mirror of
https://github.com/Rushilwiz/math381.git
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384 lines
15 KiB
TeX
384 lines
15 KiB
TeX
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\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
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}{
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\newcommand{\hmwkTitle}{Homework 3}
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\newcommand{\hmwkDueDate}{September 21, 2023}
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\newcommand{\hmwkClass}{Discrete Math}
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\newcommand{\hmwkClassTime}{Section 003}
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\newcommand{\hmwkClassInstructor}{Reese Lance}
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\newcommand{\hmwkAuthorName}{\textbf{Rushil Umaretiya}}
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%
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
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\normalsize\vspace{0.1in}\small{Tuesday/Thursday 11:00-12:15, Phillips 383}\\
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\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ - \hmwkClassTime}}
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\vspace{3in}
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}
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\author{\hmwkAuthorName\\\small{rumareti@unc.edu}}
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\date{}
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\newcommand{\Cov}{\mathrm{Cov}}
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\newcommand{\Bias}{\mathrm{Bias}}
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\renewcommand{\And}{\wedge}
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\newcommand{\Or}{\vee}
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\newcommand{\Xor}{\oplus}
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\newcommand{\Not}{\neg}
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\newcommand{\Implies}{\rightarrow}
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\newcommand{\Iff}{\leftrightarrow}
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\newcommand{\AllIntegers}{\mathbb{Z}}
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\newcommand{\AllRationals}{\mathbb{Q}}
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\newcommand{\AllReals}{\mathbb{R}}
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\newcommand{\AllComplexes}{\mathbb{C}}
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\begin{document}
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\maketitle
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\pagebreak
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\unit{1.7}
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\problem{8}
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Prove that if \(n\) is a perfect square, then \(n + 2\) is not a perfect square.\\
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\begin{proof}
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Let \(n\) be a perfect square. Given P(x) = \(\exists x \in \AllIntegers, n = x^2\) and Q(y) = \(\exists y \in \AllIntegers, n + 2 = y^2\), we will prove that \(\forall n \in \AllIntegers(P(n)\implies\neg Q(n))\) using a proof by contradiction.\\
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\textbf{Proof by Contradiction:}\\
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\begin{align*}
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\text{Assume } \exists n \in \AllIntegers(P(n) \implies Q(n)).&&\text{Premise.}\\
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P(x) = \exists x \in \AllIntegers, n = x^2&&\text{Premise.}\\
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Q(y) = \exists x \in \AllIntegers, n + 2 = y^2&&\text{Premise.}\\
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x^2 + 2 = y^2&&\text{Algebraic Substitution.}\\
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y^2 - x^2 = 2&&\text{Algebraic Equivalence.}\\
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(y + x)(y - x) = 2&&\text{Factorization.}\\
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\end{align*}
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Since \(y\) and \(x\) are integers, \(y + x\) and \(y - x\) are integers. The only integer factors of 2 are 1 and 2. Therefore,\\
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\begin{align*}
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y + x = 1\\
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y - x = 2
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\end{align*}
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Using algebraic equivalence, we can solve for \(y\).\\
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\begin{align*}
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y = \frac{3}{2}&&\bot\\
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\end{align*}
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Since \(y\) is not an integer, we have reached a contradiction. Therefore, \(\forall n \in \AllIntegers(P(n)\implies\neg Q(n))\).
