I animate and provide some explanation for classic and newer "proofs without words," which are typically diagrams without any words that indicate how a theorem could be proved.
I often pay homage to Proofs Without Words by including some animations without narration (only dramatic music :) ). I have also noticed that while there are many wonderful visual proofs available on RU-vid, most of the creators do not credit the original visual proofs. I will do my best to include links and citations for each visual proof so that you can track down the original static images, which have a beauty and a wonder all their own.
I create my animations using the amazing software Manimgl from 3Blue1Brown.
If you want to buy me a coffee, you can do that here: www.buymeacoffee.com/VisualProofs
The equality holds when x = 1. 1 is its own reciprocal, so x + 1/x = x + x when x = 1, and 1 + 1 = 2. Incidentally, the only real numbers that fit the criteria of x = 1/x are 1 and −1. 0 doesn’t simply because 1/0 is undefined, any real numbers such that 0 < |x| < 1 are such that |x| < 1/|x| (so x ≠ 1/x for such numbers), and any real numbers such that |x| > 1 are such that |x| > 1/|x| (so x ≠ 1/x for such numbers, either). (This is also true of all complex numbers, as 1/z = (a−bi)/(a²+b²) for nonzero complex z, so z = 1/z if and only if the imaginary part of z is 0 (so that bi = −bi/(a²+b²), or b = −b/y for some positive real number y, meaning that b and −b must have the same sign (positive, negative, or zero) as each other, and only 0 fits that criteria), meaning z = 1/z can only be true if z is a real number.) Also, looking at further generalizations for the formula, we exclude 0 because 1/0 is undefined, but if we use projected real numbers (which also includes an unsigned ∞ that is both less than and greater than all real numbers), 1/0 = ∞, so 0 + 1/0 = 0 + ∞ = ∞ > 2, so the inequality also holds when x = 0 under projected real numbers. (It also holds for unsigned ∞, btw, since ∞ + 1/∞ = ∞ + 0 = ∞ > 2.) For negative numbers, the inequality as such doesn’t hold. We can, however, get a different inequality: for negative x, x+1/x ≤ −2. This inequality also means that, for negative x, |x+1/x| ≥ 2, and, since |x| = x for all x ≥ 0 (including unsigned ∞), _that_ inequality also holds for positive x and, under projected real numbers, 0 and unsigned ∞. In other words, for any projected real number x, |x+1/x| ≥ 2. (We can also easily see that, under _extended_ real numbers (which includes +∞ (which is greater than all real numbers) and −∞ (which is less than all real numbers), both the original inequality and the new inequality still hold for +∞ (since +∞ + 1/+∞ = +∞ + 0 = +∞ ≥ 2, and |+∞| = +∞), and the new inequality also holds for −∞ (since −∞ + 1/−∞ = −∞ + 0 = −∞, and |−∞| = +∞ ≥ 2). However, the inequalities don’t hold for x = 0 under extended reals since 1/0 is still undefined under extended reals. Therefore, for all nonzero extended real x, |x+1/x| ≥ 2.) We can’t really extend the original inequality to complex numbers since nonreal complex numbers are not ordered (so <, >, ≤, and ≥ are undefined for nonreal complex numbers). We _can_ try to extend the new inequality, though, as |z| is a nonnegative real number for all complex z. However, it fails for at least some values of z. This can be shown by looking at the case z = ±i. See, 1/i = −i, and 1/(−i) = i, so ±i + 1/(±i) = ±i ∓ i = 0, and |0| = 0 ≱ 2. (Incidentally, these are the only complex numbers such that z + 1/z = 0.) I have also determined that the inequality should hold true for all complex z = a+bi such that either a=0 and b−1/b ≥ 2 *or* b=0, but it should also hold true for some other values of b if a ≠ 0.
Have you used other Pythagoreans angles, like [5, 12, 13]; [8,15,17] or [7,24,25]? They are very useful when you have a much larger side. There are others which ratio is larger for more squerish rectangles, like [48,55,73] you can use 1/8” , inches, mm, ft, m any measurement system to fit.
Let e be x and pi be y. And since e≈2.7 and pi≈3.1, using x,y, we ultimately get x<y. Let's substitute some number where x<y into [x^y ? y^x]. I don't know why it's possible intuitively, but I trust someone will do it for me. Let x=3, y=4. Then it is 3^4 or 4^3. 3^4 is 81, and 4^3 is 64. So, x^y is bigger, and ultimately pi^e is bigger than e^pi.
This is why i love geometry. I Never thought about calculating the area between 3 tagent circles, but the exact moment i saw this I knew in every step what the way to the answer would be. This is simply beautiful
