example 1
Theorem: every odd number can be written as a difference of squares.
Let $n = 2a +1 $
then $2a +1 = a^2 + 2a + 1 - a^2= (a+1)^2- a^2$
example 2
“The square of an odd number is odd.”
$$a = 2 \cdot n +1 $$
$$a^2 = (2 \cdot n +1)^2 $$
$$ (2 \cdot n +1)^2 = 4 \cdot n^2 + 4 \cdot n + 1$$
$$ 4 \cdot n^2 + 4 \cdot n + 1 = 2 \cdot (2 \cdot n^2 + 2 \cdot n) +1$$
Hence, $a^2$ is odd as it has the pattern $2 \cdot k +1 $
example 3
proof 1
Find a counterexample by trial and error. If n = 4, then the expression is equal to 25. Therefore, the hypothesis is false.
proof 2
Expand the expression. Then we have $n \cdot n +n+5$. Now, we take n=5 so that every term contains the factor 5 and we can factorise 5. Evaluating the expression gives
$5^2 +5+5 = 5 \cdot (5+1+1) = 5 \cdot 7 = 35$
task 1
Prove that following implications are false.
- For all natural numbers n (
$n \in \mathbb{N}$),$n^2+n+17$is prime. - For all
$n = a$,$n^2+n+a$is not a prime number.
Let n = 17. Then 17 can be factorised and we get $17 \cdot (17 + 1 + 1) = 17 \cdot 19 = 327$
$n^2+n+n = (n + 2) \cdot n $. The factors of $(n + 2) \cdot n $ are 1, n and (n+2). For n = 1, we get the factors 1 and 3, and 3 is a prime number. Therefore, the statement is only true for n > 1.
example 5
Let $x, y \in \mathbb{Z}$. If x + y is even, then x and y have the same parity.
proof
${\neg Q}$: Suppose $x$ and $y$ have opposite parity. Without loss of generality (WLOG), let $x$ be even, so $y$ is odd.
Then there exist integers a and b such that $x = 2a$ and $y = 2b+1$. Then $x+y = 2a + 2b + 1 = 2(a+b) + 1$.
Since, $a+b$ is an integer, $x + y$ is odd. Which is ${\neg P}$
So, because ${\neg Q \Rightarrow \neg P}$ we can say that ${P \Rightarrow Q}$
example 7
For all numbers a, b, where $a \geq 0$ and $b \geq 0$ the following is true:
$$a+b \geq 2 \cdot \sqrt{a \cdot b}$$
proof by contradiction (Widerspruchsbeweis)
Assume the opposite. $$a+b \lt 2 \cdot \sqrt{a \cdot b}$$ It follows that
$${\begin{align}(a+b)^2 & \lt 4 \cdot a \cdot b \newline a^2+2 \cdot a \cdot b + b^2 & \lt 4 \cdot a \cdot b \newline a^2+2 \cdot a \cdot b + b^2 - 4 \cdot a \cdot b & \lt 0\newline a^2 - 2 \cdot a \cdot b + b^2 & \lt 0\newline (a-b)^2 & \lt 0 \end{align}}$$
This is a contradiction, as for all square numbers $(a-b)^2 \geq 0$.
direct proof
$${\begin{align}(a-b)^2 & \geq 0 \newline a^2 - 2 \cdot a \cdot b + b^2 & \geq 0 \newline a^2-2 \cdot a \cdot b + b^2 + 4 \cdot a \cdot b & \geq 4 \cdot a \cdot b \newline a^2+2 \cdot a \cdot b + b^2 & \geq 4 \cdot a \cdot b \newline (a+b)^2 & \geq 4 \cdot a \cdot b \newline a+b & \geq 2 \cdot \sqrt{a \cdot b} \end{align}}$$
direct proof using the geometric mean (Höhensatz)
The geometric mean states that
“the square of the altitude of a right triangle in a semi-circle is equal to the product of the hypotenuse segments produced by the altitude.”
i.e., $h^2 = a \cdot b$. Therefore $h = \sqrt{a \cdot b}$. But the arithmetic mean of a and b is equal to the radius of the semi-circle.
$$\frac{a+b}{2} \geq \sqrt{a \cdot b}$$