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\end{proof}
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\pagebreak
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\problem{14}
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Prove that if \(x\) is rational and \(x \neq 0\), then \(1/x\) is rational.\\
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\begin{proof}
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Let \(x\) be rational and \(x \neq 0\). Given \(P(x) = \exists p,q \in \AllIntegers, p,q \neq 0, x = p/q\) and \(Q(y) = \exists r,s \in \AllIntegers, r,s \neq 0, 1/x = r/s,\)k we will prove that \(\forall x \in \AllRationals(P(x)\implies Q(x))\) using a direct proof.\\
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\textbf{Direct Proof:}
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\begin{align*}
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x = \frac{p}{q}&&\text{Premise.}\\
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\frac{1}{x} = \frac{q}{p}&&\text{Algebraic Equivalence.}\\
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\frac{1}{x} = \frac{r}{s}&&\text{Let } r = q \text{ and } s = p.\\
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s \neq 0&&\text{Since } p \neq 0.\\
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\exists r,s \in \AllIntegers, r,s \neq 0, 1/x = r/s&&\text{Premise.}\\
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\forall x \in \AllRationals(P(x)\implies Q(x))&&\text{Conclusion.}
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\end{align*}
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\end{proof}
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\pagebreak
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\problem{20}
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Prove that if \(n\) is an integer and \(3n + 2\) is even, then n is even using\\
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\begin{enumerate}[a)]
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\item a proof by contraposition\\
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\textbf{Proof by Contraposition:}
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\begin{proof}
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Given \(P(n) = \exists x \in \AllIntegers, 2x = 3n + 2\) and \(Q(n) = \exists y \in \AllIntegers, 2y = n\), we will prove that \(\forall n \in \AllIntegers(P(n)\implies Q(n))\) using a proof by contraposition. In fact, we want to show that if \(n\) is odd, then \(3n + 2\) is odd.\\
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\begin{align*}
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\text{Assume } \exists n \in \AllIntegers(\neg Q(n) \implies \neg P(n)).&&\text{Premise.}\\
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\neg Q(n) = \exists y \in \AllIntegers,n = 2y + 1&&\text{Premise.}\\
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\end{align*}
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Now we need to prove \(\neg P(n)\).\\
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\begin{align*}
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3n + 2 &= 3(2y + 1) + 2&&\text{Algebraic Substitution.}\\
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&= 6y + 5 &&\text{Algebraic Equivalence.}\\
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& = 2(3y + 2) + 1&&\text{Algebraic Equivalence.}\\
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\end{align*}
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Let \(\bar{k}=3y+2, \bar{k}\in\AllIntegers.\)
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\begin{align*}
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&= 2\bar{k} + 1&&\text{Algebraic Equivalence.}\\
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\end{align*}
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Since \(\bar{k}\) is an integer, \(2\bar{k}+1\) is odd. Therefore, \(3n + 2\) is odd and we have proved the contrapositive.\\
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\end{proof}
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\pagebreak
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\item a proof by contradiction\\
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\textbf{Proof by Contradiction:}\\
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\begin{proof}
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Given \(P(n) = \exists x \in \AllIntegers, 2x = 3n + 2\) and \(Q(n) = \exists y \in \AllIntegers, 2y = n\), we will prove that \(\forall n \in \AllIntegers(P(n)\implies Q(n))\) using a proof by contradiction. In fact, we have to show a contradiction in the statement, if \(3n+2\) is even then \(n\) is odd.\\
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\begin{align*}
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\text{Assume } \exists n \in \AllIntegers(P(n) \implies \neg Q(n)).&&\text{Premise.}\\
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P(n) = \exists x \in \AllIntegers, 2x = 3n + 2&&\text{Premise.}\\
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\neg Q(n) = \exists y \in \AllIntegers, n = 2y + 1&&\text{Premise.}\\
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\end{align*}
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Now we need to find the contradiction in the statement.\\
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\begin{align*}
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2x &= 3(2y + 1) + 2&&\text{Algebraic Substitution.}\\
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&= 6y + 5 &&\text{Algebraic Equivalence.}\\
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& = 2(3y + 2) + 1&&\text{Algebraic Equivalence.}\\
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\end{align*}
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Let \(\bar{k}=3y+2, \bar{k}\in\AllIntegers.\)
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\begin{align*}
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2x &= 2\bar{k} + 1&&\bot\\
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\end{align*}
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Here we have found the contradiction in the previous statement, since \(2x\) is even and \(2\bar{k}+1\) is odd. Therefore, we have proved the proposition.\\
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\end{proof}
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\end{enumerate}
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\pagebreak
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\problem{30}
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Prove that \(m^2 = n^2\) if and only if \(m = n\) or \(m = -n\).
