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.
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.
"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
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.
This is from my probability class in an actuarial sciences mba. Xj is a random variable of the value of a claim, N is another random variable of the number of claims in a year, S is another random variable of the total sum of claims in a year.
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