What must be the same before two radicals can be multiplied or divided?

Answer 1

The radicals must have the same indexes.

Examples when radicals have index of 2.

√3√7 = √(3*7) = √21

√2√6 = √(2*6) = √12 = √(4*3) = √4√3 = 2√3

We decided that i2 = -1, and i = √-1, to be able to find the square root of any negative radicand, which is not a real number.

For example,

√-4 = √(-1*4) = √-1√4 = 2i.

√-45 = √-1√45 = i√(9*5) = i√9√5 = 3√5 i

Also, we allow ourselves to multiply the radicals of index 2, even though one or both of the radicals are negative.

√-5√6 = = √(-5*6) = √-30 = i√30. But I do prefer to work in this way:

√-5√6 = = √-1√5√6 = = i√(5*6) = i√30. This work will eliminate the common mistake that students always do when they multiply radicals of index 2 with both negative radicands.

For example,

√-2√-3 = √(-2*-3) = √6 (wrong, a positive real number!)

√-2√-3 = (√-1√2)(√-1√3) = (i)(i)√(2*3) = i2√6 = -√6.

Examples of division:

√10/√2 = √(10/2) = √5

√-15/√3 = (√-1√15)/√3 = i√(15/3) = i√5

√21/√-7 = √21/(√-1√7) = (1/i)√(21/7) = (1/i)√3 but this result is not a representative of an imaginary number, as we decided it to be. So what to do in order to manipulate the result and write it as a proper imaginary number?

Let 1 = -(-1) = -(i2), then I will have (1/i)√3 = (-i2/i)√3 = -i√3, the right answer.

Conclusion: 1/i = -i.

√-35/√-5 = (√-1√35)/(√-1√5) = (i/i)(√35/5) = √7, the right work.

Even though, √-35/√-5 = (√(-35/-5) = √(35/5) = √7, it is wrong to work in that way.

So we solved the problem of finding the square roots of negative numbers, by using the imaginary number i, and we also called it a complex number, and so we built a new set of numbers, the set of complex numbers. Actually, all numbers are invention of our mind, even though I would prefer not to say it for the number 1.