12.4 Functions
Workshop 8
Exercise 1
(a) Let \(f\) and \(g\) be the two functions from \(\mathbb{R}\) to \(\mathbb{R}\) defined by:
Calculate \(f \circ g\) and \(g \circ f\).
\(f\circ g=f(g(x))=f(x^2-1)=3x^2-3+1=3x^2-2\)
\(g\circ f=g(f(x))=g(3x+1)=(3x+1)^2-1=9x^2+6x+1-1=9x^2+6x\)
(b) In the following examples, determine two functions \(u\) and \(v\) such that \(h = u \circ v\):
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\(h_1(x) = \sqrt{3x - 1}\)
\(u=\sqrt{x-1}\) and \(v=3x\left(x\geq\frac13\right)\)
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\(h_2(x) = \sin\left(x + \frac{\pi}{2}\right)\)
\(u=\sin x\) and \(v=x+\frac{\pi}{2}\)
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\(h_3(x) = \frac{1}{x + 7}\).
\(u=\frac1x\) and \(v=x+7\)
Exercise 2
Let \(f : I \to J\) be a function. Determine whether, from properties (a)-(d), we can deduce that:
- \(f\) is injective (but not necessarily surjective);
- \(f\) is surjective (but not necessarily injective);
- \(f\) is bijective;
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We cannot say anything about \(f\).
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\(\forall y \in J, f^{-1}(\{y\}) \neq \emptyset\)
Which means \(\forall y\in J,\exists f^{-1}(\{y\})\), then it is surjective
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\(\forall y \in J, f^{-1}(\{y\})\) contains at most one element
injective
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\(\forall y \in f(I), f^{-1}(\{y\}) \neq \emptyset\)
nothing
-
\(\forall y \in f(I), f^{-1}(\{y\})\) contains at most one element.
Just one element, injective
Exercise 3
Are the following functions injective? Surjective? Bijective?
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\(f_1 : \mathbb{Z} \to \mathbb{Z}, n \mapsto 2n\),
bijective
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\(f_2 : \mathbb{Z} \to \mathbb{Z}, n \mapsto -n\),
bijective
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\(f_3 : \mathbb{R} \to \mathbb{R}, x \mapsto x^2\),
none
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\(f_4 : \mathbb{R} \to \mathbb{R}^+, x \mapsto x^2\),
surjective
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\(f_5 : \mathbb{C} \to \mathbb{C}, z \mapsto z^2\).
none
Exercise 4
Are the following functions injective, surjective, bijective?
(a) \(f : \mathbb{N} \to \mathbb{N}, n \mapsto n + 1\),
bijective
(b) \(g : \mathbb{Z} \to \mathbb{Z}, n \mapsto n + 1\),
bijective
(c) \(h : \mathbb{R}^2 \to \mathbb{R}^2, (x, y) \mapsto (x + y, x - y)\).
bijective
Exercise 5
Is the map \(f : \mathbb{R}^2 \to \mathbb{R}^2, (x, y) \mapsto (x + y, xy)\) injective? Surjective?
Not injective, but surjective
Exercise 6
Let \(f\) and \(g\) be the functions of \(\mathbb{N}\) in \(\mathbb{N}\) defined by:
Determine \(g \circ f\) and \(f \circ g\). Are the functions \(f\) and \(g\) injective? Surjective? Bijective?
\(g\circ f=g(f(x))=x\) and \(f\circ g=f(g(x))=\begin{cases}x,\text{ x is even}\\0,\text{ x is odd}\end{cases}\)
\(f\) is bijective and \(g\) is surjective
Exercise 7
Consider 4 sets \(A, B, C, D\), and applications \(f : A \to B, g : B \to C\), and \(h : C \to D\). Show that:
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\(g \circ f\) injective \(\implies f\) injective,
Let \(f(x_1)=f(x_2)=a\), then \(g(a)=g(a)\Rightarrow g(f(x_1))=g(f(x_2))\)
Since \(g\circ f\) is injective, then \(x_1=x_2\), then we have \(f\) is injective
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\(g \circ f\) surjective \(\implies g\) surjective.
Since \(g\circ f\) is surjective. Let's take \(\forall c\in C\), then \(\exists f\left(a\right)\in B:g\left(f(a)\right)=c\)
Then let \(f(a)=b\), then we have \(\forall c\in C\), then \(\exists b\in B:g\left(b\right)=c\), then \(g\) is surjective
Show that: \(\left(g\circ f\text{ and }h\circ g\text{ are bijective}\right)\;\iff\;\left(f,g\text{ and }h\text{ are bijective}\right).\)
\(\Rightarrow\))
Since \(g\circ f\) is bijective and \(h\circ g\) is bijective, then \(g\) is bijective (use conclusion of 1)
Since \(g\circ f\) and \(g\) is bijective, then \(g^{-1}\) is bijective, then \(g^{-1}\circ g\circ f=f\) is bijective
Similarly, \(h\) is bijective
\(\Leftarrow\)) Easy
Exercise 8
Let \(E\) be a set. For \(A \in P(E)\), a subset of \(E\), we denote \(\overline{A}\) as its complement. Is the function \(\phi : P(E) \to P(E), A \mapsto \overline{A}\) injective? Surjective?
Take \(\bar A=\bar B\), then \(E-A=E-B\), then we have \(A=B\) injective
\(\forall M\in P\left(E\right),\exists\bar{M}:\phi\left(\bar{M}\right)=M\) surjective
Exercise 9 (⋆)
Let \(f : \mathbb{N}^2 \to \mathbb{N}^*, (n, p) \mapsto 2^n(2p + 1)\). Show that \(f\) is a bijection. Deduce a bijection of \(\mathbb{N}^2\) onto \(\mathbb{N}\).
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injective
Take \(2^{n_1}(2p_1+1)=2^{n_2}(2p_2+1)\), then \(2^{n_1-n_2}=1+2\frac{p_2-p_1}{2p_1+1}\)
The left is even and the right is even if unless \(n_1=n_2\), then \(p_2=p_1\)
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surjective
\(\forall k\in \N^*\), then \(k=p_1^{\alpha_1}\ldots p_{n}^{\alpha_{n}}\), let \(p_1=2\), then we have \(k=2^{\alpha_1}\ldots p_{n}^{\alpha_{n}}\), we need to prove \(p_2^{\alpha_2}\ldots p_{n}^{\alpha_{n}}\) can be expressed as \(2n+1\)
If \(2n+1\) is prime, then we are done
If \(2n+1\) is not prime, then consider \(A=\{n:2n+1\text{ can not be expressed as }p_2^{\alpha_2}\ldots p_{n}^{\alpha_{n}}\}\)
Suppose \(A=\empty\), then \(n_0\) is the first element in \(A\)
Then we know \(2n+1=ab\) since \(2n+1\) is a composite number
Thus \(a,b\) can be expressed, then \(2n+1\) must can be expressed
Contradiction!
Thus \(p_2^{\alpha_2}\ldots p_{n}^{\alpha_{n}}\) can be expressed as \(2n+1\)
It is surjective
Since \(f:\mathbb{N}^2\to\mathbb{N}^{*}\) is bijective, then we want a bijection from \(\mathbb{N}^2\) to \(\mathbb{N}\)
Let's define a bijection from \(\N^*\) to \(\N\) which is \(g(x)=x-1\)
Thus \(g\circ f\), we have a bijection from \(\mathbb{N}^2\) to \(\mathbb{N}\)