625 lines
13 KiB
Text
625 lines
13 KiB
Text
#import "@youwen/zen:0.1.0": *
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#import "@preview/mitex:0.2.5": *
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#show: zen.with(
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title: "Homework 3",
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author: "Youwen Wu",
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)
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#set heading(numbering: none)
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#set par(first-line-indent: 0pt, spacing: 1em)
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#let nonzero = $ZZ_(!=0)$
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Problems:
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1.5: $hash$ 3cdefgh, 4de, 6de, 7ab, 9, 10, 11, 12a
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1.6: $hash$ 1bd, 4abcd, 6abefik
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2.1: $hash$ 4, 5, 6abcd, 8, 11ab, 14ab, 15abcd
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= 1.5
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*3c.*
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If $x^2$ is not divisible by 4, then $x$ is odd.
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#proof[
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Suppose $x$ is not odd. We seek to show that $x^2$ is divisible by $4$.
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Then $x$ is even and $exists k in ZZ_(!=0), x = 2k$. Thus
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$x^2 = 4k^2$, which implies $4 | x^2$. So $x$ is not odd implies $x^2$ is
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divisible by 4, and therefore the contrapositive (which is our original
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statement) is also true.
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]
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*3d.*
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If $x y$ is even, then either $x$ or $y$ is even.
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#proof[
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Suppose that $not (x "or" y" is even")$. In other words, $x$ and $y$ are both
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odd. We seek to show that $x y$ is odd.
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Then $exists j,k in ZZ_(n!=0), x = 2j + 1, y = 2k + 1$. So $x y = 4 j k + 2j +
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2k + 1 = 2 (2 j k + j + k) + 1$ where $2 j k + j + k$ is an integer so $x y$
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is odd. This is the contrapositive, so the original statement is also true.
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]
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*3e.*
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If $x + y$ is even, then $x$ and $y$ have the same parity.
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#proof[
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Suppose $x$ and $y$ had different parities. We seek to show that $x + y$ is
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odd.
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Without loss of generality, let us inspect the case where $x$ is odd. Then
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$y$ must be even and
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$
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exists j,k in ZZ_(!=0), x = 2j + 1, y = 2k \
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x + y = 2j + 1 + 2k = 2(j + k) + 1 = 2n + 1, n in ZZ_(n!=0)
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$
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Repeat the same reasoning for when $x$ is even and $y$ is odd. We've shown
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that $x$ and $y$ having different parities implies that $x + y$ is odd.
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Therefore the contrapositive is also true.
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]
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*3f.*
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If $x y$ is odd, then both $x$ and $y$ are odd.
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#proof[
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Suppose that both $x$ and $y$ are odd was not true, in other words $x$ or $y$
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are even (here or is the logical $or$). We seek to show that this implies $x y
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$ is even. Then we have two cases: either $x$ or
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$y$, but not both, are even, and the case where $x$ and $y$ are both even.
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We look at the first case, without loss of generality, assume $x$ is even and
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$y$ is odd. Then
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$
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exists j,k in ZZ_(!= 0), x = 2j, y = 2k + 1 \
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x y = 2j(2k + 1) = 4 j k + 2j = 2(2 j k + j) = 2n, n in ZZ_(!=0)
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$
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So $x$ and $y$ having different parities implies $x y$ is even. The argument holds identically when instead $y$ is even and $x$ is odd. Now we turn our attention to the case when $x$ and $y$ are both even. Then
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$
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exists j,k in ZZ_(!=0), x = 2j, y = 2k \
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x y = 2j dot 2k = 4 j k = 2 (2 j k) = 2n, n in ZZ_(!=0)
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$
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So this also implies that $x y$ is even. Therefore $x$ or $y$ being even indeed
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implies $x y$ is also even, and the contrapositive is also true.
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]
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*3g.*
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If 8 does not divide $x^2 - 1$, then $x$ is even.
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#proof[
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Suppose that $x$ is odd. We seek to show that 8 does in fact divide $x^2 - 1$.
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$
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exists k in nonzero, x = 2k + 1 \
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x^2 - 1 = 4k^2 + 4k = 4(k^2 + k)
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$
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Now we need to see if it's divisible by 8. First consider the unique case $k = 0$, where $x^2 - 1 = 0$. Clearly $8 | 0$ so 8 does divide $x^2-1$.
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Now we consider all other values. Notice that for all other possible values
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of $k$, $k^2 + k$ is greater than 1. We see this by noting that $k^2 + k$ is
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a quadratic with its absolute minima at $k = 1/2$, therefore we can check the
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two non-zero integers closest to this value. For $k = 1$, $k^2 + k = 2$.
