>>8933284>3/3rds is less than one according to our math?
No. What happened is that they (essentially) had to extend the = operator so that it could handle stuff like 1/3 = 0.3333...
(Technically, it's called the "limit" operator, but it basically ended up extending the way the = operator is used.)
The reason the original = operator didn't work is because you had this unsolvable problem:
1/3 = 3/10 + 3/100 + 3/1000 + 3/10000 + ...
The right side is a sum that doesn't end. But if it doesn't end, then it makes no sense to ask questions like "what does the right side equal in the end" or "what is the final sum" or anything like that, because there's no end. All those kinds of questions are self-contradictory and hence have to be thrown out -- you simply can't justify asking something that's basically "what is the END RESULT of and ENDLESS process?". It's such a nonsensical question that mathematicians had to abandon even trying.
So to fix the problem, they essentially had to extend the = operator (via the limit) so that it yields true if it can proven that no numbers exist that are strictly between the left and the right side of the equation. (The actual extension is a little more complicated than that, but you get the basic idea.)
Mathematicians have conceded that the above equation is never true under the original (simple) meaning of the = operator. (That happened in the mid 1700s, after a few decades of heated arguments among the top guys in math.) So if one side of the equation is an infinite sum (like 0.333...) then you need to activate the new extension of the = operator that looks at whether any numbers exist between the two sides.
Here's an example of how the extension works:
1: Start with this statement: "If x approaches 3, then x^2 approaches 9."
2: Rewrite it in a standard way like this:
lim (x->3) x^2 = 9
Notice that the 2nd appearance of the word "approaches" got replaced by the equal sign in the "lim" notation. That's the extension.