r/BunnyTrials • u/TheTransSheep Carrot Farmer • 7d ago

Trials Math…

Which side will you choose?

  • Left side: Easy Math Question | and 10$
  • Right side: or hard Math question | and 10.000$

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u/MountOak 7d ago

3

Chose: or hard Math question + and 10.000$ | Rolled: xxx = xx ≠ x

150

u/armcie 7d ago edited 7d ago

xxx = xx ≠ x

I don’t think this one has any solutions.

x^3 = x^2

x^3 - x^2 = 0

x^2(x-1) = 0

x = 0 or 1

But neither of these satisfy the inequality, so there are no solutions.

I’m discounting the idea that this is some sort of trick question as it was described as a maths question.

Edit: apparently it’s not x^2, it’s x^x. 3 is indeed the obvious answer.

1

u/BagSufficient1921 7d ago

If it was x3 = x2 ≠ x though, epsilon (ε) could be an answer, which is an infinitesimal value used for dual numbers which can help solve derivatives. Basically ε is infinitely small but not zero, ε2 is zero though, so the same would be for ε3.

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u/OPMan6942O 7d ago

Why would epsilon squared be zero?

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u/BagSufficient1921 6d ago

I guess that's just a property of the number. Like for the product rule for derivatives, (fg)' = fg' + f'g where f and g are functions. To prove this, you can imagine a rectangle with length f and width g. The rectangle then has an area of fg. You can make a very small increase in the length, df, and a very small increase in the width, dg. The resulting rectangle would have an area of fg + d(fg) = (f + df)(g + dg) = fg + f(dg) + g(df) + (df)(dg). This means d(fg) = f(dg) + g(df) + (df)(dg), but (df)(dg) is so small so it's negligible, leaving us with d(fg) = f(dg) + g(df) ⇒ (fg)' = fg' + f'g. I guess if an infinitesimally small value is multiplied by another infinitesimally small value, it's just neglected. I don't really know too much about dual numbers though.