commit 6b2bf010cc5ba3f57f22769c8e809d029026c26d parent 502379c0540d55f1ca5341b69f89c68994a22f02 Author: Sebastiano Tronto <sebastiano.tronto@gmail.com> Date: Thu, 9 Sep 2021 09:09:20 +0200 Solutions hidden Diffstat:
| D | preplogic-solutions.pdf | | | 0 | |
| D | solutions/preplogic-solutions.aux | | | 5 | ----- |
| D | solutions/preplogic-solutions.fdb_latexmk | | | 79 | ------------------------------------------------------------------------------- |
| D | solutions/preplogic-solutions.fls | | | 111 | ------------------------------------------------------------------------------- |
| D | solutions/preplogic-solutions.log | | | 242 | ------------------------------------------------------------------------------- |
| D | solutions/preplogic-solutions.pdf | | | 0 | |
| D | solutions/preplogic-solutions.tex | | | 306 | ------------------------------------------------------------------------------- |
7 files changed, 0 insertions(+), 743 deletions(-)
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\item ``$7$ is odd and $2+2=5$'' - \item Every number of the form $2^{2^n}+1$, for $n=1,2,3...$, is prime. - \end{enumerate} -\end{exercise} -\begin{proof}[Solution] - \begin{enumerate} - \item \textbf{True}: regardless of when you solve this exercise, Germany - has more inhabitants than Luxembourg. - \item \textbf{True} - \item \textbf{False}: the number $2^{2^5}+1=4294967297=641\times 6700417$ - is not prime. - \end{enumerate} -\end{proof} - -\begin{exercise} - What is the negation of the sentence ``\emph{I payed attention in class and I - did not do my homework}'' ? -\end{exercise} -\begin{proof}[Solution] - ``\emph{I did not pay attention in class \textbf{or} I did my homework}'' -\end{proof} - -\begin{exercise} - Simplify the following logical expressions using the properties of logical - operations (where $A,B$ and $C$ are statements): - \begin{enumerate} - \item $A\land(A\lor B)$ - \item $A\lor (B\land A)$ - \item $(A\lor B) \land \neg A$ - \item $A \lor (\neg A\land B)$ - \item $(\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)$ - \end{enumerate} -\end{exercise} -\begin{proof}[Solution] - They are equivalent to the following (you can check with truth tables): - \begin{enumerate} - \item $A$ - \item $A$ - \item $B\land \neg A$ - \item $A\lor B$ - \item Let's do this one in more steps: - \begin{align*} - (\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)= - &(\neg A\land B)\land ((A\lor C)\land \neg C)=\\ - =&(\neg A\land B)\land ((A\land \neg C)\lor (C\land \neg C))=\\ - =&(\neg A\land B)\land ((A\land \neg C)\lor \textbf{false})=\\ - =&(\neg A\land B)\land (A\land \neg C)=\\ - =&A\land \neg A\land B\land \neg C=\\ - =&\textbf{false} - \end{align*} - \end{enumerate} -\end{proof} - -\section{Implication} - -\begin{exercise} - Fill in the following truth table: - \begin{align*} - \begin{array}{|c|c|c|c|c|} - \hline - A & B & C & \neg(A\implies B) & (A\implies B) \implies C \\ - \hline - 0 & 0 & 0 & 0 & 0 \\ - \hline - 0 & 0 & 1 & 0 & 1 \\ - \hline - 0 & 1 & 0 & 0 & 0 \\ - \hline - 0 & 1 & 1 & 0 & 1 \\ - \hline - 1 & 0 & 0 & 1 & 1 \\ - \hline - 1 & 0 & 1 & 1 & 1 \\ - \hline - 1 & 1 & 0 & 0 & 0 \\ - \hline - 1 & 1 & 1 & 0 & 1 \\ - \hline - \end{array} - \end{align*} -\end{exercise} - -\begin{exercise}[Transitivity] - Prove that the following statement is true for any statements $A,B$ and $C$: - \begin{align*} - ((A\implies B)\land (B\implies