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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, 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)\\ 
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\item[b)] a proof by contradiction.  \paragraph{Solution} Let p is the proposition that $n$ is even.  Suppose $\neg$p is true, which means ``$n$ is odd". If we add subtract an odd number from an even number, we get an odd number, so $3n + 2 - n$ = $4n + 2$ = $2(2n + 1)$ is odd. But this is not true. Therefore our supposition was wrong, and the proof by contradiction is complete.\\  \\  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 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.  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.\\