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

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?

• What does even $a=i$ mean? Jul 14, 2020 at 16:17
• The base $$a=-1^{1/2}$$ Jul 14, 2020 at 16:22
• What I wanted to point out is how do you define a triangle wirh base $i$? Jul 14, 2020 at 16:23
• Fun fact: Consider a "right-corner tetrahedron" $OPQR$ with edges $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$ meeting at right corner $O$. (Eg, put $O$ at the origin, and $P$, $Q$, $R$ at $(p,0,0)$, $(0,q,0)$, $(0,0,r)$.) Then an area-based Pythagorean relation holds: $$|\triangle OQR|^2+|\triangle POR|^2+|\triangle PQO|^2=|\triangle PQR|^2$$ Re-writing the left in terms of $a:=|QR|$, $b:=|RP|$, $c:=|PQ|$ gives Heron's formula for $|\triangle PQR|$ in terms of its sides. Nifty! But this only works for non-obtuse $\triangle PQR$ ... unless we allow one of $p$, $q$, $r$ to be imaginary. (continued)
– Blue
Jul 14, 2020 at 17:10
• (continuing) See, for instance, the "Heron's Formula for Tetrahedra" entry of Kevin Brown's MathPages. (The entire entry is interesting, but the above is discussed in the first part.)... So, "imaginary lengths" actually have their uses.
– Blue
Jul 14, 2020 at 17:13

$$i$$ 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.

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.

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 $$\sqrt 2$$