Having the proof of the cancelation law for multiplication:




$$c\cdot 1=a\cdot 1\tag{Indentity}$$


This may seem stupid, but exactly at the second step: I can multiply both sides by the inverse, but is this a consequence of the field axioms or of the equality defined as an equivalence relation? I guess it's the second one, but I'm not sure if one could prove it with the field axioms alone.

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    $\begingroup$ Are you asking why $A=B$ implies $AC=BC$? It is because the multiplication is a well defined function. $\endgroup$ – Berci Aug 29 '15 at 22:41
  • $\begingroup$ These are properties of groups. Every group has an identity element, and every element in the group has an inverse. $\endgroup$ – Marconius Aug 29 '15 at 22:42
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    $\begingroup$ You can apply a function to both sides of any equality. Even in settings where we aren't talking about numbers. Even when we're just talking about barren, structureless sets and elements of sets, we can apply a function to both sides of an equality. Of course, for any $u$ in an algebraic structure with a binary operation, there is a function $f(x):=x\cdot u$. $\endgroup$ – anon Aug 29 '15 at 22:52
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    $\begingroup$ Write your proofs in complete grammatically correct sentences, specifying exactly what all the initial assumptions are, and which direction(s) the implications go, and why each step is justified. Do not skimp on the words. Then you can see exactly what you have said and where,if any , the errors are . $\endgroup$ – DanielWainfleet Aug 30 '15 at 0:42
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    $\begingroup$ Your "operation" is a "function". And functions have this property (the "vertical line property"): if $x_1=x_2$, then $f(x_1)=f(x_2)$. $\endgroup$ – GEdgar Aug 30 '15 at 0:54

One of my teachers I think seemed to dislike "multiplying" both sides of an equation by the same element. If you share the same aversion you can almost always use something like this to avoid it.

$$c = c \cdot e = c \cdot(b \cdot b^{-1}) = (c \cdot b) \cdot b^{-1} = (a \cdot b) \cdot b^{-1} = a \cdot(b \cdot b^{-1}) = a \cdot e = a$$

All we have used here is the associative law for the product and substitution which is a much more fundamental concept than the definition of a group.

But for the record I have no issues writing,

$$ c = a \implies cb = ab $$

since all we use here is the fact that $$c= a \implies (c, b) = (a, b) \implies f(c, b) = f(a, b)$$

where $f : G \times G \to G$ is the binary product on $G$ which is a well-defined function.


Because if there were some operation you could perform on the two quantities to get different results, they would not be equal.

If two things are equal, the results of multiplying them each by $b^{-1}$, or of performing any other operation on them, must be equal, because that is what it means for two things to be equal.

  • $\begingroup$ Yes, I know. But my question concerns to the formalization of this idea. See Berci's and anon's comments. $\endgroup$ – Billy Rubina Aug 30 '15 at 0:31
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    $\begingroup$ I think this ultimately depends on the definition of "equal", which is an axiom in set theory. In hand-wavy words, if $a = b$ then $a$ and $b$ are two names that denote the same thing. $\endgroup$ – alephzero Aug 30 '15 at 4:24

Yes, you can prove it from the properties of real numbers.

Proof: If $a$, $b$, and $c$ are any real numbers, and $a = b$.

$ac$ is a unique real number $\iff$ closure property of multiplication

$ac = ac$ $\iff$reflexive property of equality

$a = b$ given

$ac = bc$ $\iff$ substitution principle


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