## Example 1

Question:

Prove that there is only one circle with $AB$ as its diameter.

Assumption:

Assume that there are 3 circles $C_1$, $C_2$, and $C_3$ passing through the points $A$ and $B$. $C_1$ and $C_2$ are concentric and $C_1$ and $C_3$ are not concentric. $C_1$ and $C_2$ have different radii and $C_3$ has any radius. Let $C_1$ be on the midpoint of $AB$ such that $AB$ is its diameter.

Contradicting Arguments:

- As $C_1$ and $C_2$  have different radii, points $A$ and $B$ cannot be on the circle $C_2$.

- As $C_3$ is not on the middle of $AB$, $AB$ cannot be its diameter.

Conclusion:

So there is only one circle $C_1$ with $AB$ as its diameter.


## Example 2

Question:

Prove that $\sqrt{2}$ is an irrational number.

Assumption:

Let $\sqrt 2$ be a rational number. So it can be represented as $\sqrt{2}=\frac{m}{n}$ where $m$ and $n$ are natural numbers without common factors other than $1$.

Contradicting Arguments:

Squaring both sides, we get
\begin{align}
2 &=\frac{m^2}{n^2}\\
m^2 &= 2n^2
\end{align}

Because $m^2$ is a multiple of $2$ then $m^2$ is an even number. Recall that

> The square of an even number is even.

it implies that $m$ is also even. Let $m=2k$ where $k$ is any natural number. Substituting   it for $m$, we get
\begin{align}
(2k)^2 &=2n^2\\
4k^2 &= 2 n^2\\
n^2 &= 2k^2
\end{align}

With the same reasoning, $n$ is even. As both $m$ and $n$ are even numbers, 2 becomes their common factors so it contradicts the assumption that they have no common factors other than 1.

Conclusion:

$\sqrt 2$ cannot be represented as a ratio of two natural numbers without common factor other than 1. It implies that $\sqrt 2$ is irrational.