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Exercise 12.1 · Q3

Q.Let ∗* be defined on R\mathbb{R} by (a∗b)=a+b+ab−7(a*b) = a+b+ab-7. Is ∗* binary on R\mathbb{R}? If so, find 3∗(−715)3*\left(\dfrac{-7}{15}\right).

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✓ Free question

Since R\mathbb R is closed under ordinary addition, multiplication and subtraction, any formula built purely from these on real inputs stays real -- so closure is automatic here; we then just substitute the given values.

Step 1. Confirm ∗* is binary on R\mathbb R. For any a,b∈Ra,b\in\mathbb R, each of a,b,ab,7a,b,ab,7 is a real number, and sums/products/differences of real numbers are real. So a+b+ab−7∈Ra+b+ab-7\in\mathbb R for every pair -- ∗* is defined everywhere and its output always lies in R\mathbb R.

Step 2. Substitute a=3, b=−715a=3,\ b=\dfrac{-7}{15} into a∗b=a+b+ab−7a*b=a+b+ab-7.

3∗(−715)=3+(−715)+3(−715)−7.3*\left(\frac{-7}{15}\right)=3+\left(\frac{-7}{15}\right)+3\left(\frac{-7}{15}\right)-7.

Step 3. Combine the whole-number terms. 3−7=−43-7=-4.

Step 4. Combine the fractional terms. −715+3(−715)=−715+−2115=−2815\dfrac{-7}{15}+3\left(\dfrac{-7}{15}\right)=\dfrac{-7}{15}+\dfrac{-21}{15}=\dfrac{-28}{15}.

Step 5. Add the two parts. −4+−2815=−6015+−2815=−8815-4+\dfrac{-28}{15}=\dfrac{-60}{15}+\dfrac{-28}{15}=\dfrac{-88}{15}.

✓Final answer

∗* is binary on R\mathbb R; 3∗(−715)=−88153*\left(\dfrac{-7}{15}\right)=\dfrac{-88}{15}.

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