Section 1.4: Composition of Functions

Solution
We have
\(gof(2) = g(f(2)) = g(3) = 7\),
\(gof(3) = g(f(3)) = g(4) = 7\),
\(gof(4) = g(f(4)) = g(5) = 11\),
\(gof(5) = g(f(5)) = g(5) = 11\).

Solution
We have
\(gof(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3 \cos^2 x\).
Similarly,
\(fog(x) = f(g(x)) = f(3x^2) = \cos(3x^2)\).

Note that \(3 \cos^2 x \ne \cos(3x^2)\) for \(x \ne 0\).
Hence, \(gof \ne fog\).

Solution
Consider an arbitrary element \(y\) of \(Y\).
By the definition of \(Y\), \(y = 4x + 3\), for some \(x\) in the domain \(\mathbb{N}\).
This shows that
\[ x = \frac{(y - 3)}{4}. \]

Define \(g : Y \rightarrow \mathbb{N}\) by
\[ g(y) = \frac{(y - 3)}{4}. \]

Now,
\[ gof(x) = g(f(x)) = g(4x + 3) = \frac{(4x + 3 - 3)}{4} = \frac{4x}{4} = x \]

and
\[ fog(y) = f(g(y)) = f\left( \frac{y - 3}{4} \right) = 4 \left( \frac{y - 3}{4} \right) + 3 = y - 3 + 3 = y. \]

This shows that \(gof = I_{\mathbb{N}}\) and \(fog = I_Y\), which implies that \(f\) is invertible and \(g\) is the inverse of \(f\).

Solution

Since \(R_1\) and \(R_2\) are equivalence relations, \((a, a) \in R_1\) and \((a, a) \in R_2\), \(\forall\, a \in A\).
This implies that \((a, a) \in R_1 \cap R_2\), \(\forall\, a\), showing \(R_1 \cap R_2\) is reflexive.
Further, \((a, b) \in R_1 \Rightarrow (a, b) \in R_1\) and \((b, a) \in R_1\), and \((a, b) \in R_2 \Rightarrow (b, a) \in R_2\).
Hence, \((a, b) \in R_1 \cap R_2 \Rightarrow (b, a) \in R_1 \cap R_2\), so \(R_1 \cap R_2\) is symmetric.
Similarly, \((a, b) \in R_1 \cap R_2\) and \((b, c) \in R_1 \cap R_2 \Rightarrow (a, c) \in R_1\) and \((a, c) \in R_2\).
This shows that \((a, c) \in R_1 \cap R_2\) is transitive.
Thus, \(R_1 \cap R_2\) is an equivalence relation.

Solution
Clearly, \((x, y) R (x, y)\), \(\forall (x, y) \in A\), since \(xy = yx\).
This shows that \(R\) is reflexive.
Further, \((x, y) R (u, v) \Rightarrow y u = v x\) and hence \((u, v) R (x, y)\).
This shows that \(R\) is symmetric.
Similarly, \((x, y) R (u, v)\) and \((u, v) R (a, b) \Rightarrow y u = v x\) and \(v a = u b \Rightarrow a u = y a\) and hence \((x, y) R (a, b)\).
Thus, \(R\) is transitive.
Thus, \(R\) is an equivalence relation.

Solution
Note that the characteristic of sets \(\{1, 4, 7\}\), \(\{2, 5, 8\}\) and \(\{3, 6, 9\}\) is that difference between any two elements of these sets is a multiple of 3.
Therefore, \((x, y) \in R_1 \Rightarrow x - y\) is a multiple of 3 \(\Rightarrow \{x, y\} \subset \{1, 4, 7\}\) or \(\{x, y\} \subset \{2, 5, 8\}\) or \(\{x, y\} \subset \{3, 6, 9\} \Rightarrow (x, y) \in R_2\).
Hence, \(R_1 \subset R_2\).
Similarly, \((x, y) \in R_2 \Rightarrow x - y\) is divisible by 3 \(\Rightarrow (x, y) \in R_1\).
Thus, \(R_2 \subset R_1 \Rightarrow R_1 = R_2\).

Solution
For every \(a \in X\), \((a, a) \in R\), since \(f(a) = f(a)\), showing that \(R\) is reflexive.
Similarly, \((a, b) \in R \Rightarrow f(a) = f(b) \Rightarrow f(b) = f(a) \Rightarrow (b, a) \in R\).
Therefore, \(R\) is symmetric.
Further, \((a, b) \in R\) and \((b, c) \in R \Rightarrow f(a) = f(b)\) and \(f(b) = f(c) \Rightarrow f(a) = f(c) \Rightarrow (a, c) \in R\), which implies that \(R\) is transitive.
Hence, \(R\) is an equivalence relation.

Solution
One-one function from \(\{1, 2, 3\}\) to itself is simply a permutation of three symbols \(1, 2, 3\).
Therefore, total number of one-one maps from \(\{1, 2, 3\}\) to itself is same as total number of permutations on three symbols \(1, 2, 3\) which is \(3! = 6\).

Solution
The smallest relation \(R_1\) containing \((1, 2)\) and \((2, 3)\) which is reflexive and transitive but not symmetric is
\[ \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}. \]

Now, if we add the pair \((2, 1)\) to \(R_1\) to get \(R_2\), then the relation will be reflexive, transitive but not symmetric.
Similarly, we can obtain \(R_3\) by adding \((3, 2)\) to \(R_1\) to get the desired relation.
However, we cannot add two pairs \((2, 1), (3, 2)\) or single pair \((3, 1)\) to \(R_1\) at a time, as by doing so, we will be forced to add the remaining pair in order to maintain transitivity and in the process, the relation will become symmetric also which is not required.
Thus, the total number of desired relations is three.

Solution
The smallest equivalence relation \(R_1\) containing \((1, 2)\) and \((2, 1)\) is
\[ \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}. \]

Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one, say (2, 3) to \(R_1\), then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1).
Thus, the only equivalence relation bigger than \(R_1\) is the universal relation.
This shows that the total number of equivalence relations containing \((1, 2)\) and \((2, 1)\) is two.

Solution
Clearly \(I_{\mathbb{N}}\) is onto.
But \(I_{\mathbb{N}} + I_{\mathbb{N}}\) is not onto, as we can find an element \(3\) in the co-domain \(\mathbb{N}\) such that there does not exist any \(x\) in the domain \(\mathbb{N}\) with \((I_{\mathbb{N}} + I_{\mathbb{N}})(x) = 2x = 3\).

Solution
Since for any two distinct elements \(x_1\) and \(x_2\) in \(\left[0, \frac{\pi}{2}\right]\),
\(\sin x_1 \neq \sin x_2\) and \(\cos x_1 \neq \cos x_2\), both \(f\) and \(g\) must be one-one.
But
\[ (f + g)(0) = \sin 0 + \cos 0 = 1 \]
and
\[ (f + g)\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 \]
Therefore, \(f + g\) is not one-one.