From 4021453266166f6973b50bb38e16b5d72b0e88d6 Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Thu, 6 Mar 2025 01:36:57 -0800 Subject: [PATCH] auto-update(nvim): 2025-03-06 01:36:57 --- .../by-course/math-8/course-notes/main.typ | 2 +- documents/by-course/math-8/pset-8/main.typ | 130 +++++++++++++++++- 2 files changed, 129 insertions(+), 3 deletions(-) diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ index 2d51412..4bd5a2d 100644 --- a/documents/by-course/math-8/course-notes/main.typ +++ b/documents/by-course/math-8/course-notes/main.typ @@ -1348,7 +1348,7 @@ function. #proof[ Suppose $x in C$. Then there exists $y in B$ such that $g(y) = x$. - Additionally, there exists $z in A$ such that $f(z) = y$. Then $g(f(x)) = (g + Additionally, there exists $z in A$ such that $f(z) = y$. Then $g(f(z)) = (g compose f)(z) = x$. So if $x in C$, there is always $z in A$ such that $(g compose f)(z) = x$. Therefore $g compose f$ is onto $C$. ] diff --git a/documents/by-course/math-8/pset-8/main.typ b/documents/by-course/math-8/pset-8/main.typ index c7bb6bf..be33ec4 100644 --- a/documents/by-course/math-8/pset-8/main.typ +++ b/documents/by-course/math-8/pset-8/main.typ @@ -23,8 +23,6 @@ Problems: - 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8 -#outline() - = 4.2 == 1 @@ -123,3 +121,131 @@ We check that $h$ and $g$ agree at their overlapping domain. $h(2) = cos 2$, but $g(2) = 4$. Without further computation note that $cos(x)$ gives values only in $[-1,1]$ so $h(2) != g(2)$. $h union g$ would contain both $(2,4)$ and $(2,cos 2)$ so it's not a function. + += 4.3 + +== 1 + +=== b + +Suppose $x in ZZ$. Let $w = -x + 1000$. We note that $w in "Dom"(f)$. Then +$f(w) = -(-x + 1000) + 1000 = x$. Hence $f$ is onto $RR$. + +=== d + +Let $x in RR$. Let $w = root(3,x)$. Then $f(w) = (root(3,x))^3 = x$. +Additionally, we note $w in RR$. Hence $f$ is onto $RR$. + +=== e + +Attempt to find $x$ such that $f(x) = -2$. Note that $f$ only produces positive +real values due to the square root. Hence no such $x$ can exist. However $-2 in +RR$ so $f$ is not onto $RR$. + +=== h + +Suppose $z in RR$. Let $w = (z,0)$. Note that $w in R times R$. Then $f(w) = +f(z,0) = z$. Hence $f$ is onto $RR$. + +== 2 + +=== b + +Suppose that $f(x) = f(z)$ for $x,z in ZZ$. Then $-x + 1000 = -z + 1000$. After +manipulation, we see that $x = z$. Hence $f$ is one-to-one. + +=== d + +Suppose that $f(x) = f(z)$ for $x,z in RR$. Then $x^3 = z^3$. Then $root(3,x^3) += root(3,z^3)$ and $x = z$. So $f$ is one-to-one. + +=== e + +Consider $f(2) = sqrt(9) = 3$ and $f(-2) = sqrt(9) = 3$. 3 has both $2$ and +$-2$ in its preimage. So it's not one-to-one. + +=== h + +Consider $f(2,1) = 2 - 1 = 1$ and $f(3,2) = 3 - 2 = 1$. Therefore 1 has both +$(2,1)$ and $(3,2)$ in its preimage and it is not one-to-one. + +== 3 + +=== a + +Suppose $x in A$. Then $I_A (x) = x$. Since we said $x in A$, $x$ always has $x$ in its preimage. So $I_A$ is onto $A$. + +Now suppose that $x,z in A$ and $I_A (x) = I_A (z)$. Then $x = z$ so $I_A$ is +one-to-one. + +=== b + +Suppose $overline(x) in ZZ_5$ (here $overline(x)$ is a representative of the +equivalence class containing $x$). Note that $x in ZZ$. Then $f(x) = +overline(x)$, so $f$ is onto $ZZ_5$. Now consider that $f(3)$ and $f(8)$. Then +$3 = 8 space (mod 5)$, hence $f(3) = f(8) = overline(3)$. So $f$ is not +one-to-one. + +== 5 + +#theorem[ + If $f : A -> B$ and $g : B -> C$ is onto $C$, then $g compose f$ is onto $C$. +] + +#proof[ + Suppose $x in C$. Then there exists $y in B$ such that $g(y) = x$. + Additionally, there exists $z in A$ such that $f(z) = y$. Then $(g compose + f)(z) = g(f(z)) = g(y) = x$. So if $x in C$, there is always $z in A$ such + that $(g compose f)(z) = x$. Therefore $g compose f$ is onto $C$. +] + +== 6 + +#theorem[ + Let $f : A -> B$ and $g : B -> C$. If $g compose f$ is one-to-one, then $f$ + is one-to-one. +] + +#proof[ + Assume $f(x) = f(z)$. Then $g(f(x)) = g(f(z))$ because $g$ is a function. + Then $(g compose f)(x) = (g compose f)(z)$. Because $g compose f$ is + one-to-one, it follows that $x = z$. Therefore $f$ is one-to-one. +] + +== 7 + +#theorem[ + If $f : A -> B$ is one-to-one, then every restriction of $f$ is one-to-one. +] + +#proof[ + Let $D subset.eq A$. Then consider the restriction $lr(f)|_D$. Suppose + $lr(f|)_D (x) = lr(f|)_D (z)$, for $x$ and $z$ in $D$. Then $x$ and $z$ are + in $A$, and $lr(f|)_D (x) = lr(f|)_D (z) = f(x) = f(z)$ so $x = z$ by the + injectivity of $f$. Hence $lr(f|)_D$ is still one-to-one. +] + +== 8 + +We're proving the following: + +Let $h : A -> C$ and $g : B -> D$ be functions. + +If $A$ and $B$ are disjoint, $h$ is onto $C$, and $g$ is onto $D$, then $h +union g : A union B -> C union D$ is onto $C union D$. + +#proof[ + First note that $h union g$ is indeed a function by a result we proved + earlier in the homework. + + Suppose $x in C union D$. Then $x in C$ or $x in D$ (here we are using the + inclusive or). Then at least one of the following must be true, by the + surjectivity of $h$ and $g$: + + + We can find $z in A$ such that $h(z) = x$. + + We can find $z in B$ such that $g(z) = x$. + + In either case, if $z in A$ then $z in A union B$ and if $z in B$ then $z in + A union B$. So we found some $z in A union B$ such that $(h union g)(z) = x$. + Therefore $h union g$ is onto $C union D$. +]