Formal proof of this simple equation?

Question

What is the formal proof of $1^{-1}=1$. Intuitively it makes a lot of sense but how would the formal proof go ?

Answer 1

By definition of $1$, $1^{-1} = 1 * 1^{-1}$. By definition of inverse $1 * 1^{-1} = 1$. Hence, $1^{-1} = 1$.

Answer 2

You can't prove it. You have to define your terms and declare your axioms first and see if it follows.

There are two approaches, both valid.

If we declare $b^n=b*b*b*.... $ a positive integer number of times, it follows by induction and associativity that $b^nb^m=b^{n+m} $. We can therefore declare axiomatically that $b^z $ for $z=1,0$ and negative integers is whatever keeps $b^nb^m=b^{n+m} $. That would mean $b^0b^n=b^n $ so $b^0=1$. $b^1b^n=b^{n+1}=b*b^n $ so $b^1=b$ so $b^{n}b^{-n}=b^0=1$ so $b^{-n}=\frac 1 {b^n} $. Hence $1 ^{-1}=1/1=1$.

Or (this is typical of abstract algebra) you can simply define that $b^{-1} $ is the symbol use to indicate the number, $x $ so that $b*x=1$. And that number for $1$ is $1*x=1\implies x=1$ so $1^{-1}=1$.

But we should note that if we take that approach than we have to prove the following statement: $(b^n)^{-1}=(b^{-1})^n $. That can not be taken for granted.

$b^n*(b^{-1})^n=b*(b*(b*....(b*b^{-1})*...*)b^{-1} =1$ so $(b^n)^{-1}=(b^n)^{-1}$. If we decide to use the symbol $b^{-n} $ for that number, and if we declare by fiat that $b^0=1$, then we have defined $b^z $ for all integers.

But we can no longer take it for granted that $b^kb^j=b^{k+j} $ if either $k $ or $j $ is non-positive.

But that is easily verified : if $k,j $ are positive then $b^{-k}b^{-j}=(b^{-1})^k (b^{-1})^j=(b^{-1})^{k+j}=b^{-k-j}$. If $j <k $ then $b^kb^{-j}=b^{k-j}b^jb^{-j}=b^{k-j}*1=b^{k-j} $. Etc.