-3 -8-1-7- 1-7-3 -8 is an example to show that

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Commutative Property of Rational Numbers

-3 /8+1/7= 1/...

Question

$\frac{- 3}{8} + \frac{1}{7} = \frac{1}{7} + \frac{- 3}{8}$ is an example to show that

addition of rational numbers is commutative.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

rational numbers are closed under addition.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

addition of rational numbers is associative.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

rational numbers are distributive under addition.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
addition of rational numbers is commutative.

For rational numbers, addition and multiplication are commutative.
Commutative law of addition: $a + b = b + a$
Commutative law of multiplication: $a \times b = b \times a$
Commutative property
1. Addition
If any two numbers are added for example $a$ and $b$ the $a + b = b + a$
So addition is the commutative property for rational numbers.
2. Subtraction
For two rational numbers $a$ and $b$ , $a - b \neq b - a$
It is not commutative for rational numbers.
3. Multiplication
For any two rational numbers $a$ and $b$ , $a * b = b * a$
Multiplication is commutative for rational numbers.
Hence, option A is correct.

Suggest Corrections

Similar questions

Question 5
$- \frac{3}{8} + \frac{1}{7} = \frac{1}{7} + (\frac{- 3}{8})$ is an example to show that
(a) addition of rational numbers is commutative.
(b) rational numbers are closed under addition.
(c) addition of rational number is associative.
(d) rational numbers are distributive under addition

$Name the multiplication property of rational numbers shown below: (i) \frac{3}{5} \times \frac{- 8}{9} = \frac{- 8}{9} \times \frac{3}{5} (i i) \frac{- 3}{4} \times (\frac{5}{7} \times \frac{- 8}{15}) = (\frac{- 3}{5} \times \frac{5}{7}) \times \frac{- 8}{15} (i i i) \frac{4}{5} \times (\frac{3}{- 8} + \frac{- 4}{7}) = \frac{4}{5} \times \frac{3}{- 8} + \frac{4}{5} \times \frac{- 4}{7} (i v) \frac{- 7}{5} \times \frac{5}{- 7} = 1 (v) \frac{8}{- 9} \times 1 = 1 \times \frac{8}{- 9} = \frac{8}{- 9} (v i) \frac{- 3}{4} \times 0 = 0$

Solve:

(i) $2 - \frac{3}{5}$

(ii) $4 + \frac{7}{8}$

(iii) $\frac{3}{5} + \frac{2}{7}$

(iv) $\frac{9}{11} - \frac{4}{15}$

(v) $\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$

(vi) $2 \frac{2}{3} + 3 \frac{1}{2}$

(vii) $8 \frac{1}{2} - 3 \frac{5}{8}$

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Join BYJU'S Learning Program

-38+17=17+-38 is an example to show that

$\frac{- 3}{8} + \frac{1}{7} = \frac{1}{7} + \frac{- 3}{8}$ is an example to show that