deletions | additions       diff --git a/untitled.tex b/untitled.tex  index be82fa2..4b4fb0d 100644  --- a/untitled.tex  +++ b/untitled.tex 
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a) $\neg$M(Chou,Koko).\\  b) $\neg$[M(Arlene,Sarah) $\vee$ T(Arlene,Sarah)].\\  c) $\neg$M(Deborah,Jose).\\  d) $\forall$xM(x,Ken).\\ $\forall$x M(x,Ken).\\  e) $\forall$x$\neg$T(x,Nina).\\ $\forall$x $\neg$T(x,Nina).\\  f) $\forall$x[M(x,Avi) $\forall$x [M(x,Avi)  $\vee$ T(x,Avii)].\\ T(x,Avi)].\\  \\  11)\\  a)$\exists$x($C_($$_x$$_)$$\wedge$$D_($$_x$$_)$$\wedge$$F_($$_x$$_)$)\\ 
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p $\rightarrow$ q \\  q $\rightarrow$ r\\  $\therefore$ p $\rightarrow$ r  \end{center} \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.  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)\\  Use a direct proof to show that the product of two odd numbers is odd.  \paragraph{Solution} Assume that $2n + 1$ is an odd number (n is a natural number). We have the product of two number is $(2n + 1)^{2}$ \\  $(2n + 1)^{2}$ = $4n^{2} + 4n +1$ = $2(2n^{2} + 2n) +1$ 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.\\  17)\\  Use a direct proof to show that the product of two odd numbers is odd.  \paragraph{Solution} Assume that $2n + 1$ is an odd number (n is a natural number). We have the product of two number is $(2n + 1)^{2}$ \\  $(2n + 1)^{2}$ = $4n^{2} + 4n +1$ = $2(2n^{2} + 2n) +1$ is an odd 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