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Since we already checked $k=-1$, let's check $k=-2$, which gives $k^2 + k =
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2$. For all values of $k$ greater than 1 or less than $-2$, $k^2 + k$ must be
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greater than 2 (because it's a quadratic).
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Therefore $4(k^2 + k)$ can be written as $8n$ for some integer $n$, so 8
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indeed divides $x^2 - 1$ for all possible $k$, and the contrapositive is also
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true.
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]
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*3h.*
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If $x$ does not divide $y z$, then $x$ does not divide $z$.
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#proof[
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Assume $x$ does divide $z$. We seek to show that $x$ does divide $y z$. If $x
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| z$ then $exists k in nonzero, k x = z$. So $y z$ can be written as $y k x$.
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But this shows $exists j in nonzero, j x = y z$, namely $j = y k$, which
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means $x | y z$, and the contrapositive is also true.
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]
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*4d.*
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If $(x+1)(x-1) < 0$, then $x < 1$.
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#proof[
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Suppose that $x >= 1$. We seek to show that $(x+1)(x-1) >= 0$. Expanding out
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factors,
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$
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(x+1)(x-1) = x^2 - 1>= 0
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$
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This quadratic is zero at exactly $x = 1$ and positive for all $x > 1$. So
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it's true for all possible values of $x$. Therefore our original statement is
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also true.
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]
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*4e.*
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If $x(x-4) > -3$, then $x < 1$ or $x > 3$.
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#proof[
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Suppose that $x >= 1$ and $x <= 3$. We seek to show $x(x-4) <= -3$. Expanding
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out factors,
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$
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x(x-4) = x^2 - 4x > -3
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$
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This quadratic has its stationary point at $x = 2$. Let's check its value at $x
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= 1$ and $x = 3$.
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At $x = 1$, $x(x-4) = -3$. So for all values greater than 1 until $x = 2$, $x^2 - 4x$ is less than $-3$. Our inequality is satisfied.
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At $x = 3$, $x(x-4) = -3$ again. So for all values less than 3 until $x = 2$,
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$x^2 - 4x$ is less than $-3$. Our inequality is satisfied for both $x >=1$
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and $x <=3$, so $x^2 -4x > -3$ is true for all possible $x$ and the
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contrapositive is also true.
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]
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*6d.*
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If $a - b$ is odd, then $a + b$ is odd.
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#proof[
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Suppose, seeking a contradiction, that if $a - b$ is even, then $a + b$ is odd.
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Then $exists k in nonzero, a - b = 2k$. Which means we can write $a + b = 2k
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+ 2b$. But we can factor this as $2(k + b)$ and so $a + b = 2n, n in
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nonzero$, implying it is even. However we assumed that $a + b$ should be odd,
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a contradiction. Therefore $a - b$ must be odd.
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]
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*6e.*
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If $a < b$ and $a b < 3$, then $a = 1$.
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#proof[
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Suppose, seeking a contradiction, that $a >= b$ and $a b >= 3$ implies $a =
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1$. Consider the specific cases $a = b$, $a b = 3$. Then $a = 3/a$, and $a =
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plus.minus sqrt(3)$. However we assumed $a = 1$ is implied, a contradiction.
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Therefore we must have $a < b$ and $a b < 3$.
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]
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*7a.*
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$a c$ divides $b$ and $b$ divides $b + 3$ if and only if $a = 2$ and $b = 3$.
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We first show the result left to right, namely, $a c | b c => a | b$.
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$
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exists k in nonzero, k a c = b c \
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k a = b
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$
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which is the definition of $a | b$.
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Now we show the right to left direction, namely, $a | b => a c | b c$.
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$
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exists k in nonzero, k a = b \
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k a dot c = b dot c
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$
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So $a c | b c$ by the definition of divisibility. Therefore the biconditional
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is true, as we have shown both directions.
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*7b.*
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The right to left direction is very easy in this case. By directly plugging in $a = 2$, $b = 3$, we see that $2 + 1 | 3$ and $3 | 3 + 3$. To show the left to right case,
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*9.*
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#proof[
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Suppose that instead $n/(n+1) <= n/(n+2)$. We can be assured the following
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operations do not flip the inequality as $n$ cannot be negative.
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$
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n(n+2) <= n(n+1) \
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n + 2 <= n + 1 \
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2 <= 1
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$
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So $n/(n+1) > n/(n+2)$.
