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This repository was archived by the owner on Jun 7, 2023. It is now read-only.
Base case: At \(n=0\), \(\sum_{j=0}^n 2(-7)^j = 2\) and \(\frac{1-(-7)^{n+1}}{4} = \frac{1-(-7)}{4} = 2\), so the base case, \(P(0)\), holds #tag:2;depends:0,1;
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Inductive Hypothesis: Suppose that \(P(n):\)\(\sum_{j=0}^n 2(-7)^j = \frac{1-(-7)^{n+1}}{4}\) holds for \(n=0,1,...,k\). #tag:3;depends:1;
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Inductive Step: We need to show that \(P(k+1):\)\(\sum_{j=0}^{k+1} 2(-7)^j = \frac{1-(-7)^{k+2}}{4}\) holds #tag:4;depends:3;
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The left hand side is \(\sum_{j=0}^{k+1} 2(-7)^j = \sum_{j=0}^k 2(-7)^j + 2(-7)^{k+1}\)#tag:5;depends:4;
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By the inductive hypothesis we have \(\sum_{j=0}^k 2(-7)^j = \frac{1-(-7)^{k+1}}{4}\). So then substituting we get \(= \frac{1-(-7)^{k+1}}{4} + 2(-7)^{k+1}\)\(= \frac{1-(-7)^{k+1} + 8(-7)^{k+1}}{4}\)\(= \frac{1+7(-7)^{k+1}}{4}\) which simplifies to \(= \frac{1-(-7)^{k+2}}{4}\)#tag:6;depends:5;
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So \(\sum_{j=0}^{k+1} 2(-7)^j = \frac{1-(-7)^{k+2}}{4}\), which was what we needed to show. #tag:7;depends:6;
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Assume $G$ and $H$ are isomorphic graphs and $G$ is 2-colorable. #tag:0;depends:;
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Let $c:V(G) \to \{red, blue\}$ be a 2-coloring of $G$. #tag: 1; depends:0;
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Let $f$ be an isomorphism $V(H) \to V(G)$ #tag: 2; depends: 0;
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Define $c':V(H) \to \{red, blue\}$ as $c'(v)=c(f(v))$ #tag:3;depends:1,2;
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Let $\langle u - v \rangle$ be an edge in $H$. (If instead there are no edges in $H$, then $H$ is trivially 2-colorable and we are done.) #tag:4;depends:0;
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$\langle f(u) - f(v) \rangle$ is an edge in $G$ #tag:5;depends:4,2;
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$c(f(u)) \ne c(f(v))$ #tag:6;depends:5,1;
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$c'(u) \ne c'(v)$ #tag:7;depends:6,3;
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$c'$ is a 2-coloring of $H$, so $H$ is 2-colorable. (end of proof) #tag:8;depends:7;
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