Exercise 6A

Date: 2023-02-06

first we prove that any number equals zero (UNRELATED AND WRONG BUT FUNNY)

$$\begin{array}{r}\text{Assume that }a=b\text{, where }a\text{ and }b\in \mathbb{R}\\ \Rightarrow ab=a^2\\ \Rightarrow ab-a^2=a^2-a^2\\ \Rightarrow a(b-a)=0\\ \Rightarrow a=0\\ \therefore \text{all real numbers are equal to 0}\\ & \text{QED}\end{array}$$

Question 1a

$\begin{array}{r}\text{Assume that }m\text{ is even and }n\text{ is even.}\\ \text{Then }m=2k\text{ and }n=2l\text{ where }k\text{ and }l\text{ are integers.}\\ \Rightarrow m+n=2k+2l\\ =2(k+l)\\ \text{Hence }m+n\text{ is even, since }k+l\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 1b

$\begin{array}{r}\text{Assume that }m\text{ is even and }n\text{ is even.}\\ \text{Then }m=2k\text{ and }n=2l\text{ where }k\text{ and }l\text{ are integers.}\\ \Rightarrow mn=2k\times 2l\\ =4kl\\ =2(2kl)\\ \text{Hence }mn\text{ is even, since }2kl\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 2

$\begin{array}{r}\text{Assume that }m\text{ is odd and }n\text{ is odd.}\\ \text{Then }m=2k+1\text{ and }n=2l+1\text{, where }k\text{ and }l\text{ are integers.}\\ \Rightarrow m+n=2k+2l+2\\ =2(k+l+1)\\ \text{Hence }m+n\text{ is even, since }k+l+1\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 3

$\begin{array}{r}\text{Assume that }m\text{ is even and }n\text{ is odd.}\\ \text{Then }m=2k\text{ and }n=2l+1\text{, where }k\text{ and }l\text{ are integers.}\\ \Rightarrow mn=4kl+2\\ =2(kl+1)\\ \text{Hence }mn\text{ is even, since }kl+1\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 4a

$\begin{array}{r}\text{Assume that }m\text{ is divisible by 3 and }n\text{ is divisible by 7.}\\ \text{Then }m=3k\text{ and }n=7l\text{, where }k\text{ and }l\text{ are integers.}\\ \Rightarrow mn=21kl\\ \text{Hence }mn\text{ is divisible by 21.}\\ & \text{QED}\end{array}$

Question 4b

$\begin{array}{r}\text{Assume that }m\text{ is divisible by 3 and }n\text{ is divisible by 7.}\\ \text{Then }m=3k\text{ and }n=7l\text{, where }k\text{ and }l\text{ are integers.}\\ \Rightarrow m^{2}n=63k^{2}l\\ \text{Hence }{m}^{2}n\text{ is divisible by 63.}\\ & \text{QED}\end{array}$

Question 5

$\begin{array}{r}\text{Assume that }m\text{ and }n\text{ are perfect squares.}\\ \text{Then }m=k^2\text{ and }n=l^2\text{, where }k\text{ and }l\text{ are integers.}\\ \Rightarrow mn=k^2{l}^2\\ =(kl{)}^2\\ \text{Hence }mn\text{ is a perfect square, since }kl\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 6

$\begin{array}{r}\text{Assume that }m\text{ and }n\text{ are integers.}\\ \text{Using the binomial expansion, we can turn }\\ (m+n{)}^2-(m-n{)}^2\\ \text{ into }\\ m^2+2mn+n^2-m^2+2mn-n^2\\ =4mn\\ \text{Hence, }(m+n{)}^2-(m-n{)}^2\text{ is divisible by 4, since }mn\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 7

$\begin{array}{r}\text{Assume }n\text{ is an even integer.}\\ \text{Then }n=2m\text{, where m is an integer.}\\ \text{Subtituting }n\text{ for }2m\text{ in the expression, we get }\\ (2m{)}^2-12m+5\\ =2(2m^2-6m)+5\\ \text{Hence, }{n}^2-6n+5\text{ is odd, since }2m^2-6m\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 8

$\begin{array}{r}\text{Assume }n\text{ is an odd integer.}\\ \text{Then }n=2m+1\text{, where m is an integer.}\\ \text{Subtituting }n\text{ for }2m+1\text{ in the expression, we get}\\ (2m+1{)}^2+16m+8+3\\ =4m^2+20m+12\\ =2(2m^2+10m+6)\\ \text{Hence, }{n}^2+8m+3\text{ is even, since }2m^2+10m+6\text{ is an integer.}\\ & \text{QED}\end{array}$

Question 9

$\begin{array}{r}\text{First assume that }n\text{ is an even number.}\\ \text{Then }n=2m\text{, where }m\text{ is an integer.}\\ \text{Subtituting }n\text{ for }2m\text{ in the expression, we get}\\ 5(2m{)}^2+6m+7\\ =20m^2+6m+7\\ =2(10m^2+3m+3)+1\\ \text{Hence, }5n^2+3n+7\text{ is odd if }n\text{ is even, since }10m^2+3m+3\text{ is an integer.}\\ \\ \text{Now assume that }n\text{ is an odd number.}\\ \text{Then }n=2m+1\text{, where }m\text{ is an integer.}\\ \text{Substituting }n\text{ for }2m+1\text{ in the expression, we get}\\ =5(2m+1{)}^2+3(2m+1)+7\\ =5((2m{)}^2+4m+1)+6m+3+7\\ =20m^2+26m+15\\ =2(10m^2+13m+7)+1\\ \text{Hence, }5n^2+3n+7\text{ is odd if }n\text{ is odd, since }10m^2+13m+7\text{ is an integer.}\\ \\ \text{Since in both cases, the expression is odd, the expression is always odd.}\\ & \text{QED}\end{array}$

Question 10

$\begin{array}{r}\text{Assume that }x\text{ and }y\text{ are positive real numbers.}\end{array}$