97 lines
3.3 KiB
Typst
97 lines
3.3 KiB
Typst
#import "@preview/cetz:0.4.0"
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#import "@preview/cetz-venn:0.1.4"
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#image("Opgaver 18-09-25-image.png")
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#image("Opgave 2.2-27.png")
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a) ${4,6}$
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b) ${0,1,2,3,4,5,6,7,8,9,10}$
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c) ${4,5,6,7,8,10}$
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d) ${0,2,4,5,6,7,8,9,10}$
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#image("opgave 2.2-29.png")
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#cetz.canvas({
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cetz-venn.venn3(
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name: "venn1",
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ab-fill: blue.lighten(60%),
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stroke: black + 1pt
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)
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import cetz.draw: *
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content("venn1.a", [A], size: 14pt)
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content("venn1.b", [B], size: 14pt)
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content("venn1.c", [C], size: 14pt)
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content((0, -3), [$A inter (B - C)$: Only the AB region (excluding ABC)], size: 12pt)
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})
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#cetz.canvas({
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cetz-venn.venn3(
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name: "venn2",
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ab-fill: green.lighten(60%),
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ac-fill: green.lighten(60%),
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abc-fill: green.lighten(40%),
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stroke: black + 1pt
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)
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import cetz.draw: *
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content("venn2.a", [A], size: 14pt)
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content("venn2.b", [B], size: 14pt)
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content("venn2.c", [C], size: 14pt)
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content((0, -3), [$(A inter B) union (A inter C)$: All regions where A overlaps with B or C], size: 12pt)
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})
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#cetz.canvas({
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cetz-venn.venn3(
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name: "venn3",
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a-fill: orange.lighten(60%),
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ab-fill: orange.lighten(50%),
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ac-fill: orange.lighten(50%),
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abc-fill: orange.lighten(40%),
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stroke: black + 1pt
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)
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import cetz.draw: *
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content("venn3.a", [A], size: 14pt)
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content("venn3.b", [B], size: 14pt)
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content("venn3.c", [C], size: 14pt)
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content((0, -3), [$(A union B) inter (A union C) = A union (B inter C)$ (distributive law)], size: 12pt)
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})
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#image("Opgave 2.2-35.png")
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We can use De morgans laws, which states: $dash(A union B)= dash(A) inter dash(B)$. We use that on the 3 parentheses.
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$
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dash((A union B)) inter dash((B union C)) inter dash((A union C))<=>\
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(dash(A)inter dash(B)) inter (dash(B) inter dash(C)) inter (dash(A) inter dash(C)) <=> \
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"We know because of the distributive laws, that in our case, the parenthases dont matter, so:"\
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dash(A) inter dash(B) inter dash(B) inter dash(C) inter dash(A) inter dash(C) <=> \
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"We can remove dupplicates:"\
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dash(A)inter dash(B) inter dash(C)
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$
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#image("Opgave 2.2-47-prerequisits.png")
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#image("Opgave 2.2-47.png")
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Yes. If A contained an element that B did not, it would mean $A plus.circle B$ now contains elements not in $B plus.circle C$ because A has an extra element not in C, so it is in $A plus.circle C$, but since B does not contain it, it is not in $B plus.circle C$. A and B must also contain the same elements of C, since if they did not, no matter what $A plus.circle C = B plus.circle C$ would not be true.
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#image("opgave 2.3-9g.png")
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$floor(1/2+ceil(3/2))<=>\ floor(1/2+2)=floor(2.5)=2$
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#image("opgave 2.3-21.png")
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a) $f(x)=2|x|+x$ - It will always be a positive even integer minus x.
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b) $f(x)=|x|$ - All positive integers are mapped to an integer, but they are all mapped to two.
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c) $underbracket(f\: ZZ arrow ZZ^+, f(x) = cases(2x "if" x > 0, -2x + 1 "if" x <= 0))$
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d) $f(x)=x^2$
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#image("opgave 2.3-27.png")
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a) For a strictly decreasing function we know: $x>y=>f(x)>f(y) space forall x,y in A$, which means that two for two different x and y, f(x) and f(y) must be different.
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b) $f(x)=cases(x "if" x>=0,-1 "if" x<=0)$
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#image("opgave 2.3-73.png")
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#image("opgave 2.3-73a.png")
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If one of $f_(A)(x) "or" f_(B)(x)$ is 0, then $f_(A inter B)(x)$ must be 0 because of the $inter$. $f_(A)(x) dot f_(B)(x)$ must also be 0 because then you are multiplying with 0, which will always be 0. |