Set of math questions focusing on rational number properties, such as associativity, identities, and inverses.

Let's solve each of these questions step by step and explain the reasoning.

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Question 6: Which of the following expressions shows that rational numbers are associative under multiplication?

Associative property of multiplication means:
> For any rational numbers $ a, b, c $,
> $ a \times (b \times c) = (a \times b) \times c $

We need to find which option shows this property — that is, changing the grouping of numbers in multiplication does not change the result.

Let’s examine each choice:

(a) $ \frac{2}{3} \times \left( \frac{-6}{7} \times \frac{3}{5} \right) = \left( \frac{2}{3} \times \frac{-6}{7} \right) \times \frac{3}{5} $

✔ This matches the associative property:
Left side: $ a \times (b \times c) $
Right side: $ (a \times b) \times c $
With $ a = \frac{2}{3}, b = \frac{-6}{7}, c = \frac{3}{5} $

So this is the correct expression for associativity.

(b) $ \frac{2}{3} \times \left( \frac{-6}{7} \times \frac{3}{5} \right) = \frac{2}{3} \times \left( \frac{3}{5} \times \frac{-6}{7} \right) $

This is showing commutativity (order of multiplication), not associativity. So ✘

(c) $ \frac{2}{3} \times \left( \frac{-6}{7} \times \frac{3}{5} \right) = \left( \frac{3}{5} \times \frac{2}{3} \right) \times \frac{-6}{7} $

This changes both order and grouping — it's a mix of commutativity and associativity, but not a clean demonstration of associativity alone. Also, the grouping is changed in a way that doesn't match the standard form. ✘

(d) $ \left( \frac{2}{3} \times \frac{-6}{7} \right) \times \frac{3}{5} = \left( \frac{-6}{7} \times \frac{2}{3} \right) \times \frac{3}{5} $

Again, this is about commutativity (swapping order of first two terms), not associativity. ✘

✔ Answer: (a)

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Question 7: Zero (0) is

(a) the identity for addition of rational numbers.
(b) the identity for subtraction of rational numbers.
(c) the identity for multiplication of rational numbers.
(d) the identity for division of rational numbers.

Identity element means: when you operate with it, the number stays unchanged.

- Addition: $ a + 0 = a $ → ✔ 0 is the additive identity.
- Subtraction: $ a - 0 = a $, but $ 0 - a = -a $, so not symmetric → no identity.
- Multiplication: $ a \times 0 = 0 $, not $ a $ → 0 is not multiplicative identity.
- Division: $ a \div 0 $ is undefined → 0 cannot be an identity.

So only (a) is correct.

✔ Answer: (a)

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Question 8: One (1) is

(a) the identity for addition of rational numbers.
(b) the identity for subtraction of rational numbers.
(c) the identity for multiplication of rational numbers.
(d) the identity for division of rational numbers.

Check:

- Addition: $ a + 1 \neq a $ → ✘
- Subtraction: $ a - 1 \neq a $ → ✘
- Multiplication: $ a \times 1 = a $ → ✔ Yes, 1 is the multiplicative identity.
- Division: $ a \div 1 = a $, but $ 1 \div a $ is not always $ a $, so not an identity.

So only (c) is correct.

✔ Answer: (c)

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Question 9: The additive inverse of $ \frac{-7}{19} $ is

Additive inverse means: a number that when added gives zero.

So we want:
$ \frac{-7}{19} + x = 0 $ → $ x = \frac{7}{19} $

So the additive inverse is $ \frac{7}{19} $

Note: Additive inverse of $ -a $ is $ a $

✔ Answer: (b) $ \frac{7}{19} $

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Question 10: Multiplicative inverse of a negative rational number is

Multiplicative inverse means: a number such that product is 1.

For example:
If $ x = -\frac{2}{3} $, then its inverse is $ -\frac{3}{2} $, because:
$ (-\frac{2}{3}) \times (-\frac{3}{2}) = 1 $

So the multiplicative inverse of a negative rational number is also negative, since negative × negative = positive.

Thus, the inverse of a negative rational number is a negative rational number.

Let’s check options:

(a) a positive rational number → ✘
(b) a negative rational number → ✔
(c) 0 → ✘ (no number has multiplicative inverse 0)
(d) 1 → ✘ (only for 1 itself)

✔ Answer: (b)

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✔ Final Answers:

6. (a)
7. (a)
8. (c)
9. (b)
10. (b)

Let me know if you'd like explanations in simpler terms or visual examples!