- #1

etotheipi

This is going to sound very silly indeed, however I've been tutoring basic algebra and it's caused me think about a few little points that I'd never really even considered before.

With an equality ##a=b##, we might define a function ##f## representing a certain operation we could perform on both sides of the equation, like for example ##f(x) = 3x##. Since, if ##a=b## then ##f(a) = f(b)##, this then formalises the various properties of equality because we can apply ##f## to both sides and obtain ##3a=3b##, etc.

If our equation was ##\frac{3}{(x+2)} = 6##, "formally" we'd use ##a=b \implies ac=bc## to do the following two steps $$\frac{3}{(x+2)}(x+2) = 6(x+2)$$ $$3 = 6(x+2)$$ Normally, I think anyone would have just skipped to the second line, though I wondered if the above example counts as two different steps - the first being applying the operation, and the second simplifying? Do these count as distinct steps? I'm aware this essentially changes nothing, though just found it a little curious. If we're working with matrices or groups, it's less common to see the intermediate step skipped, i.e. $$ab=c$$ $$a^{-1}ab = a^{-1}c \quad \text{(apply inverse operation)}$$ $$eb= a^{-1}c \quad \text{(simplify)}$$ $$b = a^{-1}c \quad \text{(simplify)}$$Thanks, and please do advise if I'm rambling!

With an equality ##a=b##, we might define a function ##f## representing a certain operation we could perform on both sides of the equation, like for example ##f(x) = 3x##. Since, if ##a=b## then ##f(a) = f(b)##, this then formalises the various properties of equality because we can apply ##f## to both sides and obtain ##3a=3b##, etc.

If our equation was ##\frac{3}{(x+2)} = 6##, "formally" we'd use ##a=b \implies ac=bc## to do the following two steps $$\frac{3}{(x+2)}(x+2) = 6(x+2)$$ $$3 = 6(x+2)$$ Normally, I think anyone would have just skipped to the second line, though I wondered if the above example counts as two different steps - the first being applying the operation, and the second simplifying? Do these count as distinct steps? I'm aware this essentially changes nothing, though just found it a little curious. If we're working with matrices or groups, it's less common to see the intermediate step skipped, i.e. $$ab=c$$ $$a^{-1}ab = a^{-1}c \quad \text{(apply inverse operation)}$$ $$eb= a^{-1}c \quad \text{(simplify)}$$ $$b = a^{-1}c \quad \text{(simplify)}$$Thanks, and please do advise if I'm rambling!

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