Finished proof techniques handout
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions,
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singlenumbering,
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unfinished
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singlenumbering
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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@ -18,7 +17,7 @@
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\let\implies\Rightarrow
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\let\rimplies\Leftarrow
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\let\iff\Leftrightarrow
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\let\notimplies\nRightarrow
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\let\notimplies\nRightarrow
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\begin{document}
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@ -3,11 +3,169 @@
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\definition{}
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The last proof technique we'll discuss in this handout is \textit{induction.} \par
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This is particularly useful when we have a \say{countable} variable, usually an integer. \par
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Let's say we're proving a statement $A$ for all positive integers $n$. \par
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We'll write the special case \say{$A$ holds for $n$} as $A_n$.
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\vspace{2mm}
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A proof by induction consists of two parts: a \textit{base case} and a \textit{inductive step}. \par
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The base case is usually fairly simple: we show that our statement holds for $n = 0$. \par
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In other words, the base case shows that $A_0$ is true.
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The inductive step is a bit more confusing: we show that if our statement holds for
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$n$, it must hold for $n = 1$. In other words, we show that $A_n \implies A_{n + 1}$.
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\vspace{2mm}
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In this way, we build an infinite implication chain: \par
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The base case proves that $A_0$. By the inductive step,
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$A_0 \implies A_1$, $A_1 \implies A_2$, and so on. \par
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We can thus conclude that $A_n$ is true for all $n \in \{0, 1, 2, 3, ...\}$
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\problem{}
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Proof by induction will make a bit more sense with an example. \par
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Read and understand the following proof.
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\begin{examplesolution}
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Show that $1 + 2 + ... + n = \frac{n(n+1)}{2}$
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\linehack{}
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\textbf{Base case:} $n = 1$ \par
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Substitute $n = 1$ into the hypothesis:
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$1 \qe \frac{1(1 + 1)}{2}$ \par
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but we have that
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$\frac{1(1 + 1)}{2} = \frac{2}{2} = 1$,
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so this is of course true.
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\vspace{2mm}
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\textbf{Inductive step:} \par
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Now, we assume our hypothesis is true for $n$, \par
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and show it is true for $n + 1$.
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\vspace{2mm}
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Write the hypothesis for $n + 1$:
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$$
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1 + 2 + ... + n + (n + 1) \qe \frac{(n+1)(n+2)}{2}
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$$
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We know that $1 + 2 + ... + n = \frac{n(n+1)}{2}$, so:
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$$
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1 + 2 + ... + n + (n + 1) = \frac{n(n+1)}{2} + (n+1)
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$$
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now we can complete the proof with some algebra:
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$$
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\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n + 2)(n+1)}{2}
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$$
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So, we've shown that the $n^\text{th}$ case implies the $(n+1)^\text{th}$ case.\par
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We're therefore done, the hypothesis is true for $n \in \{1, 2, 3, ...\}$
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\end{examplesolution}
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\vfill
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\pagebreak
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\problem{}
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Why do we need a base case when constructing a proof by induction? \par
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\hint{Try to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$}
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\begin{solution}
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Consider the following example: \par
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Say we want to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$.
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\vspace{2mm}
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\textbf{Inductive step:}
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Assume our hypothesis is true for $n$ (that is $n = n + 1$). Is this true for $n + 1$? \par
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Adding 1 to both sides, we get $n+1=n+1+1$, which means that $(n+1)=(n+1)+1$,
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which is the statement we wanted to prove. We've thus completed the inductive step!
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\vspace{2mm}
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Our problem is as follows: we've shown that $A_1 \implies A_2 \implies ...$,
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but we have no reason to believe that $A_1$ is true. If it was, our hypothesis
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would be correct---but since it isn't, this is not a complete proof.
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\end{solution}
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\vfill
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\problem{}
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Show that $1^2 + 2^2 + 3^3 + ... + n^2 = \frac{1}{6}(n)(n+1)(2n+1)$.
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\begin{solution}
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\textbf{Base case:}\par
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$1^2 = \frac{1}{6}(1)(1+1)(2 + 1) = 1$, which is true.