C))\implies (A\implies C) - \end{align*} -\end{exercise} -\begin{proof}[Solution] - Let's rewrite the first part in terms of basic logical operations: - \begin{align*} - (A\implies B)\land (B\implies C)=&(B\lor \neg A)\land(C\lor \neg B) - \end{align*} - now we can write a truth table for the two parts - \begin{align*} - \begin{array}{|c|c|c|c|c|} - \hline - A & B & C & (B\lor\neg A)\land(C\lor\neg B) & A\implies C \\ - \hline - 0 & 0 & 0 & 1 & 1 \\ - \hline - 0 & 0 & 1 & 1 & 1 \\ - \hline - 0 & 1 & 0 & 0 & 1 \\ - \hline - 0 & 1 & 1 & 1 & 1 \\ - \hline - 1 & 0 & 0 & 0 & 0 \\ - \hline - 1 & 0 & 1 & 0 & 1 \\ - \hline - 1 & 1 & 0 & 0 & 0 \\ - \hline - 1 & 1 & 1 & 1 & 1 \\ - \hline - \end{array} - \end{align*} - With the help of the truth table we see that whenever the first part - ``$(\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)$'' is true, also the - implication ``$A\implies C$'' is true. This shows that the ``big - implication'' is true. (If you are not convinced, you can add more details to - this proof, for example by writing more truth tables.) -\end{proof} - -\begin{exercise} -What is the contrapositive of ``\emph{If this table is not reserved, we sit -here}'' ? -\end{exercise} -\begin{proof}[Solution] - ``\emph{If we do not sit here, this table is reserved}''. One could also say - this in another way, for example ``\emph{We do not sit here because this - table is reserved}''. -\end{proof} - -\section{Quantifiers} - -\begin{exercise} - Write the negation of the following statements: - \begin{enumerate} - \item $\exists x\in \mathbb N,\, x^2-2=0$ - \item ``Every prime number is odd'' - \item ``Every person I have met likes pizza'' - \item ``There is at least one number greater than $7$'' - \item $\forall x\in \mathbb N,\,x\geq 0$ - \item $\forall x\in \mathbb Z,\,(\exists y\in\mathbb Z,\,x+y=0)$ - \end{enumerate} -\end{exercise} -\begin{proof}[Solution] - \begin{enumerate} - \item $\forall x\in\mathbb N,\, x^2-2\neq 0$ - \item ``There is at least one prime number which is even'' - \item ``I have met at least one person that does not like pizza'' - \item ``Every number is less or equal than $7$'' - \item $\exists x\in\mathbb N,\, x<0$ - \item $\exists x\in \mathbb Z,\, (\forall y\in \mathbb Z,\, x+y\neq 0)$ - \end{enumerate} -\end{proof} - -\begin{exercise} - There is another quantifier that we did not cover in the lecture, namely - $\exists!$ (read ``there exists exactly one''). For example, the sentence - ``\emph{there exists exactly one natural number x such that x+2=5}'' can be - written in symbols as ``$\exists!x\in \mathbb N,\,x+2=5$''. - - In this exercise, your task is to give a formal definition of this quantifier - using the logical symbols that we have defined in class. In particular, you - will need the following: - \begin{itemize} - \item the universal ($\forall$) and existential ($\exists$) quantifiers - \item the conjunction $\land$ - \item the implication $\implies$ - \end{itemize} - Moreover, you will need the equality symbol $=$ between two elements of a set - (if $a$ and $b$ are two elements of the same set, ``$a=b$'' is a mathematical - statement and it is \textbf{true} if and only if $a$ and $b$ are the same - element). - - \emph{Warning: your definition must depend on a set $S$ and on a ``variable - statement'' $A(x)$, as the existential and universal quantifiers.