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\begin{proof}
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Given \(P(m,n) = "m^2 = n^2"\) and \(Q(m,n) = "m = n" \Or "m = -n"\), we will prove that \(\forall m,n \in \AllIntegers(P(m,n)\iff Q(m,n))\) using a direct proof.\\
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\textbf{Direct Proof:}\\
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In the case of \(P(m,n) \implies Q(m,n)\),\\
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\begin{align*}
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m^2 &= n^2\\
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m^2 - n^2 &= 0\\
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(m - n)(m + n) &= 0\\
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\end{align*}
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Which yields,
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\begin{align*}
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m - n &= 0\\
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m &= n\\
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\textbf{or}\\
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m + n &= 0\\
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m &= -n\\
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\end{align*}
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Therefore, \(P(m,n) \implies Q(m,n)\).\\\\
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In the case of \(Q(m,n) \implies P(m,n)\) we can assume that \(m = n\) without the loss of generality,\\
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\begin{align*}
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m &= n\\
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m - n &= 0\\
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(m - n)(m + n) &= 0\\
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m^2 - n^2 &= 0\\
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m^2 &= n^2\\
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\end{align*}
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Therefore, \(Q(m,n) \implies P(m,n)\).
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Since \(P(m,n) \implies Q(m,n)\) and \(Q(m,n) \implies P(m,n)\), we have proved that \(\forall m,n \in \AllIntegers(P(m,n)\iff Q(m,n))\).
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\end{proof}
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\pagebreak
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\unit{1.8}
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\problem{8}
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Prove using the notion of without loss of generality that \(5x + 5y\) is an odd integer when \(x\) and \(y\) are integers of opposite parity.
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\begin{proof}
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Given \(P(x,y) = \exists k,l \in \AllIntegers, x = 2k, y = 2l+1\) and \(Q(x,y) = \exists w \in \AllIntegers, 2w + 1 = 5x + 5y\), we will prove that \(\forall x,y \in \AllIntegers(P(x,y)\implies Q(x,y)\) is odd) using a direct proof. Without loss of generality, we can assume that \(x\) is even and \(y\) is odd.\\
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\textbf{Direct proof:}
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\begin{align*}
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\text{Assume } &x = 2k, y = 2l + 1, k,l \in \AllIntegers.&&\text{Premise.}\\
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5x + 5y &= 5(2k) + 5(2l + 1)&&\text{Algebraic Substitution.}\\
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&= 10k + 10l + 5&&\text{Algebraic Equivalence.}\\
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&= 2(5k + 5l + 2) + 1&&\text{Algebraic Equivalence.}\\
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\end{align*}
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Let \(\bar{k} = 5k + 5l + 2, \bar{k} \in \AllIntegers\).
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\begin{align*}
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&= 2\bar{k} + 1&&\text{Algebraic Equivalence.}\\
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\end{align*}
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Therefore \(5x + 5y\) is an odd integer when \(x\) and \(y\) are integers of opposite parity.
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\end{proof}
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\pagebreak
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\problem{10}
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Prove that there is a positive integer that equals the sum of the positive integers not exceeding it. Is your proof constructive or nonconstructive?
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\begin{proof}
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Given \(P(n) = \exists x \in \AllIntegers, x = \sum_{i=1}^{n}i\) and \(Q(n) = \exists y \in \AllIntegers, y > 0\), we will prove that \(\forall n \in \AllIntegers(P(n)\implies Q(n))\) using a direct proof.\\
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\textbf{Constructive Proof:}
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\begin{align*}
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P(1)=\exists x \in \AllIntegers, x &= \sum_{i=1}^{1}i&&\text{Given P(1).}\\\\ &=1&&
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\end{align*}
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Using a \textbf{constructive} proof, we have shown that there exists a positive integer, 1, that equals the sum of the positive integers not exceeding it.
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\end{proof}
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\pagebreak
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\problem{32}
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Prove that there are no solutions in integers \(x\) and \(y\) to the equation \(2x^2 + 5y^2 = 14\).
