fix formatting

This commit is contained in:
Rushil Umaretiya 2023-10-23 16:38:36 -04:00
parent cfe3d4e9ab
commit 394589ce8b
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5 changed files with 12 additions and 20 deletions

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@ -1,7 +1,7 @@
# Fdb version 4
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@ -150,8 +150,8 @@
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"hw6.tex" 1698093499 10607 17031e54ea748adbb730bc9254f89a7c ""
(generated)
"hw6.aux"
"hw6.log"

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This is pdfTeX, Version 3.141592653-2.6-1.40.24 (TeX Live 2022/Debian) (preloaded format=pdflatex 2023.8.30) 23 OCT 2023 16:33
This is pdfTeX, Version 3.141592653-2.6-1.40.24 (TeX Live 2022/Debian) (preloaded format=pdflatex 2023.8.30) 23 OCT 2023 16:38
entering extended mode
restricted \write18 enabled.
file:line:error style messages enabled.
@ -479,17 +479,7 @@ LaTeX Font Info: Trying to load font information for U+msb on input line 134.
File: umsb.fd 2013/01/14 v3.01 AMS symbols B
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Underfull \hbox (badness 10000) in paragraph at lines 272--274
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Here is how much of TeX's memory you used:
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@ -497,9 +487,9 @@ Here is how much of TeX's memory you used:
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Output written on hw6.pdf (9 pages, 172285 bytes).
Output written on hw6.pdf (9 pages, 172288 bytes).
PDF statistics:
112 PDF objects out of 1000 (max. 8388607)
70 compressed objects within 1 object stream

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@ -249,7 +249,7 @@ from \(\AllReals\) to \(\AllReals\).
\end{enumerate}
\pagebreak
\prob{36} If \(f\) and \(f \circ g\) are one-to-one, does it follow that \(g\) is one-to-one? Justify your answer.\\
\prob{36} If \(f\) and \(f \circ g\) are one-to-one, does it follow that \(g\) is one-to-one? Justify your answer.
\begin{proof}
Yes, we will conduct a direct proof. Assuming \(g(a) = g(b)\),
\begin{align*}
@ -269,7 +269,9 @@ are functions from \(\AllReals\) to \(\AllReals\).
\end{align*}
\pagebreak
\prob{74} Suppose that \(f\) is a function from \(A\) to \(B\), where \(A\) and \(B\) are finite sets with \(|A| = |B|\). Show that \(f\) is one-to-one if and only if it is onto.\\\\
\prob{74} Suppose that \(f\) is a function from \(A\) to \(B\), where \(A\) and \(B\) are finite sets with \(|A| = |B|\). Show that \(f\) is one-to-one if and only if it is onto.
\parskip=0.5em
A function \(f: A \rightarrow B\) can map each element of \(A\) to a unique element of \(B\). Since \(|A| = |B|\), if \(f\) is one-to-one, then every element of \(A\) must be mapped to a different element of \(B\) and since \(|A| = |B|\), every element of \(B\) must be mapped to by an element of \(A\). Therefore, \(f\) is onto. In the other direction, in order for \(f\) to be onto it has to map every element of \(A\) to a distinct element in \(B\) since \(|A| = |B|\). Therefore, \(f\) is one-to-one. Since we have shown both directions, \(f\) is one-to-one if and only if it is onto.
\pagebreak