Real numbers and complex numbers

• B
Summary:
To find √(-2)√(-3).

Method 1.
√(-2)√(-3) = √( (-2)(-3) ) = √(6).

Method 2.
√(-2)√(-3) = √( (-1)(2) )√( (-1)(3) ) = √((-1)√(2)√(-1)√(3) = i√(2)i√(3) = (i)(i)√(2)√(3) = -1√( (2)(3) ) =-√6.

Why don't the two methods give the same answer?
Thanks for any help.

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PeroK
Homework Helper
Gold Member
2020 Award
Summary:: Why don't the answers agree?

To find √(-2)√(-3).

Method 1.
√(-2)√(-3) = √( (-2)(-3) ) = √(6).

Method 2.
√(-2)√(-3) = √( (-1)(2) )√( (-1)(3) ) = √((-1)√(2)√(-1)√(3) = i√(2)i√(3) = (i)(i)√(2)√(3) = -1√( (2)(3) ) =-√6.

Why don't the two methods give the same answer?
Thanks for any help.
Because the first one is wrong! ##\sqrt{wz} \ne \sqrt w \sqrt z##

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FactChecker
Gold Member
When you start using complex numbers, the usual simple rules for the square root need to be replaced by more sophisticated methods. You often need to consider multiple values for the square root and determine which are valid in that specific case.

DaveE
mathman
Basically the problem with square roots is that they are not single valued. Using rules doesn't change that fact. The methods you are using go through different branches. The rules might restrict the branches used, but mathematics (as an abstract being) doesn't necessarily obey man-made rules.

FactChecker
DaveE
Gold Member

I'll just add that, as a real world practicing EE that uses complex numbers all the time, you are nearly always better off thinking of them in polar coordinates along with the knowledge that the arguments can "wrap around". Yes, sometimes you need to switch to rectangular coordinates (addition, for example), but when you get confused, always check the polar notation, it's usually simpler, IMO.

WWGD, 2milehi and FactChecker
fresh_42
Mentor
Well, this wouldn't be a problem, if we would introduce ##\mathbb{C}## as ##\mathbb{R}[x]/(x^2+1).##

\begin{align*}
\sqrt{-2}&=ax+b \;(a\neq 0)\Longrightarrow -2=a^2x^2+2abx+b^2 =-a^2+2abx+b^2 \Longrightarrow b=0\wedge a=\sqrt{2}\\
\sqrt{-3}&=cx+d \;(c\neq 0)\Longrightarrow d=0\wedge c=\sqrt{3}\\
\sqrt{-2}\sqrt{-3}&=\sqrt{2}x \cdot \sqrt{3}x =\sqrt{6}x^2=-\sqrt{6}
\end{align*}

etotheipi
mfb
Mentor
##-2 =-a^2+2abx+b^2 \Longrightarrow b=0\wedge a=\sqrt{2}##

I don't see how this would follow. ##b=0\wedge a=-\sqrt{2}## is a solution, too. You still get the same ambiguity the complex square root has.

FactChecker
Gold Member
In the long run, it is worth it to get used to multi-valued functions and how to deal with them. They will occur in other situations. This example is about as simple an introduction as you can get.

fresh_42
Mentor
I don't see how this would follow. ##b=0\wedge a=-\sqrt{2}## is a solution, too. You still get the same ambiguity the complex square root has.
You are right. But this is the ambiguity in the notation of ##\sqrt{a}##, not of complex numbers. Your argument uses that ##\pm \sqrt{a}## both solve ##x^2=a## which assigns four possibilities to ##\sqrt{-2}\sqrt{-3}.## It has nothing to do with complex numbers.

I thank all those who posted their replies.

But am I correct to argue as follows?

Using only real numbers,
√( (-2)(-3) ) = √6 is correct but √(-2)√(-3) does not exist as a real number.

Hence Method 1 and Method 2 in my first post cannot be compared.

FactChecker
Mark44
Mentor
But am I correct to argue as follows?

Using only real numbers,
√( (-2)(-3) ) = √6 is correct but √(-2)√(-3) does not exist as a real number.
Yes, correct. (-2)(-3) = 6, so you are taking the square root of a positive number. In the 2nd example both square roots represent complex (pure imaginary) numbers.
Hence Method 1 and Method 2 in my first post cannot be compared.
Yes.

FactChecker
Now the math is more clear in my mind.
Thanks Mark44.

mathwonk
Homework Helper
2020 Award
"Using only real numbers,
√( (-2)(-3) ) = √6 is correct but √(-2)√(-3) does not exist as a real number."

I guess I see this differently. I agree that both individual square roots are not real, but the question asks about their product, which is real, i.e. sqrt(-2) =± i.sqrt(2) and sqrt(-3) = ±i.sqrt(3), so sqrt(-2).sqrt(-3) = ±i^2.sqrt(2).sqrt(3) = ±sqrt(6), which is a real number.

I.e. the two factors do not exist as real numbers, but their product does. So we are not disagreeing about the math, but about the meaning of the sentence. I.e. to me, the product √(-2)√(-3) does exist as a real number, namely "the" real number ±sqrt(6). (It is not unique, but it is real.)

FactChecker
Gold Member
I.e. the two factors do not exist as real numbers, but their product does. So we are not disagreeing about the math, but about the meaning of the sentence. I.e. to me, the product √(-2)√(-3) does exist as a real number, namely "the" real number ±sqrt(6). (It is not unique, but it is real.)
I am uncomfortable talking about the multiplication of numbers that do not exist. I think that a more systematic approach is required to call it proper mathematics.

pbuk
mathwonk
Homework Helper
2020 Award
sorry, i considered my post unhelpful and thought i had deleted it.

now i am tempted to make things worse, and point out that (even "real") numbers do not actually exist.

trees and plants
sorry, i considered my post unhelpful and thought i had deleted it.

now i am tempted to make things worse, and point out that (even "real") numbers do not actually exist.
Could you justify what you say? Do natural numbers exist or perhaps correspond sometimes to properties of real things?

