Addition of Rational Numbers with Examples - BYJUS

Adding Rational Numbers

Rational numbers are a set of numbers that include natural numbers, whole numbers, integers, and certain fractions that follow a specific rule. We can perform basic operations on rational numbers just like we have learned with natural numbers and whole numbers. Learn some properties and math models that will help you add rational numbers easily....Read MoreRead Less

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Rational number

Integers are the set of whole numbers that is 0, 1, 2, 3… and their opposites -1, -2, -3 …. A rational number is a number that can be written as \(\frac{P}{Q}\), where P and Q are integers and Q ≠ 0.

Adding rational numbers on number line

To add rational numbers, we have to use the same rules and principles as we used for adding integers.

 

  • When plotting on the number line, the length of each arrow is the absolute value of the number it represents.
  • You count up by moving to the right when adding a positive number on a number line. You count down by moving to the left when adding a negative number on a number line.

Follow the rules below for adding the rational number

  • Addition of rational numbers with the same signs: Add the absolute values of the rational numbers. And keep the common sign ahead of the resulting value.
  • Addition of rational numbers with different signs: Subtract the lesser absolute value from the greater absolute value. After that, use the sign of the rational number with the higher absolute value.

(Here if the lesser absolute value has a “-” sign and the higher absolute value has a “+” sign, then use the “+” sign as the sign for the sum)

  • Additive inverse property: The sum of a number and its additive inverse, or opposite, is 0. If the same numbers have different signs then it sums to zero.

Algebraic form: a + ( -a ) = 0

For example:

\(\frac{4}{5}~+~\left(-\frac{2}{5}\right)~=\left|\frac{4}{5}\right|~-~\left|~-~\frac{2}{5}\right|\)

\(=~\frac{4}{5}~-~\frac{2}{5}\)

\(=~\frac{2}{5}\) 

  

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Examples on adding rational numbers in fractional form

1)  Find \(~-\frac{9}{4}~+~\frac{6}{7}\).

Solution:

Because the signs are different and \(\left|~-\frac{9}{4}\right|~>~\left|\frac{6}{7}\right|\), so

subtract \(\left|\frac{6}{7}\right|\) from \(\left|-\frac{9}{4}\right|\).

\(\left|-\frac{9}{4}\right|-\left|\frac{6}{7}\right| \)  =  \(  \frac{9}{4} \)  \(~-~\frac{6}{7}\)                    (Find the absolute values)

\(=\frac{63~-~24}{28}\)                ( Taking 28 as denominator because 7 x 4 = 28)

\(=\frac{39}{28}\)                       ( Write the difference of the numerators over the common denominator)

Because, \(\left|-\frac{9}{4}\right|~-~\left|\frac{6}{7}\right|\), use the sign of \(~-\frac{9}{4}\).

So,\(~-~\frac{9}{4}~+~\frac{6}{7}~=~-\frac{39}{28}\)

2)  Find \(~-\frac{9}{14}~+~\frac{2}{7}\).

Solution:

Because the signs are different and \(\left|~-\frac{9}{14}\right|~>~\left|\frac{2}{7}\right|\),

so we need to subtract \(\frac{2}{7}\)  from \(~-\frac{9}{14}\).

\(\left|-\frac{9}{14}\right|-\left|\frac{2}{7}\right|\)=\(\frac{9}{14}-\frac{2}{7}\)   (Find the absolute values) 

                   

                  \(=\frac{9~-~4}{14}\)      (Taking 14 as denominator)

                     

                   \(=~\frac{5}{14}\)         ( Write the difference of the numerators over the common denominator)

Because, \(\left|-\frac{9}{14}\right|~>~\left|\frac{2}{7}\right|\), use the sign of \(~-\frac{9}{14}\).

So, \(~-\frac{9}{14}~+~\frac{2}{7}~=~-\frac{5}{14}\)

3)  Find \(~-\frac{1}{2}~+~\left(-\frac{3}{2}\right)\).

Solution:

Because the signs are same we add \(\left|-\frac{3}{2}\right|\) and \(\left|-\frac{1}{2}\right|\) and use the common sign for the sum \(\left|-\frac{3}{2}\right|~+~\left|-\frac{1}{2}\right|=\frac{3}{2}~+~\frac{1}{2}\)    (Find the absolute values) 

           

\(=\frac{3~+~1}{2}\)         ( Taking 2 as denominator )

\(=\frac{4}{2}\)              ( Write the addition of the numerators over the common denominator)

                        

= 2

So,\(-\frac{1}{2}~+~\left(-\frac{3}{2}\right)=-2\)

4)  Find \(4~+~\left(-\frac{7}{2}\right)\).

