Why is sqr(-4) not a real number again?

[alpha372]

Homework Statement

I know that the square root of a negative number is not a real number, but I can't remember the logic that gets to that.

Homework Equations

The Attempt at a Solution

I've read, "the square root of (-4) is not a real number because there is no real number you can square to get to -4,

but I'm thinking, you can square the square root of -4 to get -4:

(√-4)^2 = -4
what is the problem with this again?

 

[noumed]

True, but you'll just be negating the square root to begin with.
Plus sqrt(-4) is not a number. It's more of an expression.

 

[Citan Uzuki]

The problem is that √(-4) is not a real number, so showing that its square is -4 (which is a tautology anyway) does not establish that -4 is the square of a real number.

Now, let us escape from the circular logic. Let x be any real number. Then x is either positive, negative, or zero. In each case, what can you say about the sign of x²?

 

[Mentallic]

While $$\sqrt{-4}$$ might be an unreal number because no number squared can give -4, it doesn't mean to say you can't go about it backwards. i.e. how you suggested $$(\sqrt{-4})^2=-4$$. Now think about it. What number can you square to give -4? There are no real numbers that you can square to give a negative, so this number must be imaginary.

A little summary of how imaginary numbers work ~
let $$\sqrt{-1}=i$$

We have $$\sqrt{-4}=\sqrt{4*(-1)}=\sqrt{4} \sqrt{-1}=2i$$ could you follow this?

Now, if we take the square of 2i i.e. $$(2i)^2=4i^2$$ can you realize that while i is imaginary, the square of it becomes a real number? $$(\sqrt{-1})^2=-1$$ so we can substitute $$i^2=-1$$ and get the answer $$4i^2=-4$$ which is the result we had before

 

[Dr. Lady]

That's some circular logic that you're employing. In order to prove that sqrt(-4) is a real number, you must assume that sqrt(-4) is a real number. "because there is no REAL number that you can square to get -4.

If you need to visualize, y=+-sqrt(x) is the inverse function of y=x^2, meaning we flip it over the y=x line. so y=+-sqrt(x) is like a parabola on the x axis... no -x can be used as input to get a real number.

 

[alpha372]

That's some circular logic that you're employing. In order to prove that sqrt(-4) is a real number, you must assume that sqrt(-4) is a real number. "because there is no REAL number that you can square to get -4.

How would you go about proving that sqrt(-4) is not a real number?

 

[Mentallic]

How would you go about proving that sqrt(-4) is not a real number?

I don't think this needs a proof, rather just logic.

Take all the real numbers, the negatives, positives and zero.

For each case, square them. Can you see how the range (possible values after squaring) can only be zero or more? There can be no negative number after you have squared a real number. Thus, you can not take the square root of a negative number.

If you can, they are imaginary numbers

 

[arildno]

How would you go about proving that sqrt(-4) is not a real number?

Ok-dokey:

Let a>0, and a real number. ("a" is therefore a positive number)

Therefore, following from axioms of ordering, we may multiply each side of the inequality by "a", and retain the >-sign:
We get:
$$a^{2}>a*0$$
whereby folllows:
$$a^{2}>0$$
which implies that the square of every positive real number is necessarily some positive number.

Now, assume that a<0, i.e, a negative number.

If we multiply THIS inequality with "a", we must change the inequality sign, since "a" is negative!
We therefore get:
$$a^{2}>a*0$$
whereby folllows:
$$a^{2}>0$$
which implies that the square of any NEGATIVE real number is necessarily some positive number.


But, since the poisitives, negatives and 0 EXHAUSTS the real numbers, it follows that there cannot exist any real number so that its square is negative!.

Thus, whatever it might be, if anything, $\sqrt{-4}$ is definitely not a real number..

 

[jacksonpeeble]

Any square root of a negative number is imaginary.

 

[alpha372]

I don't think this needs a proof, rather just logic.

I didn't ask if you thought it needed a proof. I asked how it could be done.

 

[alpha372]

Ok-dokey:

Let a>0, and a real number. ("a" is therefore a positive number)

Therefore, following from axioms of ordering, we may multiply each side of the inequality by "a", and retain the >-sign:
We get:
$$a^{2}>a*0$$
whereby folllows:
$$a^{2}>0$$
which implies that the square of every positive real number is necessarily some positive number.

Now, assume that a<0, i.e, a negative number.

If we multiply THIS inequality with "a", we must change the inequality sign, since "a" is negative!
We therefore get:
$$a^{2}>a*0$$
whereby folllows:
$$a^{2}>0$$
which implies that the square of any NEGATIVE real number is necessarily some positive number.