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]
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*10.*
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#proof[
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Suppose that $sqrt(5)$ was rational. That is, $sqrt(5) = p/q$ for nonzero
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integers $p$ and $q$. Additionally, assume that $p/q$ is in its most reduced
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form, that is, $p$ and $q$ share no common factors besides 1. Then
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$
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p^2 = 5q \
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$
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implies that
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$ 5 | p^2 $
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We need to show that $5 | p^2 => 5 | p$.
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By the fundamental theorem of arithmetic, $p^2$ has 5 as one of its unique
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prime factors. If 5 was not a factor of $p$, then $p^2 = p dot p$ would not
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have 5 in its factors either. So 5 is a factor of $p$ and thus $5 | p$. Note
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that 5 appears at least twice amongst the prime factors of $p^2$. Then $5q$
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should also have at least two 5s in its prime factorization. Then $q$ has at
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least one 5 in its prime factorization. However we assumed that $p$ and $q$
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share no common factors besides 1, so this is a contradiction. Therefore
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$sqrt(5)$ is not rational.
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]
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*11.*
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#proof[
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We say that two numbers $x$ and $y$ are within $1/2$ unit from one another if
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$|x - y| < 1/2$. Consider the distance between $z$, and $y$, if it is within
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$1/2$, then we are done. Otherwise $z - y >= 1/2$.
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We know that
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$ (1 - z) + x + (y-x) + (z-y) = 1 $
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because this is the length of all the line segments partitioned by $x,y,z$,
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which is the interval 1. If $(z - y) >= 1/2$, then everything else must be
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less than $1/2$. So the maximum value of $y-x$, the distance between $x$ and
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$y$, is less than $1/2$. Therefore either $y$ and $z$ are within $1/2$ unit
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of each other or $y$ and $x$ are.
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]
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*12a.*
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F. This is a proof that $not A => B$ is a contradiction, which is not
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sufficient to show $A => B$.
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= 1.6
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*1b.*
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#proof[
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Consider $m = 1$, $n = -1$. Then $15(1) + 12(-1) = 3$.
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]
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*1c.*
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Rewrite $2m + 4n = 7$ as $2(m + 2n) = 7$. Notice that $2(m + 2n) = 2k, exists k
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in nonzero$ and therefore is an even number for any $m$, $n$. But 7 is not
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even. So no choices of $m$ and $n$ can create a number equal to 7. So no $m$ or
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$n$ exists.
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*4a.*
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#proof[
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Consider $x = 2$. Then $x^2 + x + 41 = 46$, which is not prime. So it's false.
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]
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*4b.*
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#proof[
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For any $x$, choose $y = -1 dot x$. Then $x + (-x) = x(1 - 1) = 0x = 0$. This
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is also trivially true when considering our usual field of reals because the
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existence of $y$ such that $x + y = 0$ is an axiom.
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]
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*4c.*
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Consider $x = 2$ and $y = 1$. Then $1 > 2$ is false.
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*4d.*
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Consider $a = 10$, $b = 2$, $c = 5$. Then $a | b c$ because $10 | 2 dot 5$ but
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10 does not divide either 2 or 5.
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*6a.*
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#proof[
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Rewriting the inequality,
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$
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1 / n <= 1\
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1 <= n
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$
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This is true for all $n in NN$ by the definition of $NN$.
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]
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*6b.*
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#proof[
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Rewriting,
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$
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1 / n < 0.13 \
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1 < 0.13n \
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n > 1 / 0.13
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$
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So such an $M$ is $1/0.13$, because all naturals greater than $1/0.13$ are
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greater than $1/0.13$.
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]
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*6e.*
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#proof[
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$
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forall n in NN, n + 1 > n
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$
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]
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*6f.*
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$
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forall k in ZZ, m = -k, k + m = 0 \
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forall n in NN, 0 <= 0 < n
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$
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*6i.*
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First consider $K = 10$, and therefore
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$
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forall r > 10, r^2 > 100
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$
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Then note that our inequality is equivalent to $r^2 > 100$, which is true for
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all $r$.
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$
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1 / (r^2) < 0.01 \
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r^2 > 100 \
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$
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So a $K$ exists, namely $K = 10$.
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*6k.*
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Consider $M = 51$. Then $forall r > 51, 2r > 102$ and $2r > 100$ is always
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true.
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$
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1 / (2r) < 0.01 \
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2r > 100
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$
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So such an $M$ exists, namely, $M = 51$.
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= 2.1
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*4a.*
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False.
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*4b.*
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True.
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*4c.*
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False.