Complex numbers can correspond to points of a complex plane, real numbers can correspond to points of a real line if i remember correctly from the books i have read. So, what do you want to say when you say they do not exist? In what way do you mean it?

Natural numbers can be points of a real line but discretely distributed, not continuously like the real numbers as i have concluded.

Does mathematical quantity exist or does mathematical space exist or do
they correspond sometimes to properties of real things? I think mathematical quantity sometimes correspond to properties of real things but mathematical space does not perhaps at all and i think this can be justified and understood easily, at least for myself and perhaps other mathematicians. Euclidean 3 dimensional space though can be visualised but this visualisation is not according to the definitions of its objects.

Other mathematicians obviously have way more knowledge and abilities in problem solving and concluding in mathematics and physics better than i do in my opinion, so they should be able to do this.

Math may be used for for physics, other sciences or technology but it does not mean in my opinion and as i have justified that it necessarily corresponds to properties of real things.It may be connected to properties of real things, but that is different from saying it corresponds to properties of real things.

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DaveE
Gold Member
I am compelled to address the question: Do complex numbers exist?

In language, or Math, whether something exists or not depends on what it represents.

In my world of electronics, for example, it is convenient to define a generalized impedance as a complex number. This number represents a magnitude and phase shift between the voltage and current. These are two real quantities that could be represented as a 2-D vector, if you forbade the use of complex arithmetic. But these 2-D vector quantities also satisfy the requirements of a number (identities, inverses, zeros, etc.). So we use them because the arithmetic is easier.

OTOH, if you are counting apples in a basket complex numbers make no sense and don't "exist".

"The teaching is merely a vehicle to describe the truth. Don’t mistake it for the truth itself. A finger pointing at the moon is not the moon. The finger is needed to know where to look for the moon, but if you mistake the finger for the moon itself, you will never know the real moon. The teaching is like a raft that carries you to the other shore. The raft is needed, but the raft is not the other shore. An intelligent person would not carry the raft around on his head after making it across to the other shore." - Thich Nhat Hanh

PeroK
trees and plants
I am compelled to address the question: Do complex numbers exist?

Whether something exists or not depends on what it represents.
existence ##\neq## what it represents . This can be justified in my opinion. Especially when talking about knowledge.

DaveE
Gold Member
existence ##\neq## what it represents . This can be justified in my opinion. Especially when talking about knowledge.
So, you may ask "Do you believe in God?" My answer is God exists now because you just referenced it in your question. We may not agree or completely understand what it is, but it exists because we are talking about it.

So we'll agree to disagree. Representation begets existence, at least in some sense.

pbuk
Gold Member
I.e. to me, the product √(-2)√(-3) does exist as a real number, namely "the" real number ±sqrt(6). (It is not unique, but it is real.)
The solution is unique:
\begin{align} x =& \sqrt{-2}\sqrt{-3} \\ =& \sqrt{(-1)2}\sqrt{(-1)3} \\ =& i\sqrt{2}i\sqrt{3} \\ =& i^2\sqrt{(2)(3)} \\ =& -\sqrt6 \end{align}

I am uncomfortable talking about the multiplication of numbers that do not exist. I think that a more systematic approach is required to call it proper mathematics.
Yes: 'x does not exist as a real number' is not a mathematical statement (we would say 'x is not an element of the set of real numbers', or just 'x is not a real number').
now i am tempted to make things worse, and point out that (even "real") numbers do not actually exist.
In language, or Math, whether something exists or not depends on what it represents.
None of the things we use in mathematics 'exist' until we create them from axioms: that way we don't have to make assumptions about how they are going to behave.

So we create the integer '2' as the successor of '1', not as the representation of how many things we have if we have two things (discussions elsewhere in PF about equivalence classes notwithstanding).

PeroK
mathwonk
Homework Helper
2020 Award
I also started to say that sqrt(-2) = i.sqrt(2) until I realized that sqrt(-2) means literally anything whose square is 2, so it could also be -i.sqrt(2), but I think I made that clear.

as far as talking about a product being a real number when the factors are not, do you believe that the product of some non - real complex numbers can be a real number? i.e. for example does i^2 exist as a real number? after all it does equal -1. would you say the equation i^2 = -1 makes no sense because the number i does not exist?

as to whether numbers exist, do you believe an infinite set exists? even the integers are an infinite set, but what in the universe is actually infinite? in the class i audited on this topic (constructing, or defining integers, rationals, and eventually real numbers, taught by george mackey) the first unprovable assumption was to assume an infinite set exists. I admit it surprized me at the time that this was considered unprovable.

forgive me for abetting this opinionated discussion, which will no doubt live on and on.

jim mcnamara
FactChecker
Gold Member
The history of complex numbers started with their use to give the correct answers even though they were not understood as an actual number system. They could be used as a "trick" that disappeared from the final result which was a valid real-number solution. I would hesitate to go back in history to that practice and still call it "mathematics".
That is what I feel is happening if we say that ##\sqrt {-2} \sqrt {-3} = \pm \sqrt 6## without a proper understanding of the individual factors on the left.

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Mark44
Mentor
I also started to say that sqrt(-2) = i.sqrt(2) until I realized that sqrt(-2) means literally anything whose square is 2
I'm pretty sure you meant "anything whose square is -2" ...

pbuk