deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 1e06220..b2143a1 100644  --- a/untitled.tex  +++ b/untitled.tex 
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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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\\  26)$n$ is a positive integer.\\  If $n$ is an odd number, we have $n = 2k + 1$, $5n + 6 = 10k + 11 = 10(k + 1)+1 = 2m + 1$ is an odd number (where $m = 5(k + 1)).\\  If $5n + 6$is 6$ is  an odd number, we have $5n + 6 = 2k + 1 \leftrightarrow 5(n + 1) = 2k \rightarrow 5(n + 1)$ is an even number so $n + 1$ is also an even number and $n$ is an odd number.\\ In conclusion, if $n$ is a positive integer, $n$ is odd if and only if $5n + 6$ is odd. odd.\\  27)\\  we need to prove that all these equivalent to x being even. Then x=2k( k is integer). So, 3x+2=6x+2 is even.Similarly, x+5=2x+5=2x+4+1 is odd. $x^2$=4$k^2  $ is even.\\ 
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$\hspace*{3.8cm}$= $\frac{(k+1)(k+2)(2k+3)}{6}$  \end{center}  shows that P (k + 1) is true under the assumption that P (k) is true.\\  30)\\  Prove that $2^n > 2n$ for every positive integer $n > 2$.\\  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 $n > 2$.\\  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 \\