Respuesta :
Answer:
~ is reflexive.
~ is asymmetric.
~ is transitive.
Step-by-step explanation:
~ is reflexive:
i.e., to prove [tex]$ \forall (a, b) \in \mathbb{R}^2 $[/tex], [tex]$ (a, b) R(a, b) $[/tex].
That is, every element in the domain is related to itself.
The given relation is [tex]$\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$[/tex]
Reflexive:
[tex]$ (a, b) \sim (a, b) $[/tex] since [tex]$ a^2 + b^2 = a^2 + b^2 $[/tex]
This is true for any pair of numbers in [tex]$ \mathbb{R}^2 $[/tex]. So, [tex]$ \sim $[/tex] is reflexive.
Symmetry:
[tex]$ \sim $[/tex] is symmetry iff whenever [tex]$ (a, b) \sim (c, d) $[/tex] then [tex]$ (c, d) \sim (a, b) $[/tex].
Consider the following counter - example.
Let (a, b) = (2, 3) and (c, d) = (6, 3)
[tex]$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $[/tex]
[tex]$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $[/tex]
Hence, [tex]$ (a, b) \sim (c, d) $[/tex] since [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]
Note that [tex]$ c^2 + d^2 \nleq a^2 + b^2 $[/tex]
Hence, the given relation is not symmetric.
Transitive:
[tex]$ \sim $[/tex] is transitive iff whenever [tex]$ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $[/tex] then [tex]$ (a, b) \sim (e, f) $[/tex]
To prove transitivity let us assume [tex]$ (a, b) \sim (c, d) $[/tex] and [tex]$ (c, d) \sim (e, f) $[/tex].
We have to show [tex]$ (a, b) \sim (e, f) $[/tex]
Since [tex]$ (a, b) \sim (c, d) $[/tex] we have: [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]
Since [tex]$ (c, d) \sim (e, f) $[/tex] we have: [tex]$ c^2 + d^2 \leq e^2 + f^2 $[/tex]
Combining both the inequalities we get:
[tex]$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $[/tex]
Therefore, we get: [tex]$ a^2 + b^2 \leq e^2 + f^2 $[/tex]
Therefore, [tex]$ \sim $[/tex] is transitive.
Hence, proved.
