Yes but not in terms of market share, unfortunately. Monopolies suck. The best solution would be a standard open source operating system that would run on ARM hardware with a wide range of RAM and display support, like Java apps on midrange phones in the 2000s. And a QWERTY keyboard when rotated (we need the law changed for that to be legal in US schools). And an OK, as well as a BACK or EXIT button like CASIO. On the TI, I've seen programs that use MENU, DEL, CLEAR, ALPHA, SHIFT+MENU (QUIT), ON or a menu item confirmed by SHIFT (the usual OK key in games) to exit.
There are some open source graphic calculator projects but they aren't mass-produced and none are exam-approved. Software support is lacking but TI-84 Plus emulators exist so if someone rewrites TI-OS and the ROM (or requires people to dump it like the current ones), it could run that.
Did you know the humble Nokia 3410, the first Java phone, would dither games meant for higher-resolution color screens? Of course, programs for these calculators could use semi-vector graphics like the Web, or detect screen resolution and support the common ones natively, in a single or multiple executables like PocketPC software.
it's not treating pi as a variable, though...
here's one that does work:
circumference = 2piradius,
so, pi = 2*radius/circumference... which is true... (pi is the ratio of diameter to circumference)
the meme here is just an equation that's wrong because it's wrong... pi is being treated as if it's some value that it's not in the first equation, and it's still wrong in the second equation...
Read the "1" unit side as "move left 1 unit" and the "i" side as "move up i units", and the hypotrnuse is the net distance travelled.
The imaginary line is perpendicular to the real line, so "up i unit" is equivalent to "right 1 unit". The two movements cancel out giving a net distance of zero.
Yep. A vertical line segment above A with length 𝑖 is a horizontal line segment to the left that's 1 unit long. So, the diagram needs a "not to scale" caveat like a map projection, but there's nothing actually wrong with it, and the triangle's BC side is 0 units long.
Yeah. We were making a joke about the complex plane -- you could say that measuring the hypotenuse of a triangle is equivalent to measuring the distance between points |AB| and |AC|𝑖 on the complex plane. That definition actually makes quite a bit of sense, and I think by sheer coincidence it's possible to misunderstand how to do it and wind up with a way of looking at it where the hypotenuse of a right triangle with sides 1 and 𝑖 would work out to exactly 0. Which brings it back into concordance with OP's (also wrong) Pythagorean presentation of it.
It obviously doesn't really work that way, but it's hard to see necessarily anything wrong with it, which makes it a fun math thing.
Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0
Yeah. I cheated. You have to either deliberately misunderstand how to measure vectors or else drop a minus sign for it to work my way.
(Or, from my previous example, you could just frame it as you're getting the hypotenuse by measuring between |AB| and -|AC|𝑖 instead of the way I framed it -- but that makes it more obvious that you're fishing for a particular answer.)
I get that this is just a meme, but for those who are curious about an actual mathematical argument, it is because Pythagoras's theorem only works in Euclidean geometries (see proof below). In Euclidean geometry, distances must be real numbers of at least 0.
There exists at least one ∆ABC in a 2-D non-Euclidean plane G where (AB)² + (AC)² ≠ (BC)² and m∠A = π/2
Proof: Let G be a plane of constant positive curvature, i.e. analogous to the exterior surface of a sphere. Let A be any point in G and A' the point of the furthest possible distance from A. A' exists because the area of G is finite. Construct any line (i.e. form a circle on the surface of the "sphere") connecting A and A'. Let this line be AA'. Then, construct another line connecting A and A' perpendicular to the first line at point A. Let this line be (AA')' Mark the midpoints between A and A' on this (AA')' as B and B'. Finally, construct a line connecting B and B' that bisects both AA' and (AA')'. Let this line be BB'. Mark the intersection points between BB' and AA' as C and C'. Now consider the triangle formed at ∆ABC. The measure of ∠A in this triangle is a right angle. The length of all legs of this triangle are, by construction, half the distance between A and A', i.e. half the maximum distance between two points on G. Thus, AB = AC = BC. Let us define the measure of AB to be 1. Thus, 1² + 1² = 2 ≠ 1². Q.E.D.
"um actually" I guess to properly apply the pythagoras theorem here, you'd need to consider the magnitude of the lengths of each of these vectors in complex space, both of which are 1 (for the magnitude of a complex number you ironically can use pythag, with the real and imaginary coefficients of each complex number.
So for 1 you get mag(1+0i)=root(1^2 + 0^2)
and for i you get mag(0+1i)=root(0^2 + 1^2)
Then using pythag on the magnitudes, you get hypotenuse = root(1^2 + 1^2) = root 2, as expected
I don't do CAD, but can it be done with "ruler and compass"?
Yes, as the golden ratio is (1+sqrt(5))/2 and sqrt(5) can be found in a right angled triangle with side length 1,2 and sqrt(5).
With some effort the root of any natural number can be constructed with ruler and compass by creating a spiral of right angled triangles.
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