For the addition of the rational numbers, we follow the sign rule in which we consider different cases.

**Case I:** When two rational numbers are having the same sign.

Let's consider an example:

\(\left(-2\dfrac{1}{3}\right)+\left(-2\dfrac{7}{3}\right)\)

**Step 1:** Calculate their absolute values.

\(\left|-2\dfrac{1}{3}\right|=2\dfrac{1}{3}=\dfrac{7}{3}\)

\(\left|-2\dfrac{7}{3}\right|=2\dfrac{7}{3}=\dfrac{13}{3}\)

**Step 2: **Add the absolute values.

\(\dfrac{7}{3}+\dfrac{13}{3}=\dfrac{20}{3}\)

**Step 3:** Put the common sign to the resulting number.

\(\dfrac{-20}{3}\)

Thus, \(\left(-2\dfrac{1}{3}\right)+\left(-2\dfrac{7}{3}\right)=\dfrac{-20}{3}\)

**Case II:** When two rational numbers are having the opposite sign.

Let's consider an example:

\(\left(3\dfrac{2}{5}\right)+\left(-5\dfrac{3}{4}\right)\)

**Step 1:** We calculate the absolute values of both the rational numbers.

\(\left|3\dfrac{2}{5}\right|=3\dfrac{2}{5}=\dfrac{17}{5}\)

\(\left|-5\dfrac{3}{4}\right|=5\dfrac{3}{4}=\dfrac{23}{4}\)

We should have equivalent rational numbers since, denominators of both are different.

\(\dfrac{17}{5}×\dfrac{4}{4}=\dfrac{68}{20}\)

\(\dfrac{23}{4}×\dfrac{5}{5}=\dfrac{115}{20}\)

**Step 2:** Subtract the smaller absolute value from the higher absolute value.

\(\dfrac{115}{20}-\dfrac{68}{20}=\dfrac{47}{20}\)

**Step 3:** Put the sign of the higher rational number to the resulting value.

\(\dfrac{-47}{20}\)

Thus, \(\left(3\dfrac{2}{5}\right)+\left(-5\dfrac{3}{4}\right)=\dfrac{-47}{20}\)

A \(1\dfrac{5}{18}\)

B \(2\dfrac{3}{18}\)

C \(\dfrac{39}{6}\)

D \(1\dfrac{4}{18}\)

- We can consider the subtraction problem as the addition problem after adding the additive inverse of that number which is to be subtracted from the first one.

For example: \(\dfrac{9}{4}-\dfrac{2}{3}\)

We can also write it as \(\dfrac{9}{4}+\left(\dfrac{-2}{3}\right)\)

Here, \(\dfrac{-2}{3}\) is the additive inverse of \(\dfrac{2}{3}.\)

**Note:**

Additive inverse: It means adding the number with its opposite number which gives the result as zero.

For example:

\(\dfrac{2}{5}+\left(\dfrac{-2}{5}\right)=0\)

Thus, \(\dfrac{-2}{5}\) is the additive inverse of \(\dfrac{2}{5}.\)

To understand the subtraction of rational numbers, we consider an example: \(\dfrac{-23}{19}-\left(\dfrac{-12}{17}\right)\)

**Step 1:** We take the additive inverse of \(\left(\dfrac{-12}{17}\right)\), i.e. \(\dfrac{12}{17}\).

Now, we convert the subtraction into the addition by adding additive inverse.

Thus, we can write it as, \(\dfrac{-23}{19}+\left(\dfrac{12}{17}\right)\)

**Step 2:** Now, we apply the addition rule.

First we calculate the absolute values of both the numbers.

\(\left|\dfrac{-23}{19}\right|=\dfrac{23}{19}\)

\(\left|\dfrac{12}{17}\right|=\dfrac{12}{17}\)

Here, the denominators are not same. So, we convert the given numbers into equivalent rational numbers having the same denominator.

\(\dfrac{23}{19}×\dfrac{17}{17}=\dfrac{391}{323}\)

\(\dfrac{12}{17}×\dfrac{19}{19}=\dfrac{228}{323}\)

**Step 3:** Now, we apply the sign rule. According to this,

(1) If the rational numbers have the same sign, add their absolute values and put the common sign to the result.

(2) If the rational numbers have different signs, subtract their absolute values and put the sign of the number having higher absolute value to the result.

Since, both the rational numbers having different signs, therefore subtract their absolute values.

\(\dfrac{391}{323}-\dfrac{228}{323}=\dfrac{163}{323}\)

**Step 4:** Now, put the sign of the rational number having higher absolute value.

\(\dfrac{-163}{323}\)

A \(\dfrac{44}{63}\)

B \(7\dfrac{7}{32}\)

C \(6\dfrac{55}{63}\)

D \(\dfrac{431}{63}\)

- Addition of rational numbers can be performed by using a horizontal arrow on the number line.
- Let's consider an example.

