0.2: Introduction to Proofs/Contradiction

In this section, we will explore different techniques of proving a mathematical statement "If $p$ then $q$." ($p \to q$).

Direct Proof

In this technique, we shall assume $p$ and show that $q$ is true.

Theorem $\PageIndex{1}$

Let $n$ be an integer. If $n$ is even then $n^2$ is even.

Proof

Assume that $n$ is even. Then $n=2m$ for some integer $m$.

Consider $n^2=(2m)^2=4m^2=2(2m^2).$ Since $m$ is an integer, $(2m^2)$ is an integer.

Thus $n^2$ is even.

Example $\PageIndex{1}$

Show that for all integers $n$, if $n$ is odd then $n^2$ is odd.

Answer

Assume that $n$ is odd. Then $n=2m+1$ for some integer $m$.

Consider $n^2=(2m+1)^2=4m^2+4m+1=2(2m^2+2m)+1.$

Since $m$ is an integer, $(2m^2+2m)$ is an integer.

Thus $n^2$ is odd.

Proof by Contrapositive

In this technique, we shall assume $\neg  p$ and show that $\neg q$ is true.

Theorem $\PageIndex{2}$

Let $n$ be an integer. If $n^2$ is even then $n$ is even.

Proof

We shall prove this statement by assuming $n$ is odd. Then $n=2m+1$ for some integer $m$.

Consider $n^2=(2m+1)^2=4m^2+4m+1=2(2m^2+2m)+1.$

Since $m$ is an integer, $(2m^2)+2m$ is an integer.

Thus $n^2$ is odd.

Example $\PageIndex{2}$

Show that for all integers $n$, if $n^2$ is odd then $n$ is odd.

Answer

We shall prove this statement by assuming $n$ is even. Then $n=2m$ for some integer $m$.

Consider $n^2=(2m)^2=4m^2=2(2m^2).$

Since $m$ is an integer, $(2m^2)$ is an integer. Thus $n^2$ is even.

Proof by Contradiction

In this technique, we shall assume the negation of the given statement is true and come to a contradiction.

Theorem $\PageIndex{3}$

$\sqrt{2}$ is irrational.

Proof

Assume that $\sqrt{2}$ is rational. Then $\sqrt{2}= \dfrac {b}{a}$, where $b\in \mathbb{Z}, a \in \mathbb{Z}\setminus \{0\}$, with no common factors between $a$ and $b$. Now, $\sqrt{2} a=b$. Then $2a^2=b^2$. Since $2$ divides $2a^2$, $2$ divides $b^2$. Thus $b^2$ is even. Therefore, $b$ is even (by theorem 2). Since $b$ is even, $2$ divides $b$. Therefore, $2^2$ divides $b^2$.

Since $2a^2=b^2$, $2^2$ divides $2a^2$. Therefore, $2$ divides $a^2$. Which implies $a$ is even. This contradicts the fact that $a$ and $b$ have no common factors. Thus $\sqrt{2}$ is irrational.

Proof by Counterexample

Example $\PageIndex{3}$:

Decide whether the statement is true or false and justify your answer:

For all integers $a,b,u,v$, and $u\ne 0, v \ne 0$, if $au+bv =0$ then $a=b=0.$

Solution: The statement is false.

Counterexample: Choose $a=1,b=-1, u=2,v=2$, then $au+bv =0$, but $a\ne 0. b \ne 0, a \ne b.$

Proof by induction

In mathematics, we use induction to prove mathematical statements involving integers. There are two types of induction: regular and strong. The steps start the same but vary at the end. Here are the steps. In mathematics, we start with a statement of our assumptions and intent:

Let $p(n) \forall n \geq n_0, \, n, \, n_0 \in \mathbb{Z}$ be a statement. We would show that p(n) is true for all possible values of n.

1. Show that p(n) is true for the smallest possible value of n: In our case $p(n_0)$. AND
2. For Regular Induction: Assume that the statement is true for $n = k,$ for some integer $k \geq n_0$. Show that the statement is true for n = k + 1.

OR

For Strong Induction: Assume that the statement p(r) is true for all integers r, where $n_0 ≤ r ≤ k$ for some $k ≥ n_0$. Show that p(k+1) is true.

If these steps are completed, and the statement holds, we are saying that, by mathematical induction, we can conclude that the statement is true for all values of $n \geq n_0.$

Example $\PageIndex{4}$:

Prove that $1 + 2 + ... + n = \displaystyle \frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}$.

Solution:

Base step: Choose $n = 1$. Then L.H.S =$1$ and R.H.S $= \frac{(1)(1 + 1)}{2}=1$

Induction Assumption: Assume that $1 + 2 + ... +k= \displaystyle\frac{k(k + 1)}{2}$, for $k \in \mathbb{Z}$.

We shall show that $1 + 2 + ... + k + (k + 1) = \displaystyle\frac{(k + 1)[(k + 1) + 1]}{2} = \frac{(k + 1)(k + 2)}{2}$

Consider,

$$\begin{align*}1 + 2 + ... + k + (k + 1) &= \displaystyle \frac{k(k + 1)}{2} + (k + 1)\\&= (k + 1) \left( \displaystyle\frac{k}{2} + \displaystyle\frac{1}{1}\right)\\&= (k + 1) \left( \displaystyle\frac{k + 2}{2}\right)\\&= \displaystyle \frac{(k + 1)(k + 2)}{2}.\end{align*}$$

Thus, by induction we have $1 + 2 + ... + n = \displaystyle\frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}$.

Example $\PageIndex{5}$

Prove   for .

Solution

Let . Then is true, since clearly . Thus, the statement is true for .

Assume that is true for some .

We will show that .

Consider .

Since and , we have .

Thus the statement is true for all .

By induction, for all .◻