# What is wrong with my 'proof' of $i=1$?

So I recently learnt about $$i$$ and I can't wrap my head around the concept of $$i^2=-1$$ or that $$\sqrt{-1}$$ can even exist. Today I was thinking about $$i$$ again and thought of a "proof" that $$i=1$$. $$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$ $$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$ But if $$i=1$$, then $$i^2 \neq -1$$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $$\text{"prove" }1=2$$ by accidentally dividing by $$0$$ and I suspect something similar is happening.

For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html

• In your first sentence, did you intend to write, "I can't wrap my head around the concept of $i^2=−1$ " ? – Adam Rubinson Jan 19 at 21:46
• This argument was shown million times. You cannot blindly apply the rules of real numbers to the complex. The flaw is $i^4=1\implies i=\sqrt[4]1$. Anyway, $1=\sqrt[4]1$ is still true. – Yves Daoust Jan 19 at 21:46
• "can even exist." Well, nothing in mathematics actually "exists". – fleablood Jan 19 at 21:48
• math.toronto.edu/mathnet/answers/imaginary.html – user9464 Jan 19 at 21:52
• @mrsamy Thanks a lot for the link. I just saw it and helps a lot – Silas Dyck Jan 20 at 19:23

This has nothing to do with complex numbers. By the same argument, $$-1=1$$, since$$(-1)^2=1^2\implies\sqrt{(-1)^2}=\sqrt{1^2}\implies-1=1.$$The error lies in assuming that $$\sqrt{x^2}=x$$. Actually, $$\sqrt{x^2}=|x|$$. In the case of complex numbers, it's even worst, since every complex number (other than $$0$$) has four fourth roots. So, the expression $$\sqrt[4]z$$ doesn't make sense unless and until you decide which fourth root of $$z$$ you have in mind. Even then, it will often be false that $$\sqrt[4]{z^4}=z$$.

• Thanks, I figured I must be missing something since I'd tend to assume that people who are more advanced at math than me and say $i$ is a number and the $i=\sqrt{-1}$ are probably correct. – Silas Dyck Jan 20 at 16:50
• I'm glad I could help. – José Carlos Santos Jan 20 at 17:18
• first equation should have square on both sides, for better mirroring of OP question. Unfortunately that's a 1 char change so I can't make it. – Ross Presser Jan 20 at 19:55
• @RossPresser Done. Thank you. – José Carlos Santos Jan 20 at 20:00

Square roots don't work like that. With the same argument you used, you could have done $$(-1)^2=1^2\ \ \ \implies\ \ \ -1=1.$$ It has nothing to do with complex numbers.

If this may reconcile you with complex numbers, consider this: a complex number is a pair of reals, and we define

• addition as $$(a,b)+(c,d):=(a+b,c+d)$$,

• multiplication as $$(a,b)(c,d):=(ac-bd,ad+bc)$$.

Then we accept the "shorthand" conventions $$1=(1,0)$$ and $$i=(0,1)$$. It is an easy matter to show that these rules form a consistent arithmetic with the four basic operation, where you can freely use these equivalences.

We also have the consequences

$$i^2=(0,1)(0,1)=(-1,0)=-1$$

and

$$i^4=(-1,0)(-1,0)=(1,0)=1.$$

But that means in no way that $$i=1$$. It just means that $$i$$ is one of the solutions of the equation

$$z^4=(1,0)=1.$$

All solutions are (see below)

$$(1,0)=1,\\(0,1)=i,\\(-1,0)=-1,\\(0,-1)=-i.$$

Unless you specify a convention, in the complex we don't know which solution $$\sqrt[4]1$$ denotes.

Final remark: in this discussion, we only used real numbers and the special symbol $$i$$.

$$z^4=(x,y)^4=(x^4-6x^2y^2+y^4,4yx^3-4xy^3)=(1,0)$$ requires $$x=0\lor y=0\lor x^2=y^2$$

(by cancelling the imaginary part).

