Using only the axioms of real numbers prove that 1/0 does not make sense

[Aemilius]

At the beginning, I'd like to say hello. I'm new to this forum and this is my first post.

I have been thinking about how to prove that the expression \(\displaystyle \frac{1}{0}\) does not make sense, using only the axioms of real numbers. The problem is that I don't think there is any work to do here, as one of the axioms clearly states that the multiplicative inverse of \(\displaystyle a\) denoted by \(\displaystyle \frac 1 a\) exists for all \(\displaystyle a \ne 0\). Now, if we take a look at \(\displaystyle \frac{1}{0}\), it is supposed to be the multiplicative inverse of $0$ which, based on the axiom of multiplicative inverse, does not exist. Is it enough of a "proof"? <br>
Also, I've been wondering what way of solving this one should choose when the expression in question would instead be\(\displaystyle \frac{3}{0}\) or \(\displaystyle \frac 0 0\). They are problematic because expression like \(\displaystyle frac{x}{a}\) is not in the language of a field.

 

[barrick]

At the beginning, I'd like to say hello. I'm new to this forum and this is my first post.

I have been thinking about how to prove that the expression \(\displaystyle \frac{1}{0}\) does not make sense, using only the axioms of real numbers. The problem is that I don't think there is any work to do here, as one of the axioms clearly states that the multiplicative inverse of \(\displaystyle a\) denoted by \(\displaystyle \frac 1 a\) exists for all \(\displaystyle a \ne 0\). Now, if we take a look at \(\displaystyle \frac{1}{0}\), it is supposed to be the multiplicative inverse of $0$ which, based on the axiom of multiplicative inverse, does not exist. Is it enough of a "proof"? <br>
Also, I've been wondering what way of solving this one should choose when the expression in question would instead be\(\displaystyle \frac{3}{0}\) or \(\displaystyle \frac 0 0\). They are problematic because expression like \(\displaystyle frac{x}{a}\) is not in the language of a field.

Hi Aemilius,

There are various ways to definie the real numbers axiomatically. However, as far as I know, all these definitions include the field axioms.

If, in a field, \(\displaystyle \frac{1}{0}=x\) then \(\displaystyle 1 = 0\cdot x = 0\), and this is a contradiction, because, unlike the ring axioms, the field axioms explicilty state that \(\displaystyle 1\ne 0\) (the trival ring is not a field).