Algebra is Wrong

First, a little background for where I'm coming from. Feel free to skip to the examples.

When I was taught algebra, I was taught something that is fundamentally incorrect. Despite my love of math, and success in math competitions in elementary school and middle school, it took until college to really understand algebra correctly. This is because the basics on which I build my math skills were deeply flawed, and it took me a long time to really understand this.

If you want to understand the math, as I do, this might be interesting. Perhaps I can save you from the years of misunderstanding I had, or perhaps you never really got is sorted out correctly.

# Variables

I was taught a variable is simply a stand in for a potentially unknown number. This is WRONG.

## Many values

(1)
$$x=x$$

What is the value of $x$? Asking this question is WRONG.

The correct question would be: for what values of $x$ is this equation ($x=x$) true? The answer: Any value.1

Another example where the $x$ does not represent a specific number:

(2)
$$0=0$$

For what values of $x$ is this equation true? If we replace all the $x$s (all none of them) with 7, its still $0=0$, and still true. That holds for any potential value of $x$. Once again this is true for all values of $x$.

## No values

(3)
$$x=x+1$$

What values of $x$ make this equation true? None. There are no values of $x$ that make this equation true.

# Expressions

I was taught an expression containing variables is simply a stand in for a potentially unknown number. This is WRONG.

I was taught that because the expressions of both sides of an equals sign are equal (they are NOT), you can do the same operation to each side and get an equivalent equation (you CAN'T).

(4)
$$x=x+1$$

Lets solve for $x$. If we assume the things on both sides of the "=" are equivalent, then you can subtract $x$ from the right, and $x+1$ from the left and get

(5)
$$x-x+1=x+1-x$$
(6)
$$1=1$$

Since this is true for all values of x, the original must be as well, and thus we can replace $x$ with any value and get a true equation like:

(7)
$$7=7+1$$

Right? No.

The mistake was assuming the two things on opposite sides of the equal sign were equal. That is simply not always true:

(8)
$$1=2$$

This is never true, but its still a perfectly valid equation. This happens to be what the above ends up as if $x=0$

This is a perfectly valid equation. It happens to be false. The two sides are never equal.

For two equations to be equivalent, they not only need to be true in the same cases, but also false in the same cases!

(9)
$$x+1=3$$
(10)
$$(x+1)*0=3*0$$
(11)
$$0=0$$

Here we multiply both sides by 0 to solve for $x=0$. The final equation, $0=0$ is true for all values of x, but that does not mean the initial equation has this property.

When you add 1 to both sides of an equation, you don't change the cases where its true, or the cases where its false, and thus the equation is equivalent. That is why that is a legal step in algebraic proofs. This is known as a bidirectional implication: when one is true then the other is also true, and vise versa.

Multiplying both sides by 0 does not have this property. It produces a one directional implication: in all the cases $x+1=3$ is true, $0=0$ is also true, however the reverse is not true!

Formally we can say: $x+1=3$ implies $0=0$:

(12)
\begin{align} x+1=3 \implies 0=0 \end{align}

Lets try another one:

(13)
$$1+x=2+x$$

Also perfectly valid, and also false.

(14)
$$x=x^2$$

If we apply the above (incorrect) teachings, clearly $x$ and $x^2$ both represent numbers.

Suppose we have:

(15)
$$x=1$$

we know tha

(16)
$$x=x^2$$
(17)
$$x=x^2$$

To solve this, devide both sides by $x$, you get:

(18)
$$1=x$$

This is WRONG.

(19)
$$1=2$$