Isn't it obvious that $0$ and $1$ are distinct?

Question

The field axioms for the real number system contain the following statements concerning the existence of neutral or identity elements for addition and multiplication:

(1) There exists a real number, called "zero" ($0$), such that, for all real $x,$ $x+0=0+x=x$, and
(2) There exists a real number, called "one" ($1$), such that $1\ne 0$ and, for all real $x,$ $x\cdot 1=1\cdot x=x$

Why is it necessary to include "$1\ne 0$" in (2) if we are calling these elements by different names?

Answer 1

Why is it necessary to include "$1\ne 0$" in (2) if we are calling these elements by different names?

Because "calling these elements" by different names only does not necessarily imply that they are not the same. $\quad$

For instance, the zero ring is the (unique) ring in which the additive identity $0$ and multiplicative identity $1$ coincide.

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[Added:] To elaborate the point above, suppose you have the following two axioms instead:
(a) There exists a real number, called "zero" ($0$), such that for all real $x,$ $x+0=0+x=x$, and
(b) There exists a real number, called "one" ($1$), such that for all real $x,$ $x\cdot 1=1\cdot x=x$.

Then the zero ring $\{0\}$ satisfies both (a) and (b), but you won't want a real number system like that.

Answer 2

Otherwise the "field" $\{0\}$, which is not a field regarding the definition with the usual axioms would be a field.

And it is not very practical to accept it as a field, so we usually demand $0\ne 1$.

Answer 3

1 is just the symbol for the unit element for multiplication

0 is the symbol for the unit element for addition

The elements could be the same, in fact, a trivial field {0} would have 0 = 1

Answer 4

$1\ne 0$ because if not $1=0\Rightarrow a=0$ for all $a$. So simply as this.

"$1\ne 0$" is included because if not, after verification of $0\cdot a=0$ for all $a$ we would have as said above $a=0$ for all $a$.