Would a/i=ai because of the following?
a/i=( (a/i)^{2} )^{1/2}=(a^{2}/i^{2})^{1/2}=(a^{2})^{1/2}=ai

Would a/i=ai because of the following?
a/i=( (a/i)^{2} )^{1/2}=(a^{2}/i^{2})^{1/2}=(a^{2})^{1/2}=ai
I assume that i is the imaginary unit.
If a/i = ai, then multiplying both sides by i gives a = a, so there is clearly something wrong with your proof.
The flaw is in the second step: the property a^(bc) = (a^b)^c applies if a is a positive real number, but not necessarily if otherwise.
The same flaw is also shown in the second to last step.
Just to give you some insight and not to leave you lost, I'll tell you the correct variation of your formula. If you want to figure it out on your own then sorry just don't read ahead.
Since i*i = 1, then (i*i) = 1 and (i)*i = 1 and thus 1/i = i. Multiplying both sides by a gives a/i = ai. So your formula would've been correct if you would have just a minus sign in there.
if and only if .
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This represents the danger of manytoone operations: they are not invertible. The operation of squaring is a twotoone operation, ie and square to the same value and therefore the sign information is lost. When the squareroot is taken, this is strictly a onetotwo operation and the wrong sign was chosen in that it wasn't the original sign that was squared, thus leading to the incorrect answer.
In fact, one can "prove" any sort of nonsense by neglecting this point. This can be done for example by arranging the "proof" to contain the expression:
where the is somehow obscured by being an expression that evaluates to zero. Then by dividing both sides by this zero expression, one goes from the above true expression to the nonsense:
Note that multiplying by zero is a manytoone operation and is ultimately why dividing by zero is an invalid operation.
as we know i^4 = 1
so a * 1 / i = a * i^4 / i = a * i^3
because i^3 = i^2 * i = i
then
a * i^3 = a * (i) = a * i
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