If f(x) is an increasing function on domain [a.b], then if f(x) < f(y), then x < y.

  Let F be increasing on [a.b].

(Definition of increasing: if x \(\le\) y, then f(x) \(\le\) f(y))

If f(x) <  f(y), then x < y.

By Contrapositive:  If x \(\ge\) y, then f(x) \(\ge\) f(y).

The contrapositive matches the definition of an increasing function(with the variables switched), thus proving that if f(x) < f(y), then x < y.

Let \(m,n \in \mathbb{Z}\). If \(m\) and \(n\) are both odd, then \(m^2+n^2\) is not prefect square

 By way of contradiction, assume \(m^2+n^2=p^2\) where \(\exists m,n,p \in \mathbb{Z}\). Since \(m\) and \(n\) are both odd \(m\) can be written a \(m=2k+1\) and \(n\) can be written as \(n=2j+1\). Therefore, \[\begin{align} m^2+n^2=(2k+1)^2+(2j+1)^2 \\=(4k^2+4k+1)+(4j^2+4j+1) \\=2(2k^2+2k)+1+2(2j^2+2j)+1 \\=2(2k^2+2k)+2(2j^2+2j)+2 \end{align}\].

Since \(2(2k^2+2k)\) and \(2(2j^2+2j)\) must be even, \(p^2\) must be even and therefore \(p\) must also be even (previously proved). This means \(p\) can be written as \(p=2\ell\). Thus, \(m^2+n^2=p^2\) can be written as \[\begin{align}2(2k^2+2k)+2(2j^2+2j)+2=4\ell^2 \\ (2k^2+2k)+(2j^2+2j)+1=2\ell^2 \end{align}\] 

Since \((2k^2+2k)+(2j^2+2j)+1\) is odd, and \(2\ell^2\) is even this is a contradiction, so \(m^2+n^2\) cannot be a perfect square if \(m\) and \(n\) are odd.

If \(n,m\in\mathbb{N}\) and \(k\in\mathbb{N}\), \(k < n\), then \(mn + k\) is not divisible by \(n\).

 Lemma:

pf. Suppose j|a and j|b. Then 

\(a=rj\) and \(b=sj\) \(\exists \ r,s \in \mathbb{Z}\)

\(b-a = rj-sj \ \implies \ (b-a)/j\)

\(mn=a\)

\(n|mn\)

By way of contradiction, suppose \(n|(mn+k)\) => \((mn+k)=b\)

By Lemma, \(n|(b-a)\) => \(n|((mn+k)-(mn))\) => \(n|k\)

\(n|k\) = \(np\) = k,

\(\exists \ p \in \mathbb{Z}\)

\(p\) = \(k/n\) => since \(n \ne 0\)

0 < \(k/n\) < 1

Since by way of contradiction  \(n|(mn+k)\) is possible, then \((mn+k)|n\) is also possible.

The sum and product of two rational numbers are rational.

Let \(a=\frac{m}{n}\) and \(b=\frac{p}{q}\) for some integers \(m,n,p,q\) (\(n,q\neq 0\))

(i) 

\[a+b=\frac{m}{n}+\frac{p}{q}=\frac{mq+pn}{nq}\]

Since \(m,n,p,q\) are all integers, \(a+b\) is rational.

(ii)

\[ab=\frac{m}{n}\times \frac{p}{q}=\frac{mp}{nq}\]

Since \(m,n,p,q\) are all integers, \(ab\) is also rational.

\(f(x)=x^2\) is strictly increasing on \([0, \infty) \) i.e. \(x < y \implies f(x) < f(y) \)

 Prove: f(x) = \(x^2\) is strictly increasing on [0, \(\infty\)) i.e. x < y \(\rightarrow\)  f(x) < f(y)

Case 1: x = 0

Assume x < y. 

