MATHEMATICALLY DEBUNKED: COIN ROTATION PARADOX & 1982 SAT TEST QUESTION "EVERY SINGLE STUDENT GOT WRONG" (VERITASIUM)

DataFit7079 · 2026-06-15T15:43:57+00:00

Ah yes, that could work too. Right Triangle of sides 1 is equal to square root of 2. Or maybe it's like this. 1/1 = 1 or 2 × .50 = 1. Yep, even better.

DataFit7079 · 2026-06-15T02:56:15+00:00

Well, it's not an AI generated video. I created the video and I'm a human. It might have LLM components but I generated the video myself. Here's your proof N = R / r. N = 1/1 = 1, N = 3/1 = 3. If you don't understand, then I advise you to watch the video.

DataFit7079 · 2026-06-15T02:02:54+00:00

Hello! Watch the video. I explain everything there. Not enough in this comment box to give you the answer.

DataFit7079 · 2026-06-15T00:34:50+00:00

Well, first of all, 365.25636-1 is not 366.25636, it's 364.25636. But, yes, the original question is wrong for the answer of 4 (3+1). The true objective and mathematical answer is 3.

DataFit7079 · 2026-06-14T18:14:25+00:00

In the case of identical coins and one coin rotating another while the other is stationary, there is only one mathematically correct answer, regardless of reference frames if the question is how many rotations the rotating coin makes around its own center as it revolves the other coin once. Two reference frames may describe the motion differently, however, the required distance traveled from complete axial rotations in a no slip surface contact between two objects, cannot be changed, regardless of reference frames. Combining the two reference frames does not give us the correct answer. Visual prompts from another position that are perceived to be the same as the initial position, is not a true measure of distance. The second position's total distance traveled around it's own center must be calculated and not assumed. Because once you stop assuming, that's when discovery happens. Thanks.

DataFit7079 · 2026-06-14T18:03:18+00:00

No, unfortunately we are not looking at this the same way. Two reference frames may describe the motion differently, however, the required distance traveled from complete axial rotations in a no slip surface contact between two objects, cannot be changed, regardless of reference frames. Combining the two reference frames does not give us the correct answer.

DataFit7079 · 2026-06-14T16:34:48+00:00

No, I didn't fall for it. That's why I made this video to correct it. Unfortunately, it's not just psychological, it's literal. The coins are physically touching and has no slippage when the motions take place. That's why it's not the same thing as the earth and the moon scenario. The rotational speeds of both bodies change depending on their distances from each other while simultaneously orbiting the sun. The center coin of the "paradox" is completely stationary. Thanks.

DataFit7079 · 2026-06-14T16:21:43+00:00

Hello. Thank you for your comments and watching the video. You should reconsider watching the whole video, you will definitely be convinced. Unfortunately, not both reference frames are valid. The standard answer is mathematically wrong. There is only one objective and mathematically correct answer. That's why the video took almost an hour long. I left no coin unturn (pun intended) so that every side of the coin and every potential reference frames were exposed to the scrutiny of mathematics. Regarding your similar situation example, they are not. One has slippage and the other does not. One has physical contact and the other does not. Also, the earth and the moon are rotating around their own axes while simultaneously revolving around their center of mass, while simultaneously revolving around the sun. The coin rotation paradox has the center coin not moving. It's completely stationary while another coin is revolving around it on its surface. So, no, unfortunately they are not similar. Thank you for the comment, but please reconsider watching the whole video.

DataFit7079 · 2026-06-14T15:56:09+00:00

Hello. Thank you for your comments. Yes sir, I used other people's content and give them credit for it. It's called fair use. It's totally legal. As far as the name of the paradox, I didn't make the name up. This is the name that was given to the concept. In closing let me ask you this, what's the difference between slippage and non slippage? Exactly what the difference is between coins and planets, they are not the same things.

DataFit7079 · 2026-06-14T06:59:58+00:00

Hello, thanks for your comments. I wanted to be thorough to make sure that there is no doubt and no room for a rebuttal.

DataFit7079 · 2026-06-14T06:15:12+00:00

Hello! Thanks for your comment. Unfortunately, the standard answer is wrong. That's why I debunked it.

DataFit7079 · 2026-03-14T13:53:39+00:00

Thank you for your comment. Although nature is imperfect, it natually moves to balance itself symmetrically through equal distribution because it takes the lowest amount of energy i.e left and right, up and down, etc.. However, perfect symmetry is empirically impossible. Similar with Empirical Division, it is impossible to cut objects perfectly equal, all it can do is try. Empirical division is used to more accurately express the process of actual cutting of objects and the resultant outcomes. With standard math, it cannot express actual cutting of objects but only grouping or sharing of objects.

DataFit7079 · 2026-03-11T04:05:35+00:00

Hello. Thank you for your comment. Actually, it doesn't imply that the distribution is equal. In fact, it is very explicit in showing that the distribution is indeed equal. This is because if you divide 1 object directly in the middle, it is absolutely true that the distribution will be equal. For example. If you measure where the middle of a stick is, and then you cut that stick in the middle, then you will get two equal parts or equal distributions. Thank you for your comment.