Add \(\dfrac{1}{3}\) and \(\dfrac{-1}{5}\) i.e. \(\dfrac{1}{3}+\left(\dfrac{-1}{5}\right)\)

Here, we can not add them up on the number line because both the numbers have different denominators.

Thus, convert them into their equivalent rational numbers having the same denominator.

\(\dfrac{1}{3}=\dfrac{1×5}{3×5}=\dfrac{5}{15}\)

\(\dfrac{-1}{5}=\dfrac{-1×3}{5×3}=\dfrac{-3}{15}\)

Now, \(\dfrac{5}{15}+\left(\dfrac{-3}{15}\right)\)

**Step 1:** Draw a number line where each small segment represents \(\dfrac{1}{15}\).

**Step 2: ** \(\dfrac{5}{15}\) is a positive number, thus move the arrow \(5\) units forward from zero.

**Step 3: ** \(\dfrac{-3}{15}\) is a negative number, thus move the arrow \(3\) units backward from the point \(\dfrac{5}{15}\).

**Step 4: **Thus, the answer is \(\dfrac{2}{15}\).

- Subtraction of rational numbers can be done easily on the number line with the help of the horizontal arrow.
- We can done subtraction same as addition by taking the inverse of the second term.
- Let's consider an example:

\(\dfrac{11}{5}-\left(\dfrac{7}{5}\right)\\\dfrac{11}{5}+\left(\dfrac{-7}{5}\right)\)

**Step 1:** Here, the denominators of both numbers are same so, we don't need to write their equivalent rational numbers.

Now, draw a number line where each small segment represents \(\dfrac{1}{5}\).

**Step 2:** Check whether the numbers are positive or negative. Now, start the arrow from zero and reach to the first number.

\(\dfrac{11}{5}\) is a positive number, thus move the arrow \(11\) units forward to the right.

**Step 3:** Now add \(\dfrac{-7}{5}\) to \(\dfrac{11}{5}.\)

\(\dfrac{-7}{5}\) is a negative number, thus move the arrow \(7\) units backward (to the left) from the point \(\dfrac{11}{5}\).

**Step 4:** Thus, \(\dfrac{11}{5}-\left(\dfrac{7}{5}\right)=\dfrac{4}{5}\)

So, the answer is \(\dfrac{4}{5}\).

While multiplying rational numbers, we should follow the given rules:

**Rule 1:** If both the rational numbers are having the same sign then the result has a positive sign.

To understand it in a better way, we consider an example:

\(\left(\dfrac{-1}{3}\right)\;\left(\dfrac{-1}{4}\right)\)

**Step-1:** Calculate their absolute values.

\(\left|\dfrac{-1}{3}\right|=\dfrac{1}{3},\;\left|\dfrac{-1}{4}\right|=\dfrac{1}{4}\)

**Step-2:** Multiply their absolute values.

\(\dfrac{1}{3}×\dfrac{1}{4}=\dfrac{1}{12}\)

**Step-3:** Both the rational numbers have the same sign.

So, the result is positive.

Thus, \(\left(\dfrac{-1}{3}\right)×\left(\dfrac{-1}{4}\right)=\dfrac{1}{12}\)

**Rule 2:** If both the rational numbers are having different signs then the result has a negative sign.

To understand it in a better way, we consider an example:

\(\left(\dfrac{-1}{2}\right)×\left(\dfrac{3}{8}\right)\)

**Step-1:** Calculate their absolute values.

\(\left|\dfrac{-1}{2}\right|=\dfrac{1}{2}\)

\(\left|\dfrac{3}{8}\right|=\dfrac{3}{8}\)

**Step-2:** Multiply their absolute values.

\(\dfrac{1}{2}×\dfrac{3}{8}=\dfrac{3}{16}\)

**Step-3:** Both the rational numbers have different signs.

So, the result is negative.

Thus, \(\left(\dfrac{-1}{2}\right)×\left(\dfrac{3}{8}\right)=\dfrac{-3}{16}\)

**Note:** Multiplication is a repetitive addition.

Example:

\(\left(\dfrac{-3}{2}\right)×(2)=\left(\dfrac{-3}{2}\right)+\left(\dfrac{-3}{2}\right)\)

\(=\dfrac{-6}{2}\)

A \(18\dfrac{3}{4}\)

B \(\dfrac{4}{5}\)

C \(4\dfrac{1}{5}\)

D \(\dfrac{44}{180}\)

- Division is the inverse process of multiplication which means that dividing by a number is same as multiplying with its reciprocal.
- Reciprocal of \(\dfrac{2}{3}\) is \(\dfrac{3}{2}.\)
- It is also called multiplicative inverse.

To understand it in a better way, consider an example.

