Um, I just noticed that this one doesn’t hide the division by zero as cleverly as I expected. If x=y, then you can never divide by (x-y). In that case, one can quickly see that it is invalid for all values of x. 
No, those are the same meaning. If you want to make it clearer, say “the cardinality of the set containing the Pope and I is 2” and “the cardinality of the set containing the Pope and I is 1”.
Though, of course, the cardinality of the Pope is 1, and the cardinality of myself is 0.
And I’ve also seen puzzles like this that are based on a missing constant of integration. Usually, integrals are expressed in such a way that the constant of integration is the value of the integral at 0, and so students have a tendency to assume that that’s always true, but sometimes it can’t be, like with the integral of 1/x.
There is an interesting point there and I know one such that is not obvious from a diagram. The plain fact is that Euclid’s axioms are badly incomplete. The big thing you have to add is the relation of “betweenness”. Given any three points on a line, one of them is between the other two. Then several axioms relate betweenness to the other concepts. Another problem is the undefined "principle of superposition that is used in a hand-waving fashion to “prove” SAS.
The mess of Euclid was finally cleaned up once and for all by David Hilbert around the beginning of the 20th century. It took something like 22 axioms, one of which I could abbreviate as SASAS: if two triangles have all three sides equal and two angles, they are congruent. Then all the rest, including SAS, ASA, and SSS can be rigorously proved. And the proper use of betweenness will tell you whether a point you have constructed will be inside or outside of a triangle, which will fix the problem alluded to in the quote.
Anyone want to see my simplish proof of the Four Color Map Theorem? 
In all fairness to Euclid, he wasn’t writing THE book to axiomize geometry for all time to a hyper-critical level of proof. He was writing a textbook for his students. How many modern day geometry textbooks suffer the exact same ommissions we criticize Elements for?
And in even more fairness to Euclid, it took millennia for anyone to do better.
It also makes sense spiritually, since everything is One, obviously x + x = 1 no matter what x is.
I take your point, but thank goodness that when I learned how to think early on in school, when I got “x=6” I had solved the problem, not “solved” it.
The first time a professor of mine said he was going to lecture about infinite cardinals, my first thought was, “does the Vatican know about this?”
Will it fit in the margin of this thread? 
Are the two words will and shall used the same ways in British English vs. American English? ISTM I’ve read that they are not.
In American English, “will <any verb>” is a simple future tense, indicating factually that something is going to happen. “shall <any verb>” is giving an order or mandate, indicating some action that the subject of the sentence is required to perform. It is used this way extensively in legalese, especially in the wording of laws and regulations. Actually, it’s even more nuanced than that. See here for discussion.
I think in British English (some Brit help me out here if necessary!), shall can be simply a statement of something that will happen: “I shall dream about a thousand pounds to-night, I know I shall!” from Through The Looking-Glass.
Another fertile source of fallacies, using squirts of negative numbers, involves falsely relying on the well-known rule:
(√a)(√b) = √(ab)
which doesn’t work when both a and b are negative numbers.
I have an old college-level algebra book, titled simply College Algebra by Britton and Snively, published in 1948 (older than I am!) that was laying around the house when I was a young child, which I’ve kept to this day. It is the best algebra textbook I have ever seen, and goes into detail on these kinds of things.
Their take: If, in the process of solving an equation, you ever need to multiply or divide both sides by any expression containing the unknown, you need to be careful.
If you multiply both sides by any expression containing the unknown, you might introduce a new solution that wasn’t there before. This is easy to catch: Just be sure, after you find all the solutions, to check every solution in the original equation, to discard any superfluous ones.
If you divide both sides by any expression containing the unknown, you might lose a solution that the original equation had. To find these, create an auxiliary equation in which you equate the expression (that you divided both sides by) to 0, and solve that. If any roots were lost, they will be found among the roots of this auxiliary equation.
(If anyone wants actual examples, I think I have that book someplace more-or-less accessible, and I can dig up some.)
Let x=1.999… (That is, 1.999 with 9s extending to infinity.)
Then 10x = 19.999…
10x - x = 9x
19.999… - 1.999… = 18
If 9x = 18, then 1.999… = 2
Why does step four work?
(x + y)(x - y) = y(x - y)
That suggests that y = x + y, which could only work if X = 0, in which case Y also = 0. And which also means that neither X nor Y can equal 1.
This, or something similar (0.999… = 1) was touched upon in this earlier thread, among others.
It only suggests that if you divide out the “(x - y)” factor, which is zero. This is the very division which has been noted as problematic in the next step.
Go ahead and plug in equal values for x and y; x = y = 1, say. You will indeed find that (x + y)(x - y) = y(x - y), as both sides will be zero.
In some sense, it works too well.
If x = y ( and thus, (x-y)=0 ), then the above amounts to (x + y)(0) = y(0) which is all too true no matter what x and y are. Suppose, for example that x = y = 123.
Then (x + y)(0) = y(0)
becomes (123 + 123)(0) = 123(0) which is still indubitably true and correct and valid. Now, “cancel” the factor of 0 from both sides and you get (123 + 123) = 123,
or 246 = 123 which is where the error pops up.
This is what happens when you multiply both sides of an equation by zero. For a simpler example,
2(0) = 3(0).
It’s mathematically correct, but multiplying both sides by zero, reducing the equation to
0 = 0 has eliminated whatever information the equation formerly had. The next step, “dividing out” the 0 from both sides, obscures the fact that your equation had no useful information in it, installing some new (and wrong) information that wasn’t there before.
looks around
That might be me.
(2)(0) = 1(0)
True.