So as time has gone by I've forgot some really basic algebraic rules. I'm asking you here if these are, and why they are, correct.

Question 1) Let $x$ and $y$ and $c$ be any algebraic expressions. Is it true and WHY is it true that


Question 2) True? and if it is, why?

$$-\frac xy×c=-\frac {x×c}{y}$$

That is, that if I had an equation say:

$$2x-\frac {x}{x+1}=1$$ I could multiply both sides by $x+1$, move the $x+1$ onto the numerator, and get

$$2x^2+2x-\frac {x(x+1)}{x+1}=x+1$$

After which I could cancel the $x+1$ on the numerator and the denominator and solve.

I came up with $-\frac {x}{x+1}=-1×(\frac {x}{x+1})$ after which I could use the associative property after multiplying with $x+1$ to move the $x+1$ to the numerator and cancel.

-My apologies if the formatting is bad


The rules you refer to are correct and are a direct consequence of axioms/rules for rational and real numbers.

For the equation your step is correct but you need to set $x+1\neq 0$ and then you can cancel out this term by multiplication

$$2x^2+2x-\frac {x\color{red}{(x+1)}}{\color{red}{x+1}}=x+1\iff 2x^2+2x-x=x+1\iff2x^2=1$$

from here you can solve for $x$ (with the condition $x\neq -1$).

  • $\begingroup$ Oh damn, my bad I just realized I made a mistake on question 1) I meant to ask "is $x×(−y)=−(x×y)=-x×y$ " , I assume this is correct as well? I mean I feel like it should be since it's a negative times a positive, so the order shouldn't matter. $\endgroup$ – punctuationisimportant Mar 17 '18 at 10:30
  • $\begingroup$ I think you mean $x\neq-1$ $\endgroup$ – user411437 Mar 17 '18 at 10:31
  • $\begingroup$ @punctuationisimportant Yes it is correct a way to see is that $-y=-1\cdot y$ then $x\cdot (-y)=-1\cdot x \cdot y = -(x\cdot y)$ $\endgroup$ – gimusi Mar 17 '18 at 10:32
  • $\begingroup$ @DavidP Yes of course, Thanks I fix it! $\endgroup$ – gimusi Mar 17 '18 at 10:33

We need to be somewhat careful when we say "let $x$ and $y$ and $c$ be any algebraic expressions". This is because there are different algebraic frameworks in which we could feasibly work. The one with which you are likely familiar and are indeed referring to is that of a field. For example, there is the field of rational numbers (i.e. fractions) and the field of real numbers (i.e. numbers which have a decimal expansion). As you can see in the linked Wikipedia article, the definition of a field is a list of abstract properties called axioms which any and all fields must satisfy.

Let $x$ be an element in a field. I will derive from the field axioms (look under "Classical definition") that $x\cdot 0=0$ and then $x\cdot (-1)=-x$. As gimusi suggested in their comment, you can use the commutativity of multiplication together with the latter fact to derive your first claim. We have $$x=x\cdot 1=x\cdot(1+0)=(x\cdot1)+(x\cdot 0) = x + (x\cdot 0)$$ where the equalities are by virtue of multiplicative identity, additive identity, distributive law, and multiplicative identity (in that order). Adding $-x$ to both sides, the additive inverse law gives $0=x\cdot 0$, as desired. Now for the next part! We have $$0=x\cdot 0=x\cdot (1+(-1))=(x\cdot1) +(x\cdot(-1))=x+(x\cdot(-1))$$ where we used our previous result, additive inverse, distributive law, and multiplicative identity (in that order). As before, we add $-x$ to both sides, yielding $-x=x\cdot(-1)$.

As for your second claim, you need the condition that $y\ne 0$ in order for there to be a multiplicative inverse of $y$. In other words, you cannot divide by $0$. Otherwise, your claim should follow immediately from associativity.

  • $\begingroup$ Sorry to bother you more after that well written answer, but what do you mean by "Otherwise, your claim should follow immediately from associativity"? Is the deduction in my original post correct? $\endgroup$ – punctuationisimportant Mar 17 '18 at 11:59
  • $\begingroup$ Well, it does not hold if $y=0$. If $y\ne 0$ then your deduction is true and you can prove it by using associativity. $\endgroup$ – eloiprime Mar 17 '18 at 22:11

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