Let A=-2,-1,0,1,2,3,4. Let R be a relation on A defined by x Ry if and only if 2 x+y leqslant 2. Let l be the number of elements in R. Let m and n be…

Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$A={−2,−1,0,1,2,3,4} and $xRy$xRy iff $2x + y \leqslant 2$2x+y⩽2.

For $x = -2$x=−2: $y \leqslant 6$y⩽6, all 7 elements valid (7 pairs).

For $x = -1$x=−1: $y \leqslant 4$y⩽4, all 7 elements valid (7 pairs).

For $x = 0$x=0: $y \leqslant 2 \Rightarrow y \in \{-2,-1,0,1,2\}$y⩽2⇒y∈{−2,−1,0,1,2} (5 pairs).

For $x = 1$x=1: $y \leqslant 0 \Rightarrow y \in \{-2,-1,0\}$y⩽0⇒y∈{−2,−1,0} (3 pairs).

For $x = 2$x=2: $y \leqslant -2 \Rightarrow y \in \{-2\}$y⩽−2⇒y∈{−2} (1 pair).

For $x = 3, 4$x=3,4: no valid $y$y.

So $l = 7+7+5+3+1 = 23$l=7+7+5+3+1=23.

For reflexive: $(x,x) \in R \Rightarrow 3x \leqslant 2 \Rightarrow x \leqslant 2/3$(x,x)∈R⇒3x⩽2⇒x⩽2/3. Already present for $x = -2, -1, 0$x=−2,−1,0. Missing for $x = 1, 2, 3, 4$x=1,2,3,4, so $m = 4$m=4.

For symmetric: need $(y,x) \in R$(y,x)∈R for every $(x,y) \in R$(x,y)∈R. Missing pairs: $(3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$(3,−2),(4,−2),(2,−1),(3,−1),(4,−1),(2,0), so $n = 6$n=6.

$l + m + n = 23 + 4 + 6 = 33$l+m+n=23+4+6=33.