Why this statement is false to $a \in \mathbb{C}$
$(\sqrt[n]{a} * \sqrt[k]{a} ) - (a^{\frac{n+k}{nk}})= 0$
How you can prove it with high school maths ?
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Try taking $\sqrt{-1}$. Let this equal $x$. Square both sides. What happens? – Mr Pie Mar 19 '18 at 16:20
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Related: https://math.stackexchange.com/questions/2685072/choice-of-a-square-root/2685080#2685080 , https://math.stackexchange.com/questions/2670522/order-of-operations-with-complex-numbers/2670551#2670551 – Ethan Bolker Mar 19 '18 at 16:21
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High school maths usually doesn't cover complex numbers in my experience, so even stating the problem itself is problematic given your constraints. However, if you want to do this, then first and foremost, you need to ask yourself what $\sqrt[p]{a}$ and $a^{\frac mn}$ mean for complex numbers. We can't get anywhere proving what properties those operations may or may not have or before we know exactly what they mean. And they mean different things to different people (to me they are meaningless, for instance), so in order to help you we need to know what they mean to you. – Arthur Mar 19 '18 at 16:23
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The problem arises because of $$(\sqrt{-1})^2=-1$$ and that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is not true if $a,b<0$. If it was true, we would get: $$1=\sqrt{1}=\sqrt{(-1)\cdot (-1)}=\sqrt{-1}\sqrt{-1}=-1$$
cansomeonehelpmeout
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$$(\sqrt[n]{a} * \sqrt[k]{a} ) - (a^{\frac{n+k}{nk}})= 0$$
is not always true for complex numbers because $\sqrt[n]{a}$ is not uniquely defined.
In fact in complex numbers you have $n$ different roots of $1$ for each $n$ which makes it more interesting.
Mohammad Riazi-Kermani
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