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\begin{proof}
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Given \(P(x,y) = \exists x,y \in \AllIntegers, 2x^2 + 5y^2 = 14\), we will prove that \(\forall x,y \in \AllIntegers(\neg P(x,y))\) using a proof by exhaustion.\\
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\textbf{Proof by Exhaustion:}\\
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Given \(2x^2 + 5y^2 = 14\),
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\begin{align*}
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|2x^2| &\leq 14\\
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|5y^2| &\leq 14\\
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\end{align*}
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Our possible values for \(x\) are \(\{0,1,2\}\) and our possible values for \(y\) are \(\{0,1\}\).\\
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\begin{align*}
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2(0)^2 + 5(0)^2 &= 0 + 0 = 0 &\neq 14&&(0,0)\\
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2(0)^2 + 5(1)^2 &= 0 + 5 = 5 &\neq 14&&(0,1)\\
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2(1)^2 + 5(0)^2 &= 2 + 0 = 2 &\neq 14&&(1,0)\\
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2(1)^2 + 5(1)^2 &= 2 + 5 = 7 &\neq 14&&(1,1)\\
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2(1)^2 + 5(2)^2 &= 2 + 20 = 22 &\neq 14&&(1,2)\\
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2(2)^2 + 5(0)^2 &= 8 + 0 = 8 &\neq 14&&(2,0)\\
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2(2)^2 + 5(1)^2 &= 8 + 5 = 13 &\neq 14&&(2,1)\\
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\end{align*}
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Therefore, \(\forall x,y \in \AllIntegers(\neg P(x,y))\).
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\end{proof}
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\pagebreak
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\problem{36}
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Prove that \(\sqrt[3]{2}\) is irrational.
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\begin{proof}
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Given P(x) = \(x \notin \AllRationals\), we can do a proof by contradiction to prove that \(\sqrt[3]{2}\) is irrational. Given \(\neg P(x) = \exists p,q \in \AllIntegers, q\neq 0, x = \frac{p}{q}\)\\
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\textbf{Proof by Contradiction:}\\
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\begin{align*}
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\sqrt[3]{2} &= \frac{p}{q} && \text{Given } \neg P(x)\\
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2 &= \frac{p^3}{q^3} && \text{Cubing both sides.}\\
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2q^3 &= p^3 && \text{Multiplying both sides by } q^3.\\
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p^3 &\text{ is even.} && \text{Since } 2\bar{k} \text{ is even.}\\
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p &\text{ is even.} && \text{Since } p^3 \text{ is even.}\\
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\end{align*}
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Given that p is even, Let \(\bar{k} \in \AllIntegers, p=2\bar{k}\).
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\begin{align*}
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2q^3&= (2\bar{k})^3 && \text{Algebraic Substitution.}\\
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2q^3&= 8\bar{k}^3 && \text{Algebraic Equivalence.}\\
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q^3&= 4\bar{k}^3 && \text{Algebraic Equivalence.}\\
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\end{align*}
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Since \(p\) is even, we have two cases where \(q\) is even and \(q\) is odd.\\
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\begin{enumerate}
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\item \(q\) is even.\\
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\begin{align*}
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2|p, 2|q&&\text{Since p and q are even.}\\
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\gcd(p,q) \geq 2&&\text{Since p and q are even.}
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&&\bot\\
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\end{align*}
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We find the contradiction in the statement, since \(\gcd(p,q) \geq 2\) and \(p/q\) is in lowest terms.
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\item \(q\) is odd.\\
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\begin{align*}
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\exists l \in \AllIntegers, q &= 2l + 1&&\text{Since q is odd.}\\
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(2l + 1)^3 &= 4\bar{k}^3&&\text{Algebraic Substitution.}\\
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8l^3 + 12l^2 + 6l + 1 &= 4\bar{k}^3&&\text{Algebraic Expansion.}\\
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8l^3 +12l^2 +6l-4\bar{k}^3 &= -1&&\text{Algebraic Equivalence.}\\
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2(4l^3 + 6l^2 + 3l-2\bar{k}^3) &= -1&&\text{Algebraic Equivalence.}\\
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\bar{l} \in \AllIntegers, \bar{l} &= 4l^3 + 6l^2 + 3l-2\bar{k}^3&&\text{Define } \bar{l}\\
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2\bar{l} &= -1&&\text{Algebraic Equivalence.}
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&&\bot\\
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\end{align*}
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We find the contradiction in the statement, since \(2\bar{l}\) cannot equal \(-1\).
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\end{enumerate}
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Therefore, we have proved that \(\sqrt[3]{2}\) is irrational by contradiction.
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\end{proof}
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\end{document} |