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not really. Some mathematicians have called such objects "degenerate" squares, but you can exclude those kinds of answers when they pertain to physical objects.
I remember myin's "degenerate circle" question :)
That reminds me of a question I asked about circles. Can a "circle" with radius zero still be considered a circle? (x-h)^2+(y-k)^2=r^2 (x-h)^2+(y-k)^2=0 So in other words, can a single point on a graph be seen as a circle with radius 0.
lol. We were thinking the same thing.
Yes, I borrowed Zarkon's answer because I liked it :)
whom are you calling a degenerate?
If a point has 0 area, then the point doesn't cover any area... but a point still covers *some* area right?
Like infinitesimal, but still, I don't agree with the point that a point has zero area. :p
It's not a fair point.
a square has two dimensions
I have studied something like that \(x^2=x \times x\) \(2x=x+x\)
Ohh ! im so stupid i gt it Okay :)
So 0 is not a solution for \(x\)?
in the purely mathematical sense, I would say "yes it is a solution", but as a "square" in the normal sense of the word, or as a physical object, the answer would be "no".
if a side length is zero it is not a square
Okay, I'm talking about this question. So, yes, 0 is not an answer to the word problem?
yes zero is not an answer to word problem, however it does solve the equation
Math is so idiotic.
maths is a tool,
True.^ Professional mathematicians throw out useless concepts just because they are such. They also make their own theorems and definitions depending on what they feel is called for in a given situation. For example, many great mathematicians have \(defined\) \(0^0=1\) there is no objective mathematical proof of this, but sometimes it is convenient to do such things. These guys just make it up as they go along basically :P