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*4d.*
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True.
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*4e.*
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True.
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*4f.*
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False.
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*4g.*
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True.
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*4h.*
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False.
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*4i.*
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False.
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*4j*.
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True.
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*5a.*
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True.
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*5b.*
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True.
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*5c.*
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True.
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*5d.*
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True.
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*5e.*
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False.
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*5f.*
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True.
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*5g.*
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True.
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*5h.*
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True.
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*5i.*
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False.
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*5j.*
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True.
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*5k.*
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True.
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*5L.*
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True.
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*6a.*
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$
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A = {1,2}, B = {1,2,3}, C = {1,2,4}
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$
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*6b.*
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$
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A = B = C
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$
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A particular example might be $A = {1}, B = {1}, C = {1}$.
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*6c.*
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$
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A = {1}, B = {1,2}, C = {3}
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$
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*6d.*
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$
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A = {1,2}, B = {1,2,3}, C = {5,6,7}
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$
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*8.*
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Theorem 2.1.1: $forall A, B, C, A subset.eq B and B subset.eq C => A subset.eq C$.
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#proof[
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$
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forall a in A, a in B \
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forall a in A, exists b in B, a = b \
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forall b in B, b in C \
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forall a in A, a in C => A subset.eq C
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$
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]
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*11a.*
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#proof[
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$
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{x in RR : 3 / 4 x - 2 > 10} \
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= {x in RR : x > 16 } \
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= (16, infinity)
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$
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]
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*11b.*
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#proof[
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$
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{x in RR : |x - 4| = 2|x| - 2} \
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= {x in RR : |x - 4| = 2|x| - 2} \
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$
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Now we consider various cases. Consider $x >= 4$. Then
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$
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{x in RR : x - 4 = 2x - 2} \
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= {x in RR : x - 4 = 2x - 2} \
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= {x in RR : x = -2} \
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$
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But in this case $x >= 4$. Since there is no $x >= 4$ such that $x = -2$,
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this is actually $emptyset$.
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Now consider $0 <= x < 4$.
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$
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{x in RR : 4 - x = 2x - 2} \
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= {x in RR : x = 2} \
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= {2}
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$
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Now consider $x < 0$. Then
|
|
$
|
|
{x in RR : 4 - x = -2x - 2} \
|
|
{x in RR : x = -6}
|
|
$
|
|
So the set contains $6$ when $x < 0$.
|
|
|
|
And since these inequalities span all $x in RR$, the only members of
|
|
the set are ${2, -6}$.
|
|
]
|
|
|
|
*11c.*
|
|
|
|
#proof[
|
|
$
|
|
{x in RR : 2|x+3| + x = 0}
|
|
$
|
|
For $x >= -3$, we have
|
|
$
|
|
{x in RR : 2(x + 3) + x = 0} \
|
|
= {x in RR : x = -2} \
|
|
= {-2}
|
|
$
|
|
For $x < -3$, we have
|
|
$
|
|
{x in RR : 2(3 - x) + x = 0} \
|
|
= {x in RR : x = 6 } \
|
|
= {6}
|
|
$
|
|
And since these inequalities partition $RR$ the original set is ${-2, 6}$.
|
|
]
|
|
|
|
*11d.*
|
|
|
|
$
|
|
{x in RR : |x| = 6 - |2x|}
|
|
$
|
|
|
|
Consider $x >= 0$. Then
|
|
$
|
|
{x in RR : x = 6 - 2x} \
|
|
= {x in RR : x = 2}
|
|
$
|
|
Consider $x < 0$. Then
|
|
$
|
|
{x in RR : -x = 6 + 2x} \
|
|
= {x in RR : x = -2}
|
|
$
|
|
So
|
|
$
|
|
{x in RR : |x| = 6 - |2x|} = {-2, 2}
|
|
$
|
|
|
|
*11e.*
|
|
|
|
$
|
|
{x in RR : |x + 3| <= -4x - 2}
|
|
$
|
|
|
|
Consider $x >= -3$. Then
|
|
$
|
|
{x in RR : x <= -1} \
|
|
= (-infinity, -1]
|
|
$
|
|
|
|
Now consider $x < -3$ Then
|
|
$
|
|
{x in RR : 3 - x <= -4x - 2} \
|
|
= {x in RR : x <= -5 / 3} \
|
|
= (-infinity, -5 / 3]
|
|
$
|
|
But $(-infinity, -5/3) union (-infinity, 1] = (-infinity, 1]$ so that is our
|
|
answer.
|
|
|
|
*11f.*
|