Would a right triangle with bases $a=i$ and $b=1$ have hypotenuse $c=0$?

Question

Suppose we have a right angle triangle with $a$ and $b$ as bases and $c$ as the hypotenuse, letting $$a=i$$$$b=1$$ Wouldn't the hypotenuse then be$$i^2+1=0$$ I am finding it hard to understand how this may be possible since the hypotenuse by definition is the longest side of a right angled triangle, how could this be, perhaps this exists for a non euclidean geometry?

Answer 1

$i$ is an imaginary number $i=\sqrt{-1}$

Such a right triangle with legs $a=i, \ b=1$ doesn't exist.

because side of right triangle can't be imaginary i.e. $a\ne i$

all the sides of an existing triangle must be positive real numbers.

Answer 2

It doesn't exist, because a triangle must be defined as having side lengths of positive real numbers (even $-1$ wouldn't count, let alone imaginary numbers).

If you are talking about the triangle formed by $0\leq x\leq 1$, and $0\leq y \leq i$ on the Argand Diagram, that has side lengths of $1$ and hypotenuse $\sqrt 2$

Answer 3

As Shiva Venkata said, a triangle with imaginary sides does not exist, normally. Within Euclidean space this makes sense. However, if we bend our notions of distance to accomodate alternative geometries, such as the Lorenzian/Minkowski geometry used in the special theory of relativity, then the question makes sense.

Suppose you're sitting a meter apart from your friend Marie. Creating a spacetime diagram from your reference frame, at an arbitrary starting point $t=0$ you're at the origin and she's at $(ct,x)=(0,1)$. At this moment you decide to take a picture with the flash on, sending a cascade of photons that hit Marie's eyes after $1/c$ seconds. Hence, the flash illuminates her face at the event $(ct,x)=(1,1)$.

So we have three events:

• A: The flash at $(0,0)$
• B: An unilluminated Maria at $(0, 1)$
• C: An illuminated future Maria at $(1,1)$

Let's compute the spacetime interval, i.e. distance squared, between them. For points $P,Q$ this is

$$\ell^2_{PQ} = -(P_t - Q_t)^2 + (P_x - Q_x)^2$$

• $\ell^2_{AB} = -(A_t - B_t)^2 + (A_x - B_x)^2 = -(0)^2+(1)^2 = 1$
• $\ell^2_{BC} = -(1)^2+(0)^2 = -1$
• $\ell^2_{CA} = -(1)^2+(1)^2 = 0$

This corresponds to the physically significant notion that A & B are spacelike-seprated, B & C timelike, and A & C lightlike. Taking the square-roots provides you with the actual distances between points, and therefore the lengths of the segments they define:

• $\ell_{AB} = 1$
• $\ell_{BC} = i$
• $\ell_{CA} = 0$

If we put all the events on a spacetime diagram, they would indeed form a triangle! In fact, they form a right-triangle with respect to this geometry, as $\bar{AB}$ and $\bar{BC}$ are actually still orthogonal in this non-Euclidean setting. With these segments serving as the legs, the segment $\bar{CA}$ is the hypotenuse, which does indeed have length zero.