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The area of a square is just double its side.
We have:\[\rm x^2 = 2x\]so\[x^2  2x = 0\]\[x(x  2) = 0\]The solutions are 0 and 2, but is it possible to have a square with 0 as its side?
 one year ago
 one year ago
The area of a square is just double its side. We have:\[\rm x^2 = 2x\]so\[x^2  2x = 0\]\[x(x  2) = 0\]The solutions are 0 and 2, but is it possible to have a square with 0 as its side?
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.3
not really. Some mathematicians have called such objects "degenerate" squares, but you can exclude those kinds of answers when they pertain to physical objects.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
I remember myin's "degenerate circle" question :)
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
That reminds me of a question I asked about circles. Can a "circle" with radius zero still be considered a circle? (xh)^2+(yk)^2=r^2 (xh)^2+(yk)^2=0 So in other words, can a single point on a graph be seen as a circle with radius 0.
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
lol. We were thinking the same thing.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
Yes, I borrowed Zarkon's answer because I liked it :)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
whom are you calling a degenerate?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
If a point has 0 area, then the point doesn't cover any area... but a point still covers *some* area right?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Like infinitesimal, but still, I don't agree with the point that a point has zero area. :p
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
It's not a fair point.
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.0
i don't gt why \(x^2=2x\) @ParthKohli
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
a square has two dimensions
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.0
I have studied something like that \(x^2=x \times x\) \(2x=x+x\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@jiteshmeghwal9 See the topline of my question.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
a point has no are as it is zerodimensional, and squares (at least nondegenerate ones) require two dimensions as @UnkleRhaukus said
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.0
Ohh ! im so stupid i gt it Okay :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
So 0 is not a solution for \(x\)?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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".
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
if a side length is zero it is not a square
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Okay, I'm talking about this question. So, yes, 0 is not an answer to the word problem?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
@UnkleRhaukus @Zarkon termed such objects "degenerate". I think this is a sketchy question, they should have included the condition that \(x>0\) to avoid the subtleties in the philosophical mathematical implications of a zerobyzero square.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
yes zero is not an answer to word problem, however it does solve the equation
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Math is so idiotic.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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
 one year ago
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