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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?

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  1. TuringTest
    • 3 years ago
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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.

  2. TuringTest
    • 3 years ago
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    I remember myin's "degenerate circle" question :)

  3. myininaya
    • 3 years ago
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    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.

  4. myininaya
    • 3 years ago
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    lol. We were thinking the same thing.

  5. TuringTest
    • 3 years ago
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    Yes, I borrowed Zarkon's answer because I liked it :)

  6. ParthKohli
    • 3 years ago
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    Hmm...

  7. anonymous
    • 3 years ago
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    whom are you calling a degenerate?

  8. ParthKohli
    • 3 years ago
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    If a point has 0 area, then the point doesn't cover any area... but a point still covers *some* area right?

  9. ParthKohli
    • 3 years ago
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    Like infinitesimal, but still, I don't agree with the point that a point has zero area. :p

  10. ParthKohli
    • 3 years ago
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    It's not a fair point.

  11. ParthKohli
    • 3 years ago
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    Yello.

  12. jiteshmeghwal9
    • 3 years ago
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    i don't gt why \(x^2=2x\) @ParthKohli

  13. UnkleRhaukus
    • 3 years ago
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    a square has two dimensions

  14. jiteshmeghwal9
    • 3 years ago
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    I have studied something like that \(x^2=x \times x\) \(2x=x+x\)

  15. ParthKohli
    • 3 years ago
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    @jiteshmeghwal9 See the top-line of my question.

  16. TuringTest
    • 3 years ago
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    a point has no are as it is zero-dimensional, and squares (at least non-degenerate ones) require two dimensions as @UnkleRhaukus said

  17. TuringTest
    • 3 years ago
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    no area*

  18. jiteshmeghwal9
    • 3 years ago
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    Ohh ! im so stupid i gt it Okay :)

  19. ParthKohli
    • 3 years ago
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    So 0 is not a solution for \(x\)?

  20. TuringTest
    • 3 years ago
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    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".

  21. UnkleRhaukus
    • 3 years ago
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    if a side length is zero it is not a square

  22. ParthKohli
    • 3 years ago
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    Okay, I'm talking about this question. So, yes, 0 is not an answer to the word problem?

  23. TuringTest
    • 3 years ago
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    @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 zero-by-zero square.

  24. UnkleRhaukus
    • 3 years ago
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    yes zero is not an answer to word problem, however it does solve the equation

  25. ParthKohli
    • 3 years ago
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    Math is so idiotic.

  26. ParthKohli
    • 3 years ago
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    lol. :)

  27. UnkleRhaukus
    • 3 years ago
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    maths is a tool,

  28. TuringTest
    • 3 years ago
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    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

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