fixed 2.23.26

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Rushil Umaretiya 2023-10-10 23:32:10 -04:00
parent 0198c59c2d
commit cef0c8a333
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5 changed files with 14 additions and 13 deletions

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# Fdb version 4
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@ -152,8 +152,8 @@
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Output written on hw5.pdf (10 pages, 191641 bytes).
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@ -298,12 +298,13 @@ Consider these functions from the set of students in a discrete mathematics clas
\begin{enumerate}[a)]
\item Prove that a strictly increasing function from \(\AllReals\) to itself is one-to-one.
\begin{proof}
Lets take the strictly increasing function \(f(x) = x + 2)\). We can prove that this function is one-to-one by proving that for all \(x, y \in \AllReals\), \(f(x) = f(y) \implies x = y\).
\begin{align*}
f(x) &= f(y)\\
x + 2 &= y + 2\\
x &= y
\end{align*}
Since the function is strictly increasing, we can assume that \(\forall a,b \in \AllReals, a > b \implies f(a) > f(b)\). We can prove that this function is one-to-one by proving that for all \(x, y \in \AllReals\), \(f(x) = f(y) \implies x = y\). In order to use this, we'll use cases,
\begin{enumerate}
\item \(x > y\). Since \(f\) is strictly increasing, \(f(x) > f(y)\), which contradicts \(f(x) = f(y)\). Thus, \(x > y\) is impossible.
\item \(x < y\). Since \(f\) is strictly increasing, \(f(x) < f(y)\), which contradicts \(f(x) = f(y)\). Thus, \(x < y\) is impossible.
\item \(x = y\). This is trivially true.
\end{enumerate}
Since \(x > y\) and \(x < y\) are impossible, we can conclude that given \(f(x) = f(y)\), \(x = y\), and thus \(f\) is one-to-one.
\end{proof}
\item Give an example of an increasing function from \(\AllReals\) to itself that is not one-to-one.\\
The piecewise function, \(f(x) = \begin{cases}