A few thoughts on equations:

reflection 1: For years I've been incorrectly explaining why both sides can be multiplied by the same thing

I've just realized that for many years I've been telling people something false. Whenever I explained why when solving an equation (for example x / 2 = 7) I'm allowed to multiply both sides by 2, I would say (incorrectly): "well, because when I have a balance scale that's in equilibrium, and I double the contents of both the left pan and the right pan, the scale will still be in equilibrium".

I'm not the only one who said this. I just searched the Internet for how people justify this multiplication of both sides, and at https://www.zsnienowice.pl/dlaczego-przeksztalcenia-algebraiczne-dzialaja it says:

Jeśli pomnożymy lub podzielimy obie strony przez tę samą wartość (inną niż zero), równanie pozostaje prawdziwe.

And at https://math.answers.com/math-and-arithmetic/What_allows_property_to_add_the_same_thing_to_both_sides it says:

if you have an equation, you can add the same number to both sides without changing the equality. For example, if ( a = b ), then ( a + c = b + c ) for any number ( c )

And at https://sme.goiania.go.gov.br/conexaoescola/eaja/matematica-principios/ it says:

If two numbers are equal, then adding or subtracting the same value from both numbers will result in two new numbers that are still equal

One can consider what the authors of these quotes meant, but it seems to me that the normal interpretation is that they're saying exactly what I said.

And this is faulty reasoning. Because thinking this way, one could say "when I have a balance scale that's in equilibrium, and I round down the contents of both the left pan and the right pan, the scale will still be in equilibrium" - and use this to justify the (incorrect) principle that when solving an equation, you're allowed to apply "floor" to both sides. The correct justification for multiplying both sides of an equation by two goes: "well, because when I have a balance scale that's in equilibrium, and I *reduce by half* the contents of both the left pan and the right pan, the scale will still be in equilibrium".

Because the point is that I'm doing something like this:

(1) x / 2 = 7
(2) f(x / 2) = f(7)

after which I find the solution to equation (2) and claim that it's also a solution to equation (1). But this is true not if function f preserves truth, but if the inverse function of f preserves truth. And only this explains why you're allowed to multiply both sides by 2, but you're not allowed to round both sides down or raise them to a power.

This problem is cleverly avoided by those who only say that if we multiply both sides by the same thing, we get an *equivalent* equation.

reflection 2: Why does it break down at precisely this point?

Consider this proof that 1 + 1 = -1:

(1) 1 + 1 = x
I raise both sides to the power of two:
(2) (1 + 1)² = x²
from the binomial formula:
(3) 1² + 2*1*1 + 1² = x²
(4) 1 + 2 + 1 = x²
I regroup:
(5) 1 + 1 + 2 = x²
but "1 + 1" is x, so:
(6) x + 2 = x²
(7) 0 = x² - x - 2
(8) x² - x - 2 = 0
I solve in Wolfram Alpha or in the schoolbook way using delta or by guessing and I get:
x = 2 or x = -1

It's not strange that I got a wrong result - because at the beginning I performed an illegal operation: I raised both sides to the power of two. But what's interesting is when, going from end to beginning, I check at which steps -1 is still a correct solution. At step (8) it is, at (7) it is, at (6) it is - and only at step (5) does it break down. Why did something bad happen precisely at the transition (5) → (6)? After all, this particular transition seems legal? Because truly 1 + 1 = x, we assumed that ourselves? At first when T. asked me about this, I couldn't answer, but after some thought I know: orpnhfr rdhngvbaf 1 naq 2 ner abg rdhvinyrag, fb zber oebnqyl 1 naq 5 ner abg rdhvinyrag, fb "k" va rdhngvba 1 naq "k" va rdhngvba 5 ner qvssrerag kf, fb va gur pbagrkg bs rdhngvba (1) vg'f gehr gung "1 + 1 = k", ohg va gur pbagrkg bs rdhngvba (5) guvf vf ab ybatre gehr.

idea: You can hide this squaring

In the above example, the one proving that 1 + 1 = -1, the fraud with raising to the power of two between steps (1) and (2) really stands out. I was thinking how to do this more discreetly, and I came up with this trick:

(1) 1 + 1 = x
I multiply both sides by "1 + 1":
(2) (1 + 1)(1 + 1) = x(1 + 1)
from the binomial formula:
(3) 1² + 2*1*1 + 1² = x(1 + 1)
(4) 1 + 2 + 1 = x(1 + 1)
I regroup:
(5) 1 + 1 + 2 = x(1 + 1)
but "1 + 1" is x, so:
(6) x + 2 = x * x
(7) 0 = x² - x - 2
(8) x² - x - 2 = 0
I solve in Wolfram Alpha or in the schoolbook way using delta or by guessing and I get:
x = 2 or x = -1

And now it's far less obvious where I'm cheating. Because what, you're not allowed to multiply both sides by two (which I do in the transition (1) → (2))? So far, I don't know an answer that would satisfy me.

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