Square Root of 4: Can it be +2 or -2?

[Taturana]

Consider square root of 4.

Can square root of 4 be +2 or -2?

I asked that to my math teacher so he said: NO, square root of 4 is +2!

But I can't really understand why it cannot be -2 since -2 squared is also 4. One thing I've imagined is: square root of 4 can be +2 or -2, but for CONVENTION we use +2. Is that right?

There's another question related to that: consider

i^2 = -1

squaring the two sides we obtain

i^4 = 1

so

i = 1

so: can the i (the imaginary number) be 1? I think I'm making a lot of confusion with that, could someone explain me that?

Thank you

 

[eg2333]

The square root of 4 can be +2 or -2, your math teacher is retarded. What you did with the imaginary number results in what I believe is called an extraneous solution (correct me if I'm wrong). If you substitute the 1 back into your original problem i^2=-1, you will find that 1=-1, which is not a true statement, meaning the solution is incorrect.

 

[jgens]

We get these types of questions a lot so it might be worthwhile to run a forum search on this topic to get a very thorough review of it. In fact, I might even say that this question has been discussed ad nauseum

However, for your first question the, square root of a number is often defined as the principal (or positive) root. This convention is not without reason though; if we would like to have a square root function, the square roots must necessarily be single-valued. Therefore, \sqrt{4} = 2. For a more complicated example, let x be a number such that x^2 = 4, then \sqrt{x^2} = |x| = 2 because \sqrt{x^2} is necessarily positive. Since |x| = 2, this means that x = 2 or x = -2.

For your second question, squaring equations has the unfortunate effect of adding extraneous solutions. Taking your example, clearly i^4 - 1 = 0 in which case (i^2 + 1)(i^2 - 1) = 0. By the zero product property, either i^2 + 1 = 0 or i^2 - 1 = 0 (note that only one of these need hold). From the conventional definition of i we have that i^2 + 1 = 0 and i^2 - 1 = 0 is just an extraneous solution added when we squared the equation.

Hopefully this all makes sense.

 

[Taturana]

Thank you for the help guys. Now things make more sense.

PS: next time i'll do a forum search...

 

[HallsofIvy]

The square root of 4 can be +2 or -2, your math teacher is retarded. What you did with the imaginary number results in what I believe is called an extraneous solution (correct me if I'm wrong). If you substitute the 1 back into your original problem i^2=-1, you will find that 1=-1, which is not a true statement, meaning the solution is incorrect.

If his teacher is retarded, what does that make you? His teacher is completely right. The equation x^2= 4 has two roots. But only one of them is \sqrt{4} because \sqrt{a} is defined as "the positive root of the equation x^2= a".

The reason we have to write the solutions to that equation as "x= \pm\sqrt{a} is that \sqrt{a} alone does NOT mean both solutions. If it did we would not need the "\pm".