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# pi is the most significant number.

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 Voting Style: Open Point System: Select Winner Started: 5/12/2014 Category: Miscellaneous Updated: 3 years ago Status: Post Voting Period Viewed: 605 times Debate No: 54549
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 Con π is not the most significant number. I get the thing with the circle, but there is also e, which is the limit of (1+1/n)^n, and the derivative of e^x is itself. τ is equal two 2π. Also, φ is equal to .5+√(1.25), and we have i which is √(-1), which reminds me, 1 is pretty significant. Anyway, my point is that π is not the most significant number ever. Whoever wants to challenge me can.Sources:https://en.wikipedia.org...(mathematical_constant)Report this Argument Pro Your opening statement is invalid for two reasons: You did not define significance, therefore your statement is up to interpretation You supplied no reasons confirming that pi was not significant, you only provided examples of other numbers To remain fair, I will only use the type of arguments you supplied.Pi is significant. All other numbers are not as significant as pi.Pi is significant because 3 is equal to 3. Also because A=Area.Sources:Not Wikipedia. So my argument is not based off the word of millions of Internet users. Report this Argument Con Significant meaning the most needed number, or the best number.Pi is significant in circles, but what about pentagons and pentagonal stars: then we need phi. And with squares, we need sqrt(2). I don't see why circles are more significant than squares and pentagons. And phi is significant also because it is commonly used in advertising (see sources), and have you ever seen pi used in the same way? I sure haven't. You may argue that the facts C=2pir, A=pir^2, SA=4/3*pir^2, and V=1/3*pir^3 are too coincidental, but this is a feature of circles and spheres, and pi just happens to be the number. Just like phi just happens to be the number with pentagonal stars. Sqrt(2)*n is the diagonal of a square with side length n, 2^(1/3)*n is the diagonal of a cube with side length n, sqrt(sqrt(2))*n is the diagonal of a 4-cube with side length n, and so on with the diagonal of an m-cube with side length n being 2^(1/m)*n. Now, how is pi more significant than phi and 2 (not that they are more significant, just that they are equally significant)? Also, wouldn't all whole numbers be significant, because there are infinite numbers in between all whole numbers. This means that if you randomly picked a number, including irrational numbers such as pi and e, then there would literally be a 1/infinity chance of getting a whole number.Sorry if I have not done the debate correctly, but this is my first time debating.Sources:http://www.graphicart-news.com...Report this Argument Pro Thank you for defining significance, this makes things much simpler. Since you defined significance as the most needed number then that would suggest that a significant number would be frequently used. It is well known that pi has been often used to find dimensions of circles and spheres. Since pi is so essential to finding dimensions of such shapes than it would make sense that if these shapes appeared often, pi would be used often. If pi is used often that it is needed more. If it is needed more then pi would be the most significant number. The interesting thing is that spheres and cicles are extremely common in this universe. Forming into a circular object is the natural way of things. This because of gravity and various other factors. Things such as orbits, plants, stars, Galaxys, and even black holes take on a spherical/circular shape. Now, since these shapes are so common in this universe, then it would be logical to say that pi is quite important in finding the dimensions of such objects. Of course, on our planet Earth there are shapes such as pentagons and squares but it must be noted that you'd be hard-pressed to find such shapes anywhere else besides Earth. This is due Earth's rather unique characteristic of life. So my point is that since most objects in the Universe are made up of circular/spherical shapes, that would mean pi is quite imperative to getting to know the rest of the cosmos. Without it, these vertex-ridden shapes we have now are all we'd have to compare it to.Report this Argument
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