A calculus of the absurd

5.3.3 Showing uniqueness

If we want to show that a given mathematical object is “unique” (i.e. there is at most one) we can do this by contradiction - assume that there is more than one object, and show that this implies a contradiction.

  • Example 14 Let \(x\) be a real number. Prove that there exists at most one number \(y\) such that \(x \cdot y = 1\).

This might seem obvious, but we can actually prove it from other facts about the real numbers. Let us assume that there exist more than one values which satisfy this property, \(a\) and \(b\) with the very important condition that they are distinct (that is, that \(a \ne b\)). Then, we can write that

\begin{align} & x \cdot a = 1 = a \cdot x \\ & x \cdot b = 1 = b \cdot x \label {b def} \end{align}

Then, we can consider the value of \(a\),

\begin{align} a &= a \cdot 1 \tag {By the definition of 1} \\ &= a \cdot (x \cdot b) \tag {Because of Equation \ref {b def}} \\ &= (a \cdot x) \cdot b \tag {Because of associativity} \\ &= 1 \cdot b \\ &= b \end{align}

So \(a=b\), but we specified that \(a \ne b\)! We have found a contradiction, and thus it must be the case that there is at most one inverse!

\(\Box \)