Youtube hearted comments of MC116 (@angelmendez-rivera351).

  1. The reason we use degrees, rather than hours, is because 360 hours are meant to denote not a full rotation of the Earth, but rather, a half-period of the lunar cycle. 360 hours are 15 intervals of 24 hours, which is half a lunar month. As such, one hour is 15°, which is trigonometrically a very important and fundamental constant. With all of that being said, I have learned that thinking of degrees as a unit of measurement of angles is not the conceptually appropriate way to think of them, if only because, a ratio of two quantities with the same dimensionality is dimensionless, and thus, is numerically independent of the units it is measured in: there should not exist such a thing as different units for a dimensionless quantity, mathematically speaking. So what is actually going on instead? Well, what is going on is that degrees are a scale factor for the trigonometric functions. Writing sin(1°) is the same thing as writing sin(π/180). In other words, sin(x°) = sin(π/180·x). So when you change from degrees to radians, you are not changing units of measurements. What you are doing instead is rescaling the trigonometric functions.
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  2. 0:03 - 0:12 You would need to specify what ring you are working in. In the trivial ring, 0 = 1 is true. For the sake of this video, then, I suppose we should assume we are working in a non-trivial ring, so that the proof could be conceivably incorrect. 0:32 - 0:43 This is not actually necessary, nor does it add more correctness to the proof. The restraint that x is nonzero ends up being irrelevant. 2:50 - 3:12 It should be noted that a^2 – b^2 = (a – b)·(a + b) is only true if a and b commute, which is to say, if a·b = b·a. Since x = y, x·y = y·x is indeed true, but this should be stated. 4:10 - 4:11 At this stage in the proof, we most definitely have a 0 = 0 situation. Specifically, since x = y, it follows that x – y = 0, and so (x – y)·(x + y) = 0·(x + y), while y·(x – y) = y·0 = 0·y, and so we have 0·(x + y) = 0·y. 4:12 - 4:41 This is where the proof went wrong. As I noted in my previous paragraph, the equation (x – y)·(x + y) = (x – y)·y is equivalent to 0·(x + y) = 0·y, since x = y implies x – y = 0. What the video is thus effectively doing is declaring that 0·(x + y) = 0·y implies x + y = y, which is not true. This is because, even when a is not equal to b, 0·a = 0·b = 0, and this is true in all rings. This is equivalent to just saying that 0·2 = 0·1 implies 2 = 1, which is obviously not the case. 5:13 - 5:29 This means x is idempotent with respect to addition, and since this is a ring, it implies x = 0. This contradicts the fact that x is arbitrary, though. 5:39 - 5:45 This extra restriction is not necessary, all you need is for x to be arbitrary 6:18 - 6:21 Even if it were true, it would not show the universe will end. 9:04 - 9:08 It is not that you are "not allowed to divide by 0." Rather, it is that, since we are working in a ring, as is required for distributivity to apply, and addition, subtraction, and multiplication to be well-defined, it must be the case that 0·x = y·0 = 0 for all x, y, and so 0 is not cancellable, which means that even if a is not equal to b, 0·a = 0·b.
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  3. sqrt(9·x^2 + 3·x) – 3·x = sqrt(y^2 + y) – y, where y = 3·x. Letting x —> ∞ is indeed equivalent to letting y —> ∞, so we consider lim sqrt(y^2 + y) – y (y —> ∞) = lim sqrt(y)·(sqrt(y + 1) – sqrt(y)) (y —> ∞) = lim sqrt(y)/(sqrt(y + 1) + sqrt(y)) (y —> ∞) = lim 1/(sqrt(1 + 1/y) + 1) (y —> ∞) = lim 1/(sqrt(1 + y) + 1) (y > 0, y —> 0) = 1/2.
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  4. Technically, lim sec(x) (x < –π/2, x —> –π/2) does not exist. However, given x < –π/2, x —> –π/2 implies sec(x) —> –∞, insofar as this is just an abbreviation to say that for all real numbers B < 0, there exists some real number δ > 0, such that for all real numbers –δ < x + π/2 < 0, sec(x) > B.
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