When x = 0, y > 0 we have \(x^2\) = 0 and \(y^2\) > 0

\(\therefore\) we have shown when x = 0, f(x) < f(y)

Case 2: x \(\neq\) 0

Assume x < y

Then multiply both sides of the inequality x < y by x and y

After multiplying both sides of the inequality we have: \(x^2\) < xy and yx < \(y^2\) 

Now if we combine these two inequalities together we have: \(x^2\) < xy < \(y^2\)

Using the transitive property: a < b and b < c \(\rightarrow\) a < c we get: \(x^2\) < \(y^2\)

\(\therefore\) we have shown when x \(\neq\) 0, f(x) < f(y)

In conclusion, we have shown if x < y \(\rightarrow\) f(x) < f(y)

If a,b,c are odd integers, then \(ax^2+bx+c=0\) has no integer solutions

The two cases we must prove are x is even and x is odd. We will prove \(ax^2+bx+c\) is odd for both cases to show that it cannot be equal to zero.

We will use the previously proved proposition "Two integers have the same parity if and only if their sum is even" to show that the sum of an even and an odd number is odd.

If the previous propositon is true, then the proposition "two integers have opposite parity if and only if their sum is odd". (\(P \leftrightarrow Q \equiv \neg P \leftrightarrow \neg Q\))

We will then use this statement and its negation to show that the sum of three odds is odd. this is equivilant to two odds added, which is even, added to another odd which is the opposite parity making the total sum odd.

(i) If x is even, then \(x=2k\). \[ \begin{align} ax^2+bx+c &= a(2k)^2+b \cdot 2k+c \\ &=4k^2a+2kb+c \\ &=2 \cdot(k^2a+kb)+c\\ \end{align} \] Since this takes the form \(2 \cdot int+odd\), \(ax^2+bx+c \) is odd

(ii) If x is odd, \(x=2k+1\). \[ \begin{align} ax^2+bx+c &= a(2k+1)^2+b(2k+1)+c\\ &=a(4k^2+4k+1)+2kb+b+c\\ &=(a+b+c)+2 \cdot(2k^2a+2ka+kb) \end{align} \]
Therefore \(2 \cdot(2k^2a+2ka+kb)+(a+b+c)\) is in the form \(2 \cdot int+odd\) and is odd

Two integers have the same parity if and only if their sum is even

(i) Let \(m\) and \(n\) be two integers having the same parity. Then \(m\) and \(n\) are either both even or both odd.

Case 1. (\(m\) and \(n\) both even.) If \(m\) and \(n\) are both even, then \(m=2k\) and \(n=2j\) for some integers \(k,j\). Then \[m+n=2k+2j=2(k+j)\] and hence \(m+n\) is even.

Case 2. (\(m\) and \(n\) both odd). If \(m\) and \(n\) are both odd, then \(m=2k+1\) and \(n=2j+1\) for some integers \(k,j\). Then \[m+n=(2k+1)+(2j+1)=2(k+j+1)\] and hence \(m+n\) is even.

(ii) We will prove the reverse implication, that two integers have the same parity if their sum is even, by contraposition. Hence, suppose that \(n\) and \(m\) have opposite parity. Without loss of generality we may assume that \(m\) is even and \(n\) is odd. Then \(m=2k\) and \(n=2j+1\) for some integers \(k,j\). So \[m+n=2k+(2j+1) = 2(k+j)+1,\] which proves that \(m+n\) is odd. Thus by contraposition we have proved that if their sum is even, two integers have the same parity.

An integer is even if and only if its square is even

(i) Let \(n\) be an even integer. Then \(n=2k\) for some integer \(k\). So \[n^2 = (2k)^2 = 4k^2 = 2 (2k^2) \] and since \(2k^2\) is an integer, \(n^2\) also satisfies the definition of an even integer.

(ii) To prove that \(n\) is even if \(n^2\) is even, we proceed by contraposition. Hence, suppose that \(n\) is not even and therefore odd. Then \(n=2k+1\) for some integer \(k\). So \[n^2=(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1\] and thus is also odd. Thus by contraposition, we have proved that if \(n^2\) is even, \(n\) is even.