Give examples of two functions f : N\(\to\) Z and g : Z\(\to\) Z such that g o f is injective but g is not injective.
(Hint: Consider f(x)=x and g (x= IxI )
Define f : N \(\to\) Z as f(x) = x and g : Z \(\to\) Z as g(x) = \(\mid x \mid\) .
We first show that g is not injective.
It can be observed that:
g(-1)=I-1I=1.
g(1) =I1I=1.
∴ g(−1) = g(1), but −1 ≠ 1.
∴ g is not injective.
Now, gof: N \(\to\) Z is defined as gof (x)=g(f(x))=g(x)=IxI.
Let x, y ∈ N such that gof(x) = gof(y).
\(\Rightarrow\) IxI=IyI.
Since x and y ∈ N, both are positive.
Therefore IxI=IyI \(\Rightarrow\) x=y
Hence, gof is injective
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
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