they say $$\sqrt[4]{16} = \pm 2$$

however...

$$\sqrt[4]{16(1)} \implies \sqrt[4]{16i^4} \implies \pm 2i$$

where is the fallacy?

Question

they say $$\sqrt[4]{16} = \pm 2$$

however...

$$\sqrt[4]{16(1)} \implies \sqrt[4]{16i^4} \implies \pm 2i$$

where is the fallacy?

anonymous · Answer

no they do not

anonymous · Answer

the number 16 has 4 "fourth roots" namely solutions to $x^4=16$ but only one that is represented as $\sqrt[4]{16}$ which means the positive real one, namely 2

lgbasallote · Answer

so the others are -2, +2i, -2i?

unklerhaukus · Answer

oh there are four solutions

lgbasallote · Answer

yes...that makes sense...

anonymous · Answer

we consider $y=\sqrt[4]{x}$ as a function for any given $x$ u have just one $y$ so  $\sqrt[4]{16}=2$

lgbasallote · Answer

how would you explain what i wrote then @mukushla

anonymous · Answer

there is another error in reasoning here
so say $\sqrt[4]{ab}=\sqrt[4]{a}\sqrt[4]{b}$ you are assuming that $a, b$ are positive real numbers

lgbasallote · Answer

so they're not?

anonymous · Answer

otherwise it is not true

anonymous · Answer

if you are living in the world of positive numbers, then yes
and don't forget the symbol $\sqrt{a}$ does not mean "some number whose square is $a$" but rather is means "the positive number whose square is $a$"

lgbasallote · Answer

i guess i really do have an engineer's mind and not a mathematician's =))

unklerhaukus · Answer

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