $\frac{12}{30} = 0.4$.
Five rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$ are:
$\frac{6}{30}, \frac{7}{30}, \frac{8}{30}, \frac{9}{30}, \frac{10}{30}$. — Find common denominator and select fractions between given fractions.

Question 13

If $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$, find the rational number $x$.

Accepted Answer

Solution:
$$ \frac{x}{3} + \frac{x}{5} = \frac{16}{15} $$
Find common denominator 15:
$$ \frac{5x}{15} + \frac{3x}{15} = \frac{16}{15} $$
$$ \frac{8x}{15} = \frac{16}{15} $$
Multiply both sides by 15:
$$ 8x = 16 \implies x = \frac{16}{8} = 2 $$
Hence, $x = 2$. — Combine fractions, equate numerators, solve for x.

Question 14

Let $a$ and $b$ be two non-zero rational numbers such that $a + \frac{1}{b} = 0$. Without assigning any numerical values, determine whether $ab$ is positive or negative. Justify your answer.

Accepted Answer

Given:
$$ a + \frac{1}{b} = 0 \implies a = -\frac{1}{b} $$
Multiply both sides by $b$:
$$ ab = -1 $$
Since $-1$ is negative, $ab$ is negative.
Hence, $ab < 0$.
Justification: The product of $a$ and $b$ equals $-1$, which is negative, so $ab$ must be negative. — Rearranged equation shows $ab = -1$, a negative number.

Question 15

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form $\frac{p}{10^4}$, where $p$ is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by $2^4$ or $5^4$? Give reasons.

Accepted Answer

Solution:
Since the last non-zero digit is at the 4th decimal place, the number can be expressed as $\frac{p}{10^4}$ where $p$ is an integer.
Because the last digit is non-zero, $p$ is not divisible by 10.

Regarding the denominator in lowest form:
Since $10^4 = 2^4 \times 5^4$, the denominator in lowest terms can only have factors 2 and/or 5.
It is not necessary that the denominator is divisible by $2^4$ or $5^4$ individually, but the denominator will be of the form $2^m 5^n$ where $m, n \leq 4$.

If $p$ and $10^4$ share common factors 2 or 5, they cancel out, reducing powers in denominator.
Hence, denominator in lowest form may have powers of 2 and/or 5 less than or equal to 4. — Terminating decimals correspond to denominators with only 2 and 5 factors. The exact powers depend on cancellation with numerator.

The World of Numbers

The World of Numbers — Study Notes

Introduction

Natural Numbers and Whole Numbers

Integers

Practice Questions — The World of Numbers

All 8 Chapters in Ganita Manjari (English)