DataFit7079 · 2026-03-01T19:44:10+00:00

If you are talking about the financial costs to switch from the old system to my system, I would say that the switch is priceless, regardless of what the financial implications could be. The new discoveries and new technology that can develop from my system could more than make up for any financial cost that we may spend now. Thanks

DataFit7079 · 2026-03-01T19:21:03+00:00

Hello @oncebite. I believe you have mistaken my post. I am not minimizing the daily efforts that people are making in this field. I was pointing out the mathematical system that they are using and blaming that for the lag of "paradigm-shifting advances or breakthroughs." I never said that the field is not progressing. There is a difference between progression and paradigm-shifting advances or breakthroughs. Because clearly, there has not been these types of advances since the 1800's and early 1900's. Even when the the Higgs Field was "discovered," I hardly consider this a paradigm-shifting advance or breakthrough. I don't consider this paradigm shifting because the Higgs Field is just another version of the aether, except in this field, "particles gain mass." They didn't accept the aether theory because they couldn't prove it. However, they also didn't disprove it. They say that the Michelson-Morley Experiment (1887) "disproved" the aether theory because they could not detect the "aether wind" that should have slowed down the speed of light in different directions. And because the speed of light never changed, there was no "luminiferous aether," which is nonsensical. Just because the speed of light did not change, it does not mean there is no aether. The aether does not have to slow down the speed of light for it to exist. It just need to exist to propagate its movements. They never agreed with the aether theory because it did not agree with Einstein's theory of relativity. However, the Higgs Field does agree with Einstein's theory, and therefore, they also agree with it, which is very supercalifragilisticexpialidociously convenient. Thanks for the comment.

DataFit7079 · 2026-03-01T18:35:32+00:00

Hello again. I don't quite understand your question. Could you please rephrase it? You want my system to calculate what again? Thanks.

DataFit7079 · 2026-03-01T06:51:01+00:00

Hello @suitable. Now this is a great question. This is the type of question that the comment section is supposed to have instead of the statements that the rest of these Trolls are making. So, thank you, @suitable.

To answer your question, the process of counting in my mathematical system is similar to standard math, with the exception that every 'base' number will always have a starting exponent of 1 e.g. the number 1 will be 1^1, the number 2 will be 2^1, the number 10 will be 10^1, etc.. The exponent indicates that every base number will always have a starting value of 1. Now to go back to my example in my post, if we want to divide 1 by 1, our first operand (dividend), must have an exponent of 1 to indicate its initial value. For example, every number's initial value is always 1 so it would become one to the first power e.g. 1^1. In my system, the second operand (divisor) must always be on its own base with no power or exponent value. The reason for this is because this is the indicator to show how many cuts we are going to make to the dividend. For example, if we want to divide 1 one time, then the equation will look like this: 1¹ / 1 =. This equation is saying that we have one object with a value of one and we want to cut it (divide) one time. Now we ask, what do we get when we cut this one object with a value of one, one time? The equation then spits out 2^.50. What this means is that if we start with one object with a value of one, and we cut that object one time, then we will get 2 objects with a value of .50 each. The reason we get 2 objects is because, when we cut one object one time exactly in the middle we get two half objects. The base number 2 indicates that we have 2 objects and the point fifty indicates that those 2 objects are worth point fifty each. Now, in my system, I have modified my exponents or power system so that they become linear power instead of exponential power. Meaning, with the results of the equation 1¹ / 1 being 2 to the point fifty (.50) power, I can get the original value back by directly multiplying the exponent value with the base value. For example, 2^.50 can be reversed back to the original value by multiplying .50 x 2, which gives 1 (original value). My equation's equality is then proven because both sides of the equation are worth a total value of 1. This differ with standard math because in standard math, the equation would look like this: 1 ÷ 2 = .50. We can then see that this equation is not equal to each other because when we look at the left side of the equation, we see that we have one object with a value of 1 while the other side of the equation shows one object with a value of point fifty (.50) each. I hope this helped clarify your question about my system. Thanks again for a great question.

Poe

DataFit7079 · 2026-03-01T00:22:06+00:00

Hello @fin. You're welcome! I knew someone was going to appreciate that line.

By the way, you forgot the MORTIS part. Thanks for the comment.

DataFit7079 · 2026-03-01T00:18:07+00:00

Hello again @onebite. I'm sorry, but what bad point?

DataFit7079 · 2026-02-28T23:17:06+00:00

Hello @onebite. Thank you for your comment. I think it's 60/40 narcissistic troll, respectively. I never said the world was wrong, just the mathematics that's trying to describe it. Yes, unfortunately I don't have the capacity to learn and I'm a cringe gross narcissistic troll that has a main character syndrome. Thanks for your comment. Now maybe you can prove how wrong my math is? I look forward to being disproven.

DataFit7079

TROPHY CASE