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How accurate is pythagorean theorem?

  1. Jun 6, 2015 #1
    I'm just wondering: How accurate is the pythagaroen theorem? To what exact decimal point is the calculated length of the
    hypotenus in a triangle 100% corect when using pythagarean theorem? It must not be 100% accurate to an infinite descimal point since it's called a "Theorem"
  2. jcsd
  3. Jun 6, 2015 #2


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    That isn't what theorems in Mathematics mean. Pythagoras' theorem is perfectly accurate. In fact, the equality sign = is perfect. If Pythagoras' theorem was "nearly" perfect but not quite, then it wouldn't be using an equality, it would instead have the approximate sign ##a^2+b^2\approx c^2##

    The only thing that is imperfect are our decimal approximations to irrational numbers, hence why for a right triangle with short legs of length 1, the length of the hypotenuse is

    $$\approx 1.41421356$$
    $$\approx 1.414$$

    Note where the equality is used, and where the approximation is used.
  4. Jun 6, 2015 #3
    When the space is Euclidean (is not curved), the lines are perfectly straight and the angle is a perfect right angle, the theorem is exact. There is no 'rounding' in deriving the theorem, only the assumptions have to hold (Euclidean space, straight lines, right angle, maybe something more I'm missing).

    In real life, you have all kinds of issues, space is curved, the angle is a little off, you can't measure with infinite accuracy (uncertainty principle etc).

    But for usual applications, these effects can be ignored because they have a smaller effect than your measurement device can measure, i.e. you want to calculate the diagonal of your table after measuring the sides.
  5. Jun 6, 2015 #4
    Thank you very much for the explanation! So it's the square root of numbers like 2,3,5 that's that gives us an infinite number of descimal points which is not defined,but roundede up on a calculator?
  6. Jun 6, 2015 #5


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    Yes. ##\sqrt{n}## is precisely the number that when squared, gives us n. It just turns out that to express most of these numbers in decimal form, you need an infinite number of non-repeating digits.

    Not defined is not the best word to be using here, but you seem to have the gist of it.

    Also, calculators are a bit more interesting in their approximations than the simple rule of "rounding up" which isn't necessarily correct either. As reference, I'll post ##\pi## here to 15 decimal places and see what arises from it.

    $$\pi = 3.141592653589793...$$

    Now, my standard scientific calculator which can display 10 digits at a time shows ##\pi## as

    $$\pi \approx 3.141592654$$

    Which if we compare to earlier, we can see is rounding the value up. But we can delve further into this. If I subtract 3 from the value to get 0.141592654 and then multiply by 10, I'm essentially shifting the digits up and uncovering another digit that was previously hidden. Try it on your calculator, and keep repeating it to see how much the calculator actually holds and what happens at the end of its storage compared to the actual value of ##\pi##.
  7. Jun 6, 2015 #6
    My ti89 displayet 3.14159 however if i copy and