Let A = $\{-2, -1, 0, 1, 2, 3\}$. Let R be a relation on A defined by xRy if and only if y = max{x,1}. Let $l$ be the number of elements in R. Let n and $m$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations respectively. Then $l + m + n$ is equal to
12
The set is $A = \{-2, -1, 0, 1, 2, 3\}$. The relation $R$ is defined by $xRy$ if and only if $y = \text{max}\{x, 1\}$.
We find the pairs $(x, y)$ for each element $x \in A$:
Therefore, the relation $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$.
The number of elements in $R$ is $l = |R| = 6$.
A relation $R$ on set $A$ is reflexive if $(x, x) \in R$ for all $x \in A$. We need the pairs $(-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$ to be in $R$.
Checking against the existing $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$:
The minimum number of elements required to make $R$ reflexive is $n = 3$. These are $(-2, -2), (-1, -1), (0, 0)$.
A relation $R$ is symmetric if $(x, y) \in R$ implies $(y, x) \in R$. We examine pairs $(x, y) \in R$ where $x \neq y$.
To make $R$ symmetric, we must add the pairs $(1, -2), (1, -1), (1, 0)$.
The minimum number of elements required to make $R$ symmetric is $m = 3$.
We have calculated:
The required sum is $l + m + n = 6 + 3 + 3 = 12$.
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.