Question

Let \(R\) be a relation on the set \(A=\{1,2,3,4,\ldots,10\}\) given by \(xRy\) if and only if \(x\) divides \(y\). Let \(m\) be the number of elements in \(R\) and \(n\) be the number of minimum elements required to be added in \(R\) to make it a symmetric relation. Then \(m+n\) is equal to

• A. \(16\)
• B. \(32\)
• C. \(36\)
• D. \(44\)

Hint:

Count the pairs \((x,y)\) with \(x\mid y\) by counting multiples of each \(x\); the pairs with \(x\ne y\) each need their reverse added.

Answer 1

The Correct Option is D

Here \(R=\{(x,y): x\mid y,\ x,y\in A\}\).
For each \(x\), the number of \(y\in A\) that are multiples of \(x\) is \(\left\lfloor \dfrac{10}{x}\right\rfloor\).
So \(m = 10+5+3+2+2+1+1+1+1+1 = 27\).
Out of these 27 pairs, the 10 pairs \((x,x)\) are already symmetric.
For the remaining \(27-10=17\) pairs \((x,y)\) with \(x<y\), we have \(x\mid y\) but \(y\nmid x\), so \((y,x)\notin R\). To make \(R\) symmetric we must add all 17 reverse pairs, so \(n=17\).
Hence \(m+n = 27+17 = 44\).
Answer: (4) \(44\)