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 \(\sqrt{2}= \dfrac {a}{b}\), where \(a \in \mathbb{Z}, b \in \mathbb{Z}\setminus \{0\}\), 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 \(1 + 2 + ... + k + (k + 1) \)

\(= \displaystyle \frac{k(k + 1)}{2} + (k + 1)\)

\(= (k + 1) \left( \displaystyle\frac{k}{2} + \displaystyle\frac{1}{1}\right)\)

\(= (k + 1) \left( \displaystyle\frac{k + 2}{2}\right)\)

\(= \displaystyle \frac{(k + 1)(k + 2)}{2}\).

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 .◻