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\vspace{2mm}
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\textbf{Induction:}\par
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Assume $1^2 + ... + n^2$ satisfies the equation above. \par
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$$
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1^2 + 2^2 + ... + n^2 + (n+1)^2 =
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\frac{(n)(n+1)(2n+1)}{6} + (n + 1)^2
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$$
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which is equal to
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$$
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\frac{(n)(n+1)(2n+1) + 6(n+1)^2}{6}
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$$
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now expand and factor to get
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$$
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\frac{(n+1)(n+2)(2(n+1)+1)}{6}
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$$
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\end{solution}
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\vfill
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\problem{}
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Show that $2^n - 1$ is divisible by 3 for all odd $n$ \par
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\hint{If $n$ is odd, the next odd number is $n + 2$.}
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\begin{solution}
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\textbf{Base case:} \par
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$2^2 - 1 = 3$, which is divisible by 3..
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\vspace{2mm}
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\textbf{Induction:}\par
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Assume $2^n - 1$ is divisible by 3. \par
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$2^{(n+2)} - 1 = 4(2^n) - 1 = 4(2^n - 1) + 3$ \par
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By our induction hypothesis, $(2^n - 1)$ has a factor of 3. \par
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Therefore, $4(2^n - 1) + 3$ must also have a factor of 3.
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\end{solution}
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\vfill
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\pagebreak
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\definition{}
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As you may already know, \say{n choose k} is defined as follows: \par
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$$
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\binom{n}{k} = \frac{n!}{k!(n-k)!}
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$$
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This counts the number of ways to choose $k$ things from a set of $n$,
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disregarding the order of the chosen items.
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\theorem{Pascal's Identity}
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The binomial coefficient defined above satisfies the following equality:
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$$
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\binom{n+1}{k} = \binom{n}{k-1} \times \binom{n}{k}
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$$
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\problem{}<binomsum>
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Using induction, show that
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$$
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\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n} = 2^n.
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$$
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@ -22,19 +180,63 @@ A proof by induction consists of two parts: a \textit{base case} and a \textit{i
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Note that although induction is a powerful proof technique, it usually leads to uninteresting results. \par
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If we prove a statement using induction, we conclude that it is true---but we get very little insight on
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\textit{why} that is.
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\textit{why} it is true.
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\vspace{2mm}
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Alternative proofs are take a bit more work than inductive proofs, but they are much more valuable. \par
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For example, consider the following proof of X:
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For example, see the proof of the statement in \ref{binomsum} on the next page.
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\pagebreak
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\makeatletter
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\@makeORMCbox{tmpbox}{Alternative Proof}{ogrape!10!white}{ogrape}
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\makeatother
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\begin{tmpbox}
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sdfasdf
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Consider the following problem: \par
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How many ways are there to write a number $x$ as an ordered sum of positive integers? \par
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\note{
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An \say{ordered sum} means that the order of numbers in the sum matters. \\
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For example, if $x = 5$, we will consider $4 + 1$ and $1 + 4$ as distinct sums.
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}
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\vspace{2mm}
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First, we'll think of $x$ as an array of $1$s which we want to group
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into positive integers. If $x = 3$, We have $x = 1~1~1$, which we can
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group as $1+1+1$,~ $(1+1) + 1$,~ $1 + (1+1)$,~ and $(1+1+1)$.
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\linehack{}
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\textbf{Solution 1:}\par
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One way to solve this is to use the usual \say{stars and bars} method, \par
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where we count the number of ways we can place $n$ \say{bars} between
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$x$ \say{stars}. Each bar corresponds to a \say{$+$} in the array of ones. \par
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\note[Note]{Convince yourself that there are $\binom{x-1}{n}$ ways to place $n$ bars
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between $x$ objects.}
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If we add the number of ways to write $x$ as a sum of $n \in \{1, 2, ..., x\}$ integers, we get:
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$$
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\sum_{n = 1}^{x-1} \binom{x-1}{n} = \binom{x-1}{0} + \binom{n}{1} + ... + \binom{x-1}{x-1}
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$$
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\linehack{}
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\textbf{Solution 2:}\par
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We could also observe that there are $x - 1$ places to put a \say{bar} in
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the array of ones. This corresponds to $x - 1$ binary positions, and thus
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$2^{x-1}$ ways to seperate our array of $1$s with bars.
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\linehack{}
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\textbf{Conclusion:}\par
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We've found that the number of ways to split $x$ can be written as either
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$\sum_{n = 1}^{x-1} \binom{x-1}{n}$ or $2^{x-1}$,
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and therefore $\sum_{n = 1}^{x-1} \binom{x-1}{n} = 2^{x-1}$.
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\end{tmpbox}
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