**Example:** Evaluate: \(-\dfrac{105}{148}\div\dfrac{7}{4}\)

**Step 1:** Calculate the absolute values.

\(\left|\dfrac{-105}{148}\right|=\dfrac{105}{148}\)

\(\left|\dfrac{7}{4}\right|=\dfrac{7}{4}\)

**Step 2:** Find the reciprocal of the absolute value of the divisor.

Reciprocal of \(\dfrac{7}{4}\) is \(\dfrac{4}{7}.\)

**Step 3:** Multiply the absolute value of dividend with the reciprocal of the absolute value of the divisor.

\(\dfrac{105}{148}×\dfrac{4}{7}=\dfrac{420}{1036}\)

\(=\dfrac{420}{1036}\)

Simplify: \(\dfrac{420\div28}{1036\div28}=\dfrac{15}{37}\)

**Step 4:** Apply the sign rule which is given below.

The sign rule is defined by \(2\) different cases.

**Case-I:** If both dividend and the divisor have the same sign then the quotient is positive.

Example:

\((-a)\div(-b)=\) Positive quotient

**Case-II:** If both dividend and the divisor have opposite signs then the quotient is negative.

Example:

\((a)\div(-b)=\) Negative quotient

Now, applying the sign rule,

\(\dfrac{-105}{148}\div\dfrac{7}{4}\)

Both dividend and the divisor are having opposite signs, so the quotient is negative.

\(\dfrac{-15}{37}\)

Thus, \(\left(\dfrac{-105}{148}\right)\div\left(\dfrac{7}{4}\right)=\dfrac{-15}{37}\)

A \(-42\dfrac{15}{19}\)

B \(27\dfrac{8}{17}\)

C \(-271\)

D \(\dfrac{-271}{19}\)

- We apply properties to simplify the expressions so that they become easier to solve.
- Here, we are going to learn the properties of addition and multiplication.

**(1) Associative property of addition:** It states that changing the grouping of the addends (terms) will not change the sum.

\((a+b)+c=a+(b+c)\)

**For example:**

\(\left(\dfrac{1}{5}+\dfrac{2}{5}\right)+\dfrac{4}{5}=\dfrac{1}{5}+\left(\dfrac{2}{5}+\dfrac{4}{5}\right)=\dfrac{7}{5}\)

**(2) Commutative property of addition:** It states that if we change the position of the numbers being added, the result is same.

\(a+b=b+a\)

**For example:**

\(\dfrac{13}{15}+\dfrac{14}{15}=\dfrac{14}{15}+\dfrac{13}{15}=\dfrac{27}{15}\)

**(3) ****Identity**** property of addition:** It states that when zero is added to a number, the result is the number itself.

\(a+0=0+a=a\)

**For example:**

\(\dfrac{2}{5}+0=0+\dfrac{2}{5}=\dfrac{2}{5}\)

**(4) Inverse property of addition:** It states that when a number is added to its opposite, the sum is zero.

It is also called the property of additive inverse.

\(a+(-a)=(-a)+a=0\)

**For example:**

\(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)=\left(\dfrac{-4}{5}\right)+\dfrac{4}{5}=0\)

**(1) Associative property of multiplication:** It states that it does not matter how we group the numbers (i.e. where we put parentheses) but on multiplying them the product remains the same.

\((a.b).c=a.(b.c)\)

**For example:**

\(\left(\dfrac{1}{3}\cdot\dfrac{1}{4}\right)\cdot\dfrac{1}{5}=\dfrac{1}{3}\cdot\left(\dfrac{1}{4}\cdot\dfrac{1}{5}\right)=\dfrac{1}{60}\)

**(2) Commutative property of multiplication:** It states that while multiplying if we change the position of the numbers, the result remains the same.

\(a.b=b.a\)

**For example:**

\(\dfrac{1}{3}\cdot\dfrac{1}{7}=\dfrac{1}{7}\cdot\dfrac{1}{3}=\dfrac{1}{21}\)

**(3) Identity property of multiplication:** It states that when a number is multiplied by \(1,\) the result is the number itself.

\(a\cdot1=1\cdot a=a\)

**For example:**

\(\dfrac{11}{12}\cdot1=1\cdot\dfrac{11}{12}=\dfrac{11}{12}\)

**(4) Inverse property of multiplication:** It states that when a number is multiplied by its reciprocal, the result will always be \(1.\)

\(b\cdot\dfrac{1}{b}=\dfrac{1}{b}\cdot b=1\)

where \(b\neq0\)

**For example:**

\(\dfrac{15}{2}\cdot\dfrac{2}{15}=\dfrac{2}{15}\cdot\dfrac{15}{2}=1\)

**Distributive property of multiplication over addition**

- The property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the product together.

