# Relation $R$ in a set $N$ of natural numbers defined as $R=\{(x,y):y=x+5$ and $x<4\}$. Which of the following is true about $R$?

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Toolbox:
• A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
• A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
• A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
Given $N$ is a set of natural numbers and $R=\{(x,y):y=x+5 and x<4\}$:
Since $x<4$ and $y=x+5 \Rightarrow x = \{1,2,3,4\}$ and $y = \{6,7,8,9\}$
$\Rightarrow R = \{(1,6), (2,7), (3,8), (4,9)\}$
For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be reflexive, $(a,a) \in R\;$ for every $\; a\in\;A$
$\Rightarrow$ if $y=x, x = x+5 \Rightarrow 0 \neq 5$. Therefore $R$ is not reflexive.
We can verify this w/ a simple substitution:
If $x=1, y=1, y = x+5 \rightarrow 1 = 6$, which is not correct. Therefore $R$ is not reflexive.
For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be symmetrical $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
$\Rightarrow R \{x,y\}: y = x+5 \rightarrow y - x = 5$
$\Rightarrow R \{y,x\}: x = y+5 \rightarrow y - x = -5$
Therefore, since $(x,y) \in R,$ but $(y,x) \not \in R, \; R$ is not transitive
We can verify this w/ a simple substitution:
If $x=1, y=6, R \{x,y\}: y = x+5 \rightarrow y = x+5 \rightarrow 6 = 6$
However, $R \{y,x\}: x = y+5 \rightarrow 1 = 6+5 \rightarrow 1 = 11$ which is not correct. Hence, $R$ is not symmetric.
For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be transitive, $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
Let's take the first ordered pair $(1,6) \in R$. Here $x=1, y=6$.
However, since $x<4$ in $R = \{x,y\}$, there can exisit no ordered pair for $(y,z)$ where $y$ can be equal to 6.
Therefore $R$ is not transitive.