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Mathematics

State whether each of the following statements is true or false:

(i) The difference of two rational numbers is always a rational number.

(ii) 23−35=35−23\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{3}{5} - \dfrac{2}{3}

(iii) 1 is the additive identity for rational numbers.

(iv) Subtraction is commutative on rational numbers.

(v) There exists a rational number which is its own additive inverse.

(vi) 19−5+−311=195+311\dfrac{19}{-5} + \dfrac{-3}{11} = \dfrac{19}{5} + \dfrac{3}{11}

Rational Numbers

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Answer

(i) Rational numbers are closed under subtraction, i.e. if ab and cd\dfrac{a}{b} \text{ and } \dfrac{c}{d} are two rational numbers, then ab−cd\dfrac{a}{b} - \dfrac{c}{d} is also a rational number.

Hence, the given statement is true.

(ii) Solving L.H.S.,

23−35=2×5−3×315=10−915=115\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{2 \times 5 - 3 \times 3}{15} = \dfrac{10 - 9}{15} = \dfrac{1}{15}

Solving R.H.S.,

35−23=3×3−2×515=9−1015=−115\dfrac{3}{5} - \dfrac{2}{3} = \dfrac{3 \times 3 - 2 \times 5}{15} = \dfrac{9 - 10}{15} = \dfrac{-1}{15}

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.

(iii) For every rational number ab\dfrac{a}{b}, we have ab+0=ab=0+ab\dfrac{a}{b} + 0 = \dfrac{a}{b} = 0 + \dfrac{a}{b}, so 0 is the additive identity and 1 is the multiplicative identity.

Hence, the given statement is false.

(iv) For any two rational numbers ab and cd\dfrac{a}{b} \text{ and } \dfrac{c}{d}, we have ab−cd≠cd−ab\dfrac{a}{b} - \dfrac{c}{d} ≠ \dfrac{c}{d} - \dfrac{a}{b}. For example, 23−35=115\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{1}{15} but 35−23=−115\dfrac{3}{5} - \dfrac{2}{3} = \dfrac{-1}{15}.

Hence, the given statement is false.

(v) The additive inverse of the rational number 0 is 0 itself, as 0 + 0 = 0.

Hence, the given statement is true.

(vi) Solving L.H.S.,

⇒19−5+−311⇒−195+−311⇒(−19)×11+(−3)×555=−209−1555=−22455\Rightarrow \dfrac{19}{-5} + \dfrac{-3}{11} \\[1em] \Rightarrow \dfrac{-19}{5} + \dfrac{-3}{11} \\[1em] \Rightarrow \dfrac{(-19) \times 11 + (-3) \times 5}{55} = \dfrac{-209 - 15}{55} = \dfrac{-224}{55}

Solving R.H.S.,

⇒195+311⇒19×11+3×555=209+1555=22455\Rightarrow \dfrac{19}{5} + \dfrac{3}{11} \\[1em] \Rightarrow \dfrac{19 \times 11 + 3 \times 5}{55} = \dfrac{209 + 15}{55} = \dfrac{224}{55}

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.

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