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Suppose $A$ is a set equipped with a binary operation $+$.

How I can prove if $a=b$ then $a+c=b+c$? Why we can add an element on both sides like this?

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  • $\begingroup$ You can prove it by induction. $\endgroup$
    – Kenny Lau
    Aug 2, 2016 at 12:26
  • $\begingroup$ a=b iff (a,c)=(b,c) only if +(a,c)=+(b,c) $\endgroup$ Aug 2, 2016 at 14:08
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – yago
    Aug 2, 2016 at 14:20
  • $\begingroup$ @KennyLau This has nothing to do with induction except when $A = \mathbb{N}$. $\endgroup$
    – lisyarus
    Aug 2, 2016 at 15:24
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    $\begingroup$ Why are people downvoting this? It's a good question. $\endgroup$
    – D_S
    Aug 2, 2016 at 16:09

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A simple explanation is that saying $a=b$ literally says that $a$ and $b$ are the exact same mathematical object. Hence 'adding' $c$ to $a$ on the right is one the same as 'adding' $c$ to $b$ on the right (adding in quotes because what I am really referring to is the given binary operation).

Note that the converse, $a+c=b+c \Rightarrow a=b$, is not guaranteed to be valid unless every element of the set $A$ has unique right inverses, and in such a case, we say that $c$ has right inverse $c_R^{-1}$ and we can apply that right inverse to both sides to obtain $a+c+c_R^{-1}=b+c+c_R^{-1}\Rightarrow a+e=b+e$ where $e$ is an identity element (required for the notion of a right inverse to make sense) and thus $a+e=b+e\Rightarrow a=b$.

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    $\begingroup$ Actually, $a + c = b + c \implies a = b$ just requires $c$ to be right cancellable, not necessarily invertible. $\endgroup$
    – yago
    Aug 2, 2016 at 14:18
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This also confused me when I first started studying math. A binary operation $+$ on $A$ is a function $+: A \times A \rightarrow A$. By $a+b$, we mean $+(a,b)$.

The question is why if $b, c \in A$ with $b =c$, then $+(a,b) = +(a,c)$?

The naive way to do this is to just say: it's obvious; two things are "equal" if they are the same. If $b$ is the same thing as $c$, then $+(a,b)$ is obviously the same thing as $+(a,c)$, because $b$ is the same thing as $c$. There is no difference in the expressions. If I add $3$ to the number $2$, that's obviously the same thing as adding $3$ to the number $1+1$, because $1+1$ is the same thing as $2$.

Another way if you want to think more carefully about it: fix $a \in A$. Then the mapping $x \mapsto +(a,x)$ is a function $f: A \rightarrow A$.

So this reduces the problem to showing the following: if $f$ is a function from $A$ to $A$, and $b, c \in A$ with $b = c$, then $f(b) = f(c)$. Can you see why this is tre?

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  • $\begingroup$ I like the idea of explaining $+$ as a function $+:X\times X \rightarrow X$. I think, however, you can omit the part where you fix $a$, provided that you have suitable rules for determining the equality of ordered pairs. So +1 to this answer for introducing $+$ as a function at the very beginning, +1 to goblin's answer for applying the binary function directly. $\endgroup$
    – David K
    Aug 2, 2016 at 16:49
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For the binary operation to be well-defined, it must be the case that for any elements $a, b, c, d$ with $a=b$ and $c=d$, $a+c = b+d$. It is not at all clear in your question what further conditions there are on the binary operation $+$, but being well-defined is absolutely necessary for what you've written to make sense.

The converse is not necessarily true, however. Since you have not mentioned the existence of inverses for your binary operation, there is no reason to assume that $a+c=b+c \implies a=b$. Taking an example from ring theory, replacing your '+' with multiplication in $\mathbb{Z}_6$, we have $2 \times 3 = 4 \times 3 = 0$, but $2 \neq 4$.

See this thread for a more concise explanation of what I've said.

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Because this is what equality means.

From a very formal perspective, equality is just some relation between mathematical objects. It becomes interesting if we restrict it's behavior via some axioms, like this:

$$x=y \Rightarrow \left[\phi(x) \leftrightarrow \phi(y)\right]$$

for any one-argument proposition $\phi$.

That is, we require equal elements to behave exactly the same. If something is true for $x$, it is true for $y$, and vice versa.

You particular case corresponds to $\phi(x)=x+c$ for a fixed $c$.

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Let be $a=b$ now add c to a: $$a+c$$ Now:$$a+c = a+c$$ but $a=b$, so we can replace $a$ in the second member, then $=> a+c = b+c$

To prove that $a = a$, $0+a = a$ and $a+0 = a$ then, $a+0 = 0+a => a=a$.

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