But, since the poisitives, negatives and 0 EXHAUSTS the real numbers, it follows that there cannot exist any real number so that its square is negative!.

Thus, whatever it might be, if anything, $\sqrt{-4}$ is definitely not a real number..

Thanks!

 

[alpha372]

Any square root of a negative number is imaginary.

You didn't really understand what I was asking.

 

[alpha372]

The problem is that √(-4) is not a real number, so showing that its square is -4 (which is a tautology anyway) does not establish that -4 is the square of a real number.

Now, let us escape from the circular logic. Let x be any real number. Then x is either positive, negative, or zero. In each case, what can you say about the sign of x²?

The top half is very helpful to my issue- thanks, I'm making a note of it.

The bottom half was irrelevant to the question.

 

[alpha372]

True, but you'll just be negating the square root to begin with.
Plus sqrt(-4) is not a number. It's more of an expression.

How would you prove that sqrt(-4) is not a number?

I argue that it is in fact a number -- a complex number to be exact. Complex numbers are a subset of what we call, "numbers", therefore, I find your statement, "Plus sqrt(-4) is not a number. It's more of an expression." to be false.

 

[HallsofIvy]

I don't think this needs a proof, rather just logic.

Take all the real numbers, the negatives, positives and zero.

For each case, square them. Can you see how the range (possible values after squaring) can only be zero or more? There can be no negative number after you have squared a real number. Thus, you can not take the square root of a negative number.

If you can, they are imaginary numbers

That is a proof, not "just logic" because you use the properties of real numbers (the product of two positive numbers is positive and the product of two negative numbers is negative) which are not "just logic" (i.e. the rules of logic).

 

[HallsofIvy]

Alpha372, you can prove, from the properties of real numbers, that the product of any two positive numbers is positive and that the product of any two negative numbers is positive. Of course, it is also true that 0*0= 0.

Now, as you say in your first post, "you can square the square root of -4 to get -4:

(√-4)^2 = -4". The "difficulty with that" is that you haven't shown that √-4 is a real number. If it were it would have to be positive, or negative, or 0. It cannot be 0 because 0²= 0, not -4. It can't be positive because then (√-4)² would be the product of two positive numbers and so would be positive, not -4. Finally, it cannot be negative because then (√-4)² would be the product of two negative numbers and so would be positive, not -4. Since those are the only three possibilities, it follows that √-4 is not a real number.

That, of course, depends on the fact that the product of two positive numbers is positive and the product of two negative numbers is negative. To prove that, remember the properties of an "ordered field".
1) If a> b then a+ c> b+ c.
2) If a> b and c> 0, then ac> bc.
3) If a and b are any two numbers in the ordered field than one and only one of these is true:
i) a= b
ii) a> b
iii) b> c

Now, if a is a positive number and b are both positive numbers then a> 0 and b> 0 so, by ii ab> 0 and ab is a positive number.

If a is a negative number and b is a negative number, then 0> a so, adding -a to each side, i tells us that 0+ -a> a+ -a or -a> 0. Also, 0> b so ii says that (-a)(0)> (-a)(b) or 0> -ab. Finally, adding ab to both sides, by i, 0+ ab> -ab+ ab or ab> 0. That is, ab is positive.

 

[Avodyne]

the product of any two negative numbers is negative

I think you were typing a bit too fast there, Halls :)

 

[HallsofIvy]

Thanks for catching that. I have editted it.

 

[alpha372]

To answer my own question directly, and save everyone some time from trying to explain what doesn't need to be explained (i.e, I don't care about how things look on a graph; I wanted to know the error in the logic, which has been pointed out several times already anyway)

the problem is, I'd have to turn "the square root of (-4) is not a Real Number because there is no real number you can square to get to -4," (1)into:

" (√-4) is a real number because you can square the real number (√-4) to get to -4." (2)

Which means that I would have to prove that (√-4) is a real number, by definition of what a real number is. Or, to save time, I could simply say, it is not a real number, because it is a complex number via definition of a complex number, therefore, it cannot be a real number, meaning that statement (2) is false. (This was most directly noted previously by Citan Uzuki, and some what by Mentallic , and also noted by Dr. Lady, and later restated again by Halls of Ivy. Extra proof that it was not a real number was shown by arildno and again by Halls of Ivy.)

It is also interesting to note the following:

(√-4)^2 = √((-4)^2) = √(4) = 2 (1)

and at the same time:

(√-4)^2 = (-4) ^ 2/2 = -4 (2)


the first statement breaks down because (√-4)^2 = √((-4)^2) can only work if -4 is a positive number (this property of radicals only holds for positive numbers), and clearly, -4 is not positive.