$\iota \equiv \pm 3, \pmod{10}$

Question

I was reading up on the properties modulo function, when I saw the property:

$$-a \equiv (10-a) \space \pmod{10}$$

Which means

$$-1 \equiv (10-1) \equiv 9 \space \pmod{10}$$

Now: $$\iota = \sqrt{-1}$$

Substituting $-1 \equiv 9 \pmod{10}$:

$$\iota \equiv \sqrt{9} \pmod{10}$$ $$\iota \equiv \pm 3 \pmod{10}$$

Where did I go wrong?

Answer 1

The confusion in the question is in thinking that $\sqrt{-1}$ always denotes the same number. In the end, the expression "$\sqrt{-1}$" denotes a number whose square is $-1$. So the meaning of the symbol $\sqrt{-1}$ depends on the multiplication operation - if we switch to a different multiplication operation, we can get a different meaning for some square roots.

• In the real numbers, there is no value for $\sqrt{-1}$ - there is no real number that is the square root of $-1$ under multiplication of real numbers.

• In the complex numbers, there are two possible values for the square root of $-1$ under multiplication of complex numbers. These are usually called $i$ and $-i$.

• If we work with integers modulo $5$, $2 = \sqrt{-1}$, because $2^2 = 4 \equiv -1 \pmod{5}$. But $4$ is not a square root of $-1$, modulo $5$, because $4^2 = 16 \equiv 1 \not \equiv -1 \pmod{5}$. The number $3$ is the other possible value for $\sqrt{-1}$, modulo $5$.

• If we work with integers modulo $17$, now we have $4 = \sqrt{-1}$, because $4^4 = 16 \equiv -1 \pmod{17}$. The other square root of $-1$, modulo $17$, is $13$, which is $-4 \pmod{17}$.

• If we work modulo 3, there is no square root of $-1$, because $1^2 \equiv 2^2 \equiv 1 \not \equiv -1 \pmod{3}$.

There is nothing special about square roots here. For example, modulo $5$ we have $1/2 = 2^{-1} \equiv 3 \pmod{5}$, because $3$ multiplied by $2$ gives $1$, modulo $5$. Of course the real numbers $1/2$ and $3$ are different, but when we work with a different multiplication operation the meaning of fractions can be different.

Answer 2

As others already mentioned, the problem is that $i$ is not an integer, so the typical modular arithmetic doesn't really make sense here. However, there's a way to get something similar, by working with the gaussian integers $\mathbb Z[i]$, which is the set of complex numbers with integer comoponents together with the usual addition and multiplication. Suppose we take the principal ideal $(10)\subseteq\mathbb Z[i]$, then we look at the image of the elements of $\mathbb Z[i]$ under the natural ring homomorphism $\phi:\mathbb Z[i]\to\mathbb Z[i]/(10)$ (here, we are "looking at the gaussian integers mod $10$"). We do indeed have $$ i^2=-1=9=3^2=7^2\pmod{10}. $$ But the problem with concluding that $i=3$ or $7$ is that in general, $x^2=y^2$ does not imply $x=y$. It does not even imply $x=\pm y$. That's only true in the field of real numbers. To take a much easier example, note that $$ 1^2=3^2=5^2=7^2=1\pmod{8},$$ but obviously, $1,3,5,7$ are not equal.