# What's wrong with this proof?

#### murshid_islam

what's wrong with the following proof:

$$i = \sqrt\left(-1\right)$$
$$i.i = \sqrt{\left(-1\right)}.\sqrt{\left(-1\right)}$$
$$i^2 = \sqrt\left(-1\right).\left(-1\right)$$
$$i^2 = \sqrt 1$$
$$i^2 = 1$$

but we know that i2 = -1

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#### matt grime

Homework Helper
Oh dear god not again.

There must be one of these a week.

Given you've put a 'winking' thing there we assume you know what is wrong with it (branches of square roots). Which makes this thread bloody annoying.

#### murshid_islam

sorry, the winking thing was UNintended. i meant to give the "question mark". i really dont know whats wrong with this. so any help will be appreciated.

#### matt grime

Homework Helper
I told you: branches. Square roots don't behave like that. There is no reason to suppose they do. This demonstrates that they don't. It's only confusing if you think it must be true and there is no reason to suppose that.

#### murshid_islam

can you please explain what you mean by branches?

$$i = \left(-1\right)^{1 \over 2}$$
$$i.i = \left(-1\right)^{1 \over 2}.\left(-1\right)^{1 \over 2}$$
$$i^2 = \left(\left(-1\right).\left(-1\right)\right)^{1 \over 2}$$ (since apbp = (ab)p)
$$i^2 = 1^{1 \over 2}$$
$$i^2 = 1$$

G

#### grant9076

When you square a complex number, you are actually multiplying it by its conjugate. If the number is a+bi then its conjugate is a-bi. In this case, a=0 and b=1. If you substitute them into your equation, the error should be abundantly clear.

#### HallsofIvy

From a different viewpoint: It is not sufficient to define "$i= \sqrt{-1}$" because there are two complex numbers whose square is -1. Since the complex numbers do not form an ordered field, we cannot distinguish between them by saying "the positive root" as we do in real numbers.

Better definition of complex numbers: the set of ordered pairs of real numbers (a, b) with addition defined "coordinate wise"- (a,b)+ (c,d)= (a+c, b+d)- and multiplication defined by (a,b)(c,d)= (ac- bd,ad+bc). Identify the real number, a, with (a, 0) to make the real numbers a subset and then i= (0, 1). i*i= (0,1)(0,1)= (0*0-1*1,0*1+1*0)= (-1, 0). With those definitions, it is not in general true that $\sqrt{a}\sqrt{b}= \sqrt{ab}$.

#### murshid_islam

so the 3rd line of the following is wrong???

$$i = \left(-1\right)^{1 \over 2}$$
$$i.i = \left(-1\right)^{1 \over 2}.\left(-1\right)^{1 \over 2}$$
$$i^2 = \left(\left(-1\right).\left(-1\right)\right)^{1 \over 2}$$ (since apbp = (ab)p)
$$i^2 = 1^{1 \over 2}$$
$$i^2 = 1$$

G

#### grant9076

Yes. The third line is wrong.

#### matt grime

Homework Helper
grant9076 said:
When you square a complex number, you are actually multiplying it by its conjugate

No, you are not.

#### murshid_islam

HallsofIvy said:
Identify the real number, a, with (a, 0) to make the real numbers a subset and then i= (0, 1). i*i= (0,1)(0,1)= (0*0-1*1,0*1+1*0)= (-1, 0). With those definitions, it is not in general true that $\sqrt{a}\sqrt{b} = \sqrt{ab}$
can you please give me a specific example where $\sqrt{a}\sqrt{b}$ is not equal to $\sqrt{ab}$?

Homework Helper

#### murshid_islam

matt grime said:
so $\sqrt{\left(-1\right)}.\sqrt{\left(-1\right)}$ is NOT equal to $\sqrt{\left(-1\right).\left(-1\right)}$. is that what you mean?

#### Swapnil

Essentially, the confusion is due to the fact that $$\sqrt{a}\sqrt{b} = \sqrt{ab}$$ is only true when $$a,b \geq 0$$. So, yes, in your third step you violated this essential rule.

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#### Data

murshid_islam said:
so $\sqrt{\left(-1\right)}.\sqrt{\left(-1\right)}$ is NOT equal to $\sqrt{\left(-1\right).\left(-1\right)}$. is that what you mean?
Yes, that is what he means. Safest is to only assume sqrt(ab) = sqrt(a)sqrt(b) when a, b are both positive reals.

We can write any complex number $z$ in a unique way as $z=re^{i\theta}$ where $r>0$ and $\theta \in [0, 2\pi)$.

When $x$ is a nonnegative real define $\sqrt{x}$ to be the unique nonnegative real number $y$ such that $y^2 = x$. You can show using this definition that $\sqrt{a}\sqrt{b}$ holds when $a,b$ are nonnegative reals. Then for any complex number $z$ written in the form I mentioned, we define $\sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.$

So let's try out this definition. If $z=-1$, then we write $z = e^{i\pi}$, and we get $\sqrt{z} = e^{i\pi / 2} = i$. As you might expect.

So what's $\sqrt{-1}\sqrt{-1}$? Well, it's $e^{i\pi/2}e^{i\pi /2} = e^{i\pi} = -1$. As you've already shown, $\sqrt{(-1)(-1)} = 1$, so we conclude that $\sqrt{a}\sqrt{b}$ does not hold in general for complex $a,b$.

(This has all been posted in previous threads, by myself and others!)

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#### murshid_islam

thanks very much for your explanation, Data. it made a lot of things clear to me.

#### HallsofIvy

grant9076 said:
When you square a complex number, you are actually multiplying it by its conjugate. If the number is a+bi then its conjugate is a-bi. In this case, a=0 and b=1. If you substitute them into your equation, the error should be abundantly clear.
No, the square of a number is the number multiplied by itself. You are talking about getting the square of the modulus which is a completely different thing.

"What's wrong with this proof?"

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