\(p\cdot(q+r)=p\cdot q+p\cdot r\)

**For ****example:**

\(\dfrac{5}{2}\left(\dfrac{2}{5}+\dfrac{4}{5}\right)=\dfrac{5}{2}\left(\dfrac{2}{5}\right)+\dfrac{5}{2}\left(\dfrac{4}{5}\right)\)

\(=1+2=3\)

**Distributive property of multiplication over subtraction**

- The distributive property of multiplication over subtraction is like the distributive property of multiplication over addition.
- In this case, either we find out the difference first and then multiply or we first multiply with each number and then subtract.

\(p\cdot(q-r)=p\cdot q-p\cdot r\)

**For ****example:**

\(\dfrac{3}{7}\left(\dfrac{1}{6}-\dfrac{1}{9}\right)=\dfrac{3}{7}\left(\dfrac{1}{6}\right)-\dfrac{3}{7}\left(\dfrac{1}{9}\right)\)

\(=\dfrac{1}{14}-\dfrac{1}{21}\)

\(=\dfrac{7}{294}=\dfrac{1}{42}\)

**Note:**

Opposite of a sum is the sum of its opposites.

It means that if we have a sum and want to take its opposite.

**For ****example:**

\(\left(\dfrac{-7}{11}+\dfrac{(-2)}{11}\right)=\dfrac{-9}{11}\)

Now, opposite of \(\dfrac{-9}{11}\) is \(\dfrac{9}{11}\) ... (1)

So, the sum of the opposites of the numbers being added is,

\(\dfrac{7}{11}+\dfrac{2}{11}=\dfrac{9}{11}\) ... (2)

Here, we can see that the results of both (1) and (2) are same.

A Commutative property of addition

B Property of additive inverse

C Associative property of addition

D Property of additive identity

- In real world problems, observations and analyses are as important as calculations.
- To solve a real world problem, consider the following steps:

**Step 1:** First of all, observe the problem to pick out the given data and identify what is being asked to calculate.

**Step 2:** Identify the relevant keywords.

Here, we are mentioning some of the keywords, through which we can identify which operations are to be performed.

**(1) For addition****:**

In all, increased by, altogether, sum, combined, total, added to, both, together etc.

**(2) For subtraction:**

Take away, left, change, difference, fewer, reduce, exceed by, remain, decreased by, deduct, etc.

**(3) For multiplication:**

Twice (double), thrice (triple), times, of, product etc.

**(4) For division:**

Separated, distribute, per, into, equally, split, each, shared, divided, every, fractions

(**Example:** half, forth) etc.

**Note:**

Emphasis on opposite words that represent the positive and negative numbers.

above - below

low - high

south - north

give - take

**Step 4:** Identify what is being asked and move ahead to compute it.

- To understand these steps, consider an example given below:

**For ****example:**

Winter vacations are going on.

Jessica and Carl are making snowballs to build a fort. Together they can make \(28\) snowballs in an hour, but \(\dfrac{1}{7}\) of the total snowballs made in one hour get melted in every \(16\) minutes. If they need \(260\) snowballs to build the fort then how long will it take them to build that fort (calculate in hours)?

Observe the problem and pick out the given data.

Given:

Total snowballs made in one hour \(=28\)

Total snowballs needed to build the fort \(=260\)

Number of snowballs get melted in every \(16\) minutes \(=\dfrac{1}{7}\) of the total snowballs made in one hour

Here, "of" means multiply.

So, number of snowballs get melted in every \(16\) minutes \(=\dfrac{1}{7}×28\) \(=4\)

**Step 2:** Identify what is being asked.

Since, we are asked "how long will it take to build that fort?", therefore, we need to compute the time required to make \(260\) snowballs (in hours).

**Step 3:** Convert minutes into hours.

\(1\) minute \(=\dfrac{1}{60}\) hours

\(16\) minutes \(=\dfrac{1}{60}×16\)

\(=\dfrac{4}{15}\) hours

\(\therefore\) Number of snowballs get melted in \(\dfrac{4}{15}\) hours \(=4\)

Let the number of snowballs get melted in one hour \(=x\)

Now, solving the proportion to calculate \(x.\)

\(4×1=x×\dfrac{4}{15}\)

\(x=15\)

**Step 4:** Number of snowballs Jessica and Carl have in one hour = Snowballs made in one hour – snowballs melted in one hour

\(=28-15\)

\(=13\)

Let it takes them \(y\) hours to build that fort.

\(13\) snowballs made in \(1\) hour and \(260\) snowballs made in \(y\) hours.

\(\therefore\) Solving the proportion to calculate \(y.\)

\(y×13=260×1\)

\(y=20\) hours

Therefore, they will take \(20\) hours to build that fort.

A \(17\dfrac{3}{4}\) feet

B \(118\dfrac{1}{2}\) feet

C \(\dfrac{66}{8}\) feet

D \(71\dfrac{1}{4}\) feet