-1

Multiplying $2$ times a real number gives $n \cdot n$ positive:

$2^2 =2\cdot2 =4$

But multiply $2$ times $i$ gives $-1$:

$i^2 =i\cdot i=-1$

Shouldn't $i \cdot i $ give $i^2$?

Does $i \cdot i = -1 \cdot 1$?  

Caleb Stanford
  • 45,895
molo32
  • 151
  • What is your definition of $i$? – Andrei Mar 24 '20 at 18:17
  • 1
    Yes, $i\cdot i=i^2$ and $i^2=-1$. –  Mar 24 '20 at 18:17
  • By definition, $i = \sqrt{-1}.$ And $ i^2 = (\sqrt{-1})^2 = -1.$ Note, $i$ is NOT a real number. If it were, $i=\sqrt{-1}$ would be undefined. $i$ is a complex number, not real. – amWhy Mar 24 '20 at 18:20
  • $x^2 = -1$ has no solutions in reals, and $,2x = 1$ has no solutions in integers. In both cases we can extend these number systems (rings to obtain another number system of similar algebraic structure (rings) which has solutions of these equations, viz. $\Bbb C = $ complex numbers, $,\Bbb Q = $ rationals or fractions, and this proves fruitful: working in these extended numbers systems often simplifies results in our original number system. – Bill Dubuque Mar 24 '20 at 18:34

5 Answers5

5

multiplying $2$ times a real number gives $n \cdot n$ positive.

$2^2 =2 \cdot 2 =4$

This is called squaring, not multiplying by $2$. Multiplying by $2$ would be, for example: $2 \cdot 5 = 10$, $2 \cdot 3 = 6$.

Squaring is different than multiplying by $2$: $5^2 = 5 \cdot 5 = 25$, $3^2 = 3 \cdot 3 = 9$.

but multiply $2$ times $i$ gives $-1$

$i^2 =i \cdot i=-1$

Yes, squaring $i$ gives $-1$. Isn't that strange? This is why $i$ is a new number: all the previous numbers square to get a positive number, and $i$ squares to get a negative number.

shouldn't $i \cdot i $ give $i^2$?

Yes it does! But $i^2$ is also equal to $-1$, just like $3^2$ is also equal to $9$.

Caleb Stanford
  • 45,895
  • Isn't squaring rather a special case of multiplication? – PCeltide Mar 24 '20 at 18:46
  • @PratikApshinge Yes, that's true – Caleb Stanford Mar 24 '20 at 19:13
  • @PratikApshinge In some contexts, squaring can be interpreted as a special case of multiplication. On the other hand, it can also be seen as a special case of exponentiation, where the exponential function is defined in more abstract ways (e.g. $u(x) = \mathrm{e}^x$ is the unique solution to the BVP $u' = u$ with $u(0) = 1$; or $\mathrm{e}^x$ can be defined in terms of a series expansion). In any event, when you square a number, you are not multiplying by $2$; you are multiplying a number by itself. – Xander Henderson Mar 30 '20 at 02:20
2

Since, by definition, $x^2$ means $x\times x$, yes, $i^2=i\times i$.

And $i^2=-1$ because we define $i$ so that $i^2=-1$.

Finally, the fact that the product of a real number by itself is always positive is not relevant here, since $i$ is not a real number.

José Carlos Santos
  • 427,504
2

The number $i$ was defined such that $i^2=-1$, so as a consequence, $i=\sqrt{-1}$. Because we are in the complex plane, we can no longer use the rules used in the real plane.

$i^2 \neq 1$, because $i=\sqrt{-1}$ $\notin$ $\mathbb R$, because $\forall$ $x$ $\in$ $\mathbb R$, $x^2 \ge 0$.

Darius Chitu
  • 447
1

The most interesting question is "why $i^2=-1 $ ?".

It turns out that mathematicians were a little "frustrated" that some polynomial equations like

$$x^2+1=0$$

seem to have no solutions. Then they "arbitrarily" decided that this equation had a solution, called it $i$, so that

$$i^2+1=0$$

(if you prefer, $i^2=-1$).

From this they discovered that using $i$ and reals, you can in fact solve all polynomials in the world. They also discovered many other interesting and useful properties and that gave birth to the powerful theory of complex numbers.

  • 1
    This is an oversimplification. Mathematicians where not bothered at all by quadratic equations with no solutions. Instead they were very much concerned with cubic equations with solutions, but where Cardano’s formula required to use square roots of negative numbers. – egreg Mar 25 '20 at 09:04
  • 1
    @egreg: yes, this is willingly an oversimplification and historical nonsense. The pedagogical goal is to (try to) give the OP a taste of why we use complex numbers. –  Mar 25 '20 at 09:09
1

As much as we have that: $$\underbrace{a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}=a^n\tag{1.1}$$ (which is the general rule you are using), we also have the corollary (by replacing $a$ with $a^\frac1n$): $$\underbrace{a^\frac1n\cdot a^\frac1n\cdot \ldots\cdot a^\frac1n}_{n\text{ times}}=a\tag{1.2}$$

The imaginary unit $i$ is defined by $i^2=-1\implies i=(-1)^\frac12$

and so inserting $a=-1, n=2$ into $(1.2)$ yields your answer

Rhys Hughes
  • 12,842