Solutions to Discrete Mathematics Assignment

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\title{\textbf{Solutions to Discrete Mathematics Assignment}}
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\textbf{QUESTION ONE}\\
Given, $$f(x) = \frac {x+1}{x-1}$$
a.\\ (i) Domain of $f(x)$ is $\{x\in \mathbb{R} -\{1\}\}$. This is an implication that the function
is defined for all real numbers in exception of 1.\\
(ii) For range,\\
$$f(x)=\frac{2+(x-1)}{x-1}$$
$$ =\frac{2}{x-1} + 1$$
\hspace{30mm}$f(x)\ne1$ since $\frac{2}{x-1}\ne$ for $ x\in \mathbb{R}$\\
\hspace{10mm}The range of $\frac{2}{x-1}$ is $\{\mathbb{R} -\{0\}\} $\\
Thus Range of the function is $\{\mathbb{R} -\{1\}\} $\\
b. Proof of whether f(x) is invertible and calculation of its inverse\\
$$f^{-1} = \frac {x-1}{x+1}$$
such that $$f(x)*f^{-1}=1$$\\
$$f\circ f^{-1}(x)=f(\frac {x-1}{x+1})$$\\
$$=\frac {\frac {x-1}{x+1}+ 1}{\frac {x-1}{x+1}-1}$$
$$=\frac {\frac{x-1 + (x+1)}{x+1}}{\frac{x-1 - (x+1)}{x+1}}= -x$$
Thus $$f\circ f^{-1}(x)=-x$$
Proof that $$f\circ f^{-1}(x)=f^{-1} (\frac {x+1}{x-1})$$
$$=\frac {\frac {x+1}{x-1}- 1}{\frac {x+1}{x-1}+1}$$
$$=\frac {\frac{(x+1) - (x-1)}{x-1}}{\frac{(x+1) + (x-1)}{x-1}}= \frac {1}{x}$$
Thus $$f^{-1}\circ f(x)=\frac {1}{x}$$
\textbf{QUESTION TWO}\\
Using the concept of the golden ratio,$f(n)$ is the closed form solution of $n^{th}$ fibonacci
number. But, $n^{th}$ fibonacci number is given by:\\
$$F_n = \frac{\varphi^n-\psi^n}{\sqrt{5}}$$
where the golden ratio is $$\varphi=\frac{1+\sqrt{5}}{2}\approx 1.618$$
The domain of this problem is a natural number$\ge$ 2 with the co-domain being a set of
fibonacci numbers $\ge$ 2.\\
A bijective function is both injective and surjective i.e. $f: A\to B$. This implies that $\forall b$
in the co-domain B there is exactly one element a in the domain A such that $f(a)=b$ and vice-
versa.\\
Since $f(2)=1$, is missing in the co-domain of this example and that some elements in the co-
domain are also not fibonacci numbers, the function is not bijective.\\
\textbf{QUESTION THREE}\\

Let $\mathbb{Q}$ denote the set of rational numbers and $\mathbb{Z}$ the set of integers.
Because $\mathbb{N}$ is countable $\Rightarrow \mathbb{N}\times\mathbb{N}$ is also
countable.\\
\hspace{60mm} Defining $\mathbb{Q}= \mathbb{Q}+ \cup\{O\}\cup\mathbb{Q}$\\
$$f:\mathbb{Q}^{+}\to\mathbb{N}\times\mathbb{N}$$
$$f(p,q) = (p,q)\hspace{3mm} p,q\in v^{+}, (p,q)\in\mathbb{N}\times\mathbb{N}$$\\
$\Rightarrow$ f is one to one but not onto.\\
Since$ |\mathbb{Q}^{+}|\le|\mathbb{N}\times\mathbb{N}|$, $\mathbb{Q}^{+}$ is countable.
Similarly, $\mathbb{Q}^{-}$ is countable too.\\
Thus, $$|\mathbb{Q}|=|\mathbb{N}|$$ \hspace{90mm}...equation (i)\\
Now, $g:\mathbb{Z}\to\mathbb{N}$ defined by $g(n)=2n + 1$ if $n\in\mathbb{Z}^{+}$\\
$$g(+n)=-2n \hspace{3mm} if \hspace{1mm} n\in\mathbb{Z}^{-}$$\\
Because this is one to one and onto mapping, \\
$$|\mathbb{Z}|=|\mathbb{N}|$$ \hspace{90mm}...equation (ii)\\
Combining equations (i) and (ii),
$$|\mathbb{Q}|=|\mathbb{N}|=|\mathbb{Z}|$$
$$\Rightarrow |\mathbb{Q}|=|\mathbb{Z}|$$
\textbf{QUESTION FOUR}\\
Yes. We can infer that the cardinality of $|X|=|Y|$\\
Let,\\
X and Y be two infinite set in which X$\subseteq Y$\\
Also assume that$ f: X\to$ Y is a surjective function\\
Suppose $|X|\ne|Y|$,\\
$$As \hspace{2 mm}X\subset Y, |X|<|Y|$$\\
$\Rightarrow$ for the given function $ f: X\to Y,\exists y\in Y$ such that the function has no
pre-image in X\\
$\Rightarrow$ the given function, f is not a surjective map which contradicts $|X|<|Y|$\\
$$\therefore |X|\nless|Y|$$\hspace{90mm}...equation (i)\\
Also $$X\subseteq Y \Rightarrow |X|\le|Y|$$ \hspace{90mm}...equation (ii)\\
From (i) and (ii), $$|X|=|Y|$$
\textbf{QUESTION FIVE}\\
Given, $$Y\subset X$$\hspace{90mm}...equation 1\\
The function $ f: X\to$ Y is injective i.e. for every element in X, there is a unique element in Y \\
Therefore, $$X\subset Y$$\hspace{90mm}...equation 2\\
From (i) and (ii), $$|Y|=|X|$$\hspace{90mm}...hence proved\\
\textbf{QUESTION SIX}\\
Defining $\mathbb{R}^{+}=\{a\in\mathbb{R}:a>0\}$\\
Let, $$f:\mathbb{R}\to\mathbb{R}^{+}$$\\
$$f(x)=e^x$$
If $f(a) = f(b)$, then $e^a = e^b$ $\Rightarrow a=b$\\
$$\Rightarrow f\hspace{1mm} is \hspace{1mm}one-one$$
For any positive real number $y\in\mathbb{R}^{+}$\hspace{1mm}$\exists\hspace{1mm} \
ln(y)\in \mathbb{R}$ such that $e^{\ln(y)}= y$\\