92 lines
2.5 KiB
TeX
92 lines
2.5 KiB
TeX
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Abstract Algebra
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\bigskip
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{\bf Problem A.} Prove the following corollary, using the lemma:
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There is no integer between 0 and 1.
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\medskip
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{\bf Corollary.} Let $a\in Z$. Then there is no integer
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between $a$ and $a+1$.
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\bigskip
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{\bf Proof.} Let $a\in Z$.
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Suppose there is a $b\in Z$ such that $a<b<a+1$.
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Then by Lemma 2.11 (Unit 7) we can add $-a$ to obtain
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$0<b-a<1$.
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We conclude that $b-a$ is an integer by closure.
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However, by the lemma ``there is no integer between 0 and 1,''
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we also conclude that $b-a$ does not exist.
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This is a contradiction, therefore the premise must be false.
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Hence there is no integer between $a$ and $a+1$
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\bigskip
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This proof uses the corollary ``1 is the smallest element of $N$.''
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\medskip
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{\bf Proof.} Let $a\in Z$.
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Suppose there is a $b\in Z$ such that $a<b$ and $b<a+1$.
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By the first less-than we have $b-a\in N$.
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Adding $-a$ to the second less-than yields $b-a<1$.
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Then by Corollary 1.2, ``1 is the smallest element of $N$,''
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we have $b-a\not\in N$.
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This is a contradiction, therefore the premise must be false.
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Hence there is no integer between $a$ and $a+1$
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\bigskip
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\bigskip
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{\bf Problem B.} Prove that
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$$2+4+\cdots+2n=n(n+1)$$
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for any $n\in N$.
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\bigskip
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{\bf Proof.}
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(i) Substitute 1 for $n$ to obtain $2=1(1+1)$ hence the equality is
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true for $n=1$.
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(ii)
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Assume that the equality holds for a natural number $n$.
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Does it also hold for $n+1$?
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For $n+1$ we have
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$$\eqalign{
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2+4+\cdots+2n+2(n+1)&=(n+1)((n+1)+1)\cr
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&=(n+1)(n+2)\cr
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&=n^2+3n+2\cr
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}$$
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Subtract $2n+2$ from both sides of the equation.
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$$\eqalign{
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2+4+\cdots+2n&=n^2+3n+2-2n-2\cr
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&=n^2+n\cr
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&=n(n+1)\cr
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}$$
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This result is the original equality.
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Therefore when the equality is true for $n$ then it is also true for $n+1$.
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Hence by induction the equality is true for all $n\in N$.
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\vfill
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\eject
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{\bf Problem C.} Prove that
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$$1+3+9+\cdots+3^n={3^{n+1}-1\over 2}$$
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for all $n\in N$.
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\bigskip
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{\bf Proof.} (i) Substitute 1 for $n$ to obtain
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$$1+3^1={3^2-1\over 2}=4$$
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hence the equality is true for $n=1$.
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(ii) Assume the equality is true for a natural number $n$.
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Is it also true for $n+1$?
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For $n+1$ we have
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$$
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1+3+9+\cdots+3^n+3^{n+1}={3^{(n+1)+1}-1\over 2}
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={3\cdot3^{n+1}-1\over 2}
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$$
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Subtract $3^{n+1}$ from both sides to obtain
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$$\eqalign{
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1+3+9+\cdots+3^n
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={3\cdot3^{n+1}-1-2\cdot3^{n+1}\over 2}
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={3^{n+1}-1\over 2}
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}$$
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This result is the original equality.
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Therefore when the equality is true for $n$ then it is also true for $n+1$.
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Hence by induction the equality is true for all $n\in N$.
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\end
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