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
Let \( A = \{0,1,2,\ldots,9\} \). Let \( R \) be a relation on \( A \) defined by \((x,y) \in R\) if and only if \( |x - y| \) is a multiple of \(3\). Given below are two statements:
Statement I: \( n(R) = 36 \).
Statement II: \( R \) is an equivalence relation.
In the light of the above statements, choose the correct answer from the options given below.