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ParthKohli
 3 years 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?
ParthKohli
 3 years 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?

This Question is Closed

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3I remember myin's "degenerate circle" question :)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2That 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.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2lol. We were thinking the same thing.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3Yes, I borrowed Zarkon's answer because I liked it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whom are you calling a degenerate?

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

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0It's not a fair point.

jiteshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0i don't gt why \(x^2=2x\) @ParthKohli

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0a square has two dimensions

jiteshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0I have studied something like that \(x^2=x \times x\) \(2x=x+x\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@jiteshmeghwal9 See the topline of my question.

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

jiteshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh ! im so stupid i gt it Okay :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0So 0 is not a solution for \(x\)?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3in 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".

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0if a side length is zero it is not a square

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

TuringTest
 3 years ago
Best 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.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0yes zero is not an answer to word problem, however it does solve the equation

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3True.^ 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
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