By plugging these in the real part, we get $$y^4=1\lor x^4=1\lor -4x^4=-4y^4=1.$$

As $$x,y$$ are real, the only options are

$$(\pm1,0),(0,\pm1).$$

Think of the complex numbers as a pair of real numbers $$\mathbb{R}\times\mathbb{R}$$, but with a really useful property: the product $$(x, y)\cdot (0,1) = (-y, x)$$. Which shows the rotation characteristic of the number i.

Thinking that way is clear that the number $$(0,1)$$ has the property of $$(0,1)^2 = (-1,0)$$.

The thing is that if $$m= n$$ then we can conclude $$f(m) = f(n)$$.

But if $$f(m) = f(n)$$ we can not conclude $$m = n$$ unless $$f$$ is one to one.

If $$f$$ is not one to one is perfectly possible to have $$m \ne n$$ but $$f(m) = f(n)$$.

A simple example is $$f(x) = x^2 - 3x +6$$ If $$x=2$$ we get $$f(2) = 4 - 6 + 6 = 4$$. And if $$x=1$$ we get $$f(1)=1-3 + 6 = 4$$. So $$f(1) = f(2) =4$$ but $$1\ne 2$$.

An even easier example is $$f(x) = x^4$$ and $$f(-1) = f(1) = f(i) = f(-i)$$ will give us $$1^4 = 1$$ and $$(-1)^4 = 1$$ and $$i^4 = (i^2)^2 =(-1)^2 =1$$ and $$(-i)^4 = ((-i)^2)^2=((-1)^2i^2)^2 = (1*(-1))^2 = 1$$.

the problem is when sloppy teachers give to impressionable students this INCORRECT definition:

THIS IS WRONG: $$\sqrt[k]{m}$$ is equal to the $$x$$ so that $$x^k = m$$.

The problem is that there are $$k$$ different $$x$$ that give $$x^k = m$$ and so that isn't actually a definition of a single value.

Ex. $$\sqrt{16}$$ iss the $$x$$ so that $$x^2= 16$$. Well $$4^2 = 16$$ and $$(-4)^2 = 16$$. So which is it? is $$\sqrt{16} = 4$$ of is $$\sqrt{16} = -4$$.

Well, the answer is we define that $$\sqrt{m}$$ is the positive value $$x$$ so that $$x^2 = m$$.

But two things come about from this.

One $$\sqrt{x^2} \ne x$$. That is just wrong. $$\sqrt{x^2} =|x|$$ because we don't know that $$x$$ is positive.

And 2) $$i^2 = 1$$ is a property of $$i$$ but that does not mean $$\sqrt{-1} = i$$. Because $$i$$ is neither positive nor negative.

Anyway.....

tl;dr

There are four values, all different of $$x$$ so that $$x^4 = 1$$. They are $$1^4 =1; (-1)^4 = 1; (-i)^4=1$$

$$\sqrt[4]{x^4} \ne x$$. IF $$x^4$$ is a positive real number then $$\sqrt[4]{x^4} = |x|$$ and indeed $$|1| = |-1| = |i| = |-i|$$ but $$\sqrt[4]{x^4} \ne x$$.

And if $$x^4$$ is not a positive real number we don't actually have a definition for $$\sqrt[4]{x^4}$$.

• Sorry, I don't like this explanation much, because you take for granted that $x^4=1$ has four solutions, when the OP is still at a stage that he knows about the two real solutions $\pm1$. Strictly speaking, $i$ is just surrealistic, unless you introduce the imaginary and complex numbers with some formalism. – Yves Daoust Jan 19 at 22:20
• Point taken. I was trying explain the error in $\sqrt[4]{x^4} = x$ so $i^4 = 1^4 \implies 1=i$. I was pointing out that $x^4=1$ has more than $1$ solution and not claiming that I was going to prove that it has exactly four in any way. But I did demonstrate that there are at least four. Point taken that $i^2 = -1$ is something tossed in from no-where but I didn't think that was really the issue of the question. – fleablood Jan 19 at 23:21
• My comment was influenced by the first sentence of the OP, but no worries. – Yves Daoust Jan 20 at 9:03