} -\end{exercise} -\begin{proof}[Solution] - The idea is that we want to write ``\emph{there is $x\in S$ such that $A(x)$ - is true, \textbf{and}, for any other $y\in S$, $A(y)$ is false}. From this - we see that the structure of the statement is - \begin{align*} - \exists x\in S,\,(\text{``something''}\land\text{``something else''}) - \end{align*} - The ``something'' part is just $A(x)$. The ``something else'' part can be - written in different ways, for example - \begin{align*} - \forall y\in S,\,(y\neq x\implies \neg A(y)) \quad \text{or} - \quad \forall y\in S,\,(A(y)\implies y=x) - \end{align*} - (notice that the two implications above are one the contrapositive of the - other); or also - \begin{align*} - \forall y\in S\setminus \{x\},\, \neg A(y) - \end{align*} - So in conclusion, one way to define ``$\exists!$'' is the following: - \begin{align*} - \exists!x\in S,\,A(x)\quad:=\quad\exists x\in S,\,(A(x)\land - (\forall y\in S,\,(A(y)\implies y=x))) - \end{align*} - -\end{proof} - -\section{Proofs} - -\begin{exercise} - Prove by induction that - \begin{align*} - \forall n\in\mathbb N,\quad \sum_{k=1}^n(2k-1)=n^2 - \end{align*} - (here $\sum_{k=1}^n(2k-1)$ means $1+3+5+\cdots+ (2n-1)$). -\end{exercise} -\begin{proof}[Solution] - \textbf{Base case:} for $n=0$ the sum is empty, so we have $0=0$ which is - true. (If you do not think that the formula makes sense for $n=0$, you can do - prove it for $n\geq 1$ and fo $n=1$ as a base case.) - - \textbf{Inductive step:} we can assume that the formula works for a generic - (but fixed) $n\in\mathbb N$ and prove that then it also works for $n+1$. So: - \begin{align*} - \sum_{k=1}^{n+1}(2k-1)&=\left(\sum_{k=1}^{n}(2k-1)\right)+2(n+1)-1=\\ - &=n^2 +2(n+1)-1=\\ - &=n^2+2n+1=\\ - &=(n+1)^2 - \end{align*} - which is the formula for $n+1$. -\end{proof} - -\begin{exercise} - If $n\in \mathbb N$ the \emph{factorial} of $n$, denoted by $n!$ is defined - as follows: - \begin{align*} - n!=\begin{cases} - 1&\text{if } n=0,\\ - n\times (n-1)! & \text{if } n> 0. - \end{cases} - \end{align*} - Prove by induction that if $n\geq 4$ then $n!\geq 2^n$. -\end{exercise} -\begin{proof}[Solution] - \textbf{Base case:} here the base case is $n=4$, and we have - $4!=24\geq 16=2^4$. - - \textbf{Inductive step:} we can assume that the inequality is true for $n$, - and prove that it is also true for $n+1$. We have: - \begin{align*} - (n+1)!=(n+1)\times n!\geq (n+1)\times 2^n\geq 2^{n+1} - \end{align*} -\end{proof} - -\begin{exercise} - Is the following statement true or false? Give a proof of your answer. - \begin{align*} - \forall n\in \mathbb N,\, n^2 -4n +5>n - \end{align*} -\end{exercise} -\begin{proof}[Solution] - The statement is \textbf{false}. Since it starts with a universal - quantifier, in order to prove that it is false we just need to provide one - example of $n\in \mathbb N$ which makes it false. In other words, we need to - prove that - \begin{align*} - \exists n\in \mathbb N,\, n^2-4n+5\leq n - \end{align*} - and a proof of this fact is very simple: for $n=2$ we have - $2^2-4\times 2+5=1\leq 2$. -\end{proof} - -\begin{exercise} - Do the last point of Exercise 1.1 again, but this time give a proof of your - answer. -\end{exercise} -\begin{proof}[Solution] - Again, since we have to prove that the statement is false, we just need to - show one counterexample. for example, the number - $2^{2^5}+1=4294967297=641\times 6700417$ is not prime. -\end{proof} - - - - -\end{document}