Is i<0?

True or False?

$i=\sqrt{-1}<0$

Why some people say it's true: $-1$ is less than $0.$

Why some people say it's false: Square roots are positive.

The statement is neither true nor false, but rather $\color{#D61F06}{\textbf{not well-defined}}$.

The answer is ambiguous but can be called false as $i$ cannot be defined as less than $0,$ but it can be called true in that it cannot be defined as more than $0$ or equal to $0.$ However, in itself, $i$ in the imaginary world is equal to $1$ in the real world, as all imaginary numbers are based off of $i$ like real numbers based off of $1.$

Short answer: Complex numbers are not totally ordered and hence the question is not well-defined. To be explicit, we can introduce various orderings on the complex numbers. In fact, in the three most common expansions of the notion of "inequality," the statement $i < 0$ is false.

Long answer: See below.

$i$ is a value on the complex plane. It can be thought of as either a counterclockwise or a clockwise rotation of 90 degrees $\big($or $\frac{\pi}2$ radians$\big)$ from the real number line.

Note that performing the same rotation twice will get $-1.$ While we can thus think of $i$ defined as the value such that $i^2 = - 1,$ we have to be careful: while the counterclockwise and clockwise directions are not distinguishable at first, after one direction is picked, the direction cannot be changed. That is, we can't say $i = \sqrt{-1}$ at the same time as $i = -\sqrt{-1} ;$ this allows proofs like $1 = -1$ (as shown below; the third step is the one that allows both rotations at the same time):

$$\begin{aligned} &-1 \\ =&i \times i \\ =&-\sqrt{-1} \times \sqrt{-1} \\ =&-i \times i \\ =&1. \end{aligned}$$

$\big($This is incidentally why Brilliant generally picks the direction $i = \sqrt{-1}$ throughout all explorations, as a general rule.$\big)$

In any case, this means numerically $i$ is defined by two values. We can either use one indicating horizontal and one indicating vertical distance $($that is, the coefficients in the representation $0 + 1i )$ or by distance and rotation in a format like $\big(1, \frac{\pi}{2}\big).$ The latter form has the drawback that representations are not unique, since the angle can be rotated as many full circles as desired, or distance can be written as negative $($with the angle rotated a half circle $\pm$ as many full circles as desired$).$

While one can do things like writing $i$ as part of Euler's formula and translating it to being just the rotation, this is generally irrelevant as it loses information and no longer represents the point $i$ on the complex plane.

Inequality is quite specifically defined in mathematics because it relates to fundamental axioms. It can even be baked into the fundamental set of axioms, if desired. (For instance, in Tarski's Introduction to Logic from 1941.)

The important thing to note is that inequality requires a well-ordered property; that every nonempty subset has a least element. Also, adding new elements to a subset will not change the ordering. In practical terms, if you choose two terms $x$ and $y,$ where $x < y ,$ then the addition of a new element $z$ means that it will still always be the case $x < y .$

If we have points represented by two values, the well-ordering property no longer applies unambiguously. We have to choose some meaning to $<$ to restore well-ordering.

The common possible choices are as follows:

Absolute-value ordering: Taking the distance to the origin. In this case, since $i$ is a distance of 1 from the origin, and $0$ has a distance of 0, $0 < i .$ Due to potential confusion the actual absolute value symbols are usually used, which means inequality does not need any extra definition: $|0| < |i| .$

Lexiographical ordering: First ordered by real value; if the real values are the same, then the imaginary part is used. With this ordering, $3 - 2i < 3 + 3i ,$ and $0 + 0i < 0 + 1i .$ $($In other words, $0 < i. )$

Polar lexiographical ordering: First ordered by absolute value; if the absolute values are the same, then by polar angle using the range $(-\pi, \pi].$ With this ordering, applying the absolute value, $0 < i .$

Note that while the above three are the most commonly used and $0 < i$ in all the cases, this doesn't mean we can say it's unilaterally true; the question "is $i < 0 ?$" is meaningless without some extra definition.