Solution:

Because the signs are same and

\(\left|4\right|>\left|-\frac{7}{2}\right|\), we subtract \(\left|-\frac{7}{2}\right|\) from |4|.

\(\left|4\right|~+~\left|-\frac{7}{2}\right|~=~4-\frac{7}{2}\)      (Subtract the absolute values) 

        

         \(=\frac{8~-~7}{2}\)                      (Taking 2 as denominator )

         

         \(=\frac{1}{2}\)                             (Write the difference of the numerators over the common denominator)

                      

Because, \(\left|4\right|~>`\left|-\frac{7}{2}\right|\), use the sign of |4|.

So, \(\left|4\right|~+~\left|~-\frac{7}{2}\right|~=~0.5\)

Examples on adding rational numbers in decimal form

1)  Find – 0.76 + (- 1.6).

 

Solution:

 

Because the signs are the same, add |– 0.76| add – |1.6| |- 0.76| + |– 1.6| = 0.76 + 1.6 (Find the absolute values)

                        = 2.36         

 

Because -0.76 and -1.6 are both negative, use a negative sign in the sum.

So, – 0.76 + (- 1.6) = – 2.36

 

2)  Find – 3.3 + (- 2.7).

 

Solution:

 

Because the signs are the same, add |– 3.3| and |– 2.7|

 

|- 3.3| + |– 2.7| = 3.3 + 2.7 (Find the absolute values)

                     

                      = 6         

 

Because -3.3 and -2.7 are both negative, use a negative sign in the sum.

 

So, -3.3 + (-2.7) = – 6

 

Check:

 

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3)  Find – 32.306 + (- 24.884).

 

Solution:

 

Because the signs are the same, add |– 32.306| and |– 24.884|

 

|- 32.306| + |– 24.884| = 32.306 + 24.884 (Find the absolute values)

 

                                    = 57.19                   

 

Because -32.306 and -24.884 are both negative, use a negative sign in the sum.

 

So, – 32.306 + (- 24.884) = – 57.19 

 

4)  Find 1.65 + (- 0.9).

 

Solution:

 

Because the signs are different, subtract |– 0.9| from |1.65|

 

|1.65| – |– 0.9| = 1.65 – 0.9  (Find the absolute values)

                      = 0.75           

 

Because 1.65 is having a positive sign, use a positive sign in the sum.

 

So, 1.65 + (- 0.9) = 0.75

Examples on adding rational numbers in real life situations

1)  Positive numbers indicate gains, while negative numbers indicate losses. Jay Jays Hot Dogs recorded their income. Can you help Jay Jay figure out if the hot dog stand is profiting or losing? Amounts(in thousands) is – 1.8, – 4.76, 1.8, 0.9, 3.5 represents income over a five month period.

 

money

 

 

Solution: 

 

Find the total income = – 1.8 + (- 4.76) + 1.8 + 0.9 + 3.5 

                         

                                   = – 1.8 + 1.8 + (- 4.76) + 0.9 + 3.5 

                     

                                   = 0 + (- 4.76) + 0.9 + 3.5               (Used the additive inverse property)

                     

                                   = – 4.76 + 4.4                                 

                      

                                   = – 0.36   

                                       

This means that the company has lost $0.36 thousand or $360 over the last five months.

 

2)  Your bank account balance is -30.85. You deposit 25.50. What is your new balance?

 

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Solution:

 

New balance = -$30.85 + $25.50 (We have to write the sum from the given data)

                     

                      = – $5.35                  (subtract $25.50 from $30.85)

 

The |– 30.85| > |25.50| so use the negative sign to the result.

 

Therefore, the new bank account balance = – $5.35.

Frequently Asked Questions on Adding Rational Numbers

Absolute value of a rational number is the value of the number ignoring its sign. Absolute value of a is represented as a=a. For example, absolute value of – 3 is |– 3| = 3

The following are the properties of rational number addition:

  • A rational number is formed by adding two rational numbers (Closure property). 
  • You can add three rational numbers in any order(Associative property) 
  • Two rational numbers can be internally rearranged without affecting their addition(Commutative property) 
  • Any rational number has an additive identity of 0.