Assignment -1 Question - None

Question

The relation R defined in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} is

Accepted Answer

Given, R = {(x, y) : y is divisible by x]
and A = {1, 2, 3, 4, 5, 6}
Reflexive: Let x $\in$ A be any arbitrary element.
We know that, x is divisible by x.
[$\because$ every real number except zero is divisible by itself]
$\Rightarrow$ (x, x) $\in$ R
Since, x $\in$ arbitrary element, therefore (x, x) $\in$ R, $\forall$ x $\in$ A. So, R is reflexive.
Symmetric: Clearly, 2, 4 $\in$ A and 4 is divisible by 2, but 2 is not divisible by 4.
$ herefore$ (2, 4) $\in$ R but (4, 2) $ otin $ R
So, R is not symmetric.
Transitive: Let x, y, z $\in$ A such that (x, y) $\in$ R and (y, z) $\in$ R.
Now, as (x, y) $\in$ R, therefore y is divisible by x.
i.e. $\frac{y}{x}$ = k₁ (say) ...(i)
where, k₁ is a natural number
and as (y, z) $\in$ R, therefore z is divisible by y.
i.e. $\frac{z}{y}$ = K₂ (say) ...(ii)
where, k₂ is a natural number.
On multiplying Eqs. (i) and (ii), we get
$\frac{y}{x} imes \frac{z}{y}$ $=k_1 k_2 \Rightarrow \frac{z}{x}=k_1 k_2$
where, k₁k₂ is a natural number.
$ herefore$ z is divisible by x.
Thus, (x, z) $\in$ R, for {x, y), (y, z) $\in$ R,
i.e. (x, y) $\in$ R, (y, z) $\in$ R $\Rightarrow$ (x, z) $\in$ R
Hence, R is transitive.

Question:

Step-by-Step Solution

Master Assignment -1 with Mathbee