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\hline   \end{tabular}   \end{center}  \\  2) Negation:\\  a) Steve does not has more than 100 GB free disk space on his laptop.\\  b) Zach does not blocks e-mails and texts from Jennifer.\\  c) 7.11.13 $\ne$ 999.\\  d) Diane did not ride her bicycle 100 miles on Sunday.\\  \\  3)\\  a) x=2\\  b) x=1\\  c) x=2\\  d) x=2\\  e) x=2\\  \\  4)\\  a) converse:"I will ski tomorrow only if it snows today"\\  Contrapositive:"If i don't ski tomorrow, then it will not have snowed today."\\ 
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Contrapositive:"If i don't come to class, then there won't be a quiz"\\  c) Converse:"A positive integer is a prime if it has no divisors other than 1 and itself"  Contrapositive:"If a positive integer has a divisor other than 1 and itself, then it is not prime"\\  \\  5)\\  Let p, q and r be the propositions \\  p : You have the flu. \\ 
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You will not pass the course if you have the flu or you miss the final examination.  \item[f)] (p $\wedge$ q) $\vee$ ($\neg$q $\wedge$ r) \\  You have the flu and you miss the final examination or you do not miss the final examination and you pass the course.\\  \\  6)\\  a) r $\wedge$$\neg$q\\  b) p $\wedge$ q $\wedge$ r\\ 
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d) p $\wedge$ $\neg$q $\wedge$ r\\  e) r $\rightarrow$ (p $\wedge$ q) \\  f) r $\leftrightarrow$ (q$\vee$p)\\  \\  7)\\  a)11000\\  b)01101\\  \\  8) parts (a) and (b)\\  \begin{table}   \begin{tabular}{ |c| |c| |c| |c| |c| |c| } 
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= $\neg$($\neg$p $\vee$ ($\neg$q $\wedge$ r))\\  = p $\wedge$ $\neg$($\neg$q $\wedge$ r)\\  = p $\wedge$ (q $\vee$ $\neg$r)\\  \\  11) a$\rightarrow$e.\\  12) (e$\wedge$p)$\rightarrow$m\\  13)\\ 
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b) Every body is a funny comedian.\\  c) If there is a person who is a comedian, then he or she is funny.\\  d) Some comedians are funny.\\  \\  7) \\  a)$\exists$x($P_($$_x$$_)$$\wedge$$Q_($$_x$$_)$)\\  b)$\exists$x($P_($$_x$$_)$$\wedge$$\neg$$Q_($$_x$$_)$)\\  
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d) $\forall$x M(x,Ken).\\  e) $\forall$x $\neg$T(x,Nina).\\  f) $\forall$x [M(x,Avi) $\vee$ T(x,Avi)].\\  \\  11)\\  a)$\exists$x($C_($$_x$$_)$$\wedge$$D_($$_x$$_)$$\wedge$$F_($$_x$$_)$)\\  b)$\forall$x($C_($$_x$$_)$$\wedge$$D_($$_x$$_)$$\wedge$$F_($$_x$$_)$)\\ 
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q $\rightarrow$ r\\  $\therefore$ p $\rightarrow$ r  \end{center}\\  \\  14)Thee premise is "Someone is happy", it is not said the exactly the name of that person so we can not say that is Lola or someone else, and we do not know that Lola is in the domain or not. not.\\  16)\\  Let a and b be odd integers. By definition of odd we have that a = 2n + 1 and b = 2m + 1. Consider the sum a + b = (2n + 1) + (2m +1) = 2n + 2m +2 = 2k, where k = n + m + 1 is an integer. Therefore by definition of even we have shown that a + b is even.\\  17)\\ 
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$(2n + 1)^{2}$ = $4n^{2} + 4n +1$ = $2(2n^{2} + 2n) +1$ is an odd number.\\  18)Use a direct proof to show that every odd integer is the difference of two squares.\\  Assume that $2n+1$ is an odd integer. We have that $2n+1$ = $n^2 + 2n + 1 - n^2$ = $(n+1)^2 - n^2$. So every odd integer is the difference of two squares.\\  \\  19) we will use direct proof.\\  Let n + m = 2k ( k$\in$ Z)\\  Let n + p = 2t ( t$\in$ Z)\\  we have m + p = 2(k+t-n) so m+p is even number.\\    20)\\  Let r and w be two rational numbers. By definition of rational r= $\frac{a}{b}$ and w=$\frac{c}{d}$ where a,b,c,d are integers and b$\ne$0, d$\ne$0.  Consider r+w= $\frac{a}{b}$ + $\frac{c}{d}$ = $\frac{ad+bc}{bd}$ where bd$\ne$0 is an integer and ad +bc is an 
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By algebra we see that r + (-p) = a/b - c/d = (ad - cb)/bd is a rational number. \\  r + (-p) = p + q - p = q. We have q is an irrational number. So r + (-p) is rational is incorrect.  In conclusion, r is irrational.\\  \\  22)We have $x$ is irrational. Suppose that $1/x$ is rational so that $1/x$ = $a/b$ where $a$ and $b$ are integers and $b$ $\ne$ $0$ $\rightarrow$ $x = b/a$ is a rational number ($x$ can write in fraction form), but $x$ is a irrational number. So $1/x$ is rational is incorrect.\\  In conclusion, if $x$ is irrational, then $1/x$ is irrational.\\  \\  23)\\  Assume that x$<$1 and y$<$1 so x+y$<$2 so the hypothesis statement is true,as required.\\  24)\\ 
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When $n = 3 \rightarrow 2^3 = 8 > 2*3 = 6$ so when $n = 3$ the statement is right.\\  Assume that when $n = k$ we have $2^k > 2k$.\\  When $n = k + 1 \rightarrow 2^k+1 = 2^k*2 > 2k+2 = 2(k+1)$, the statement is also right with $n = k + 1$.\\  So $2^n > 2n$ for every positive inreger integer  $n > 2$.\\ 32)\\  when n = 1, $6^1$ - 1 = 6 - 1 = 5   Assume $6^{n}$ - 1 divisible by 5, which means $6^{n}$ - 1 = 5a, for some integer a and so $6^{n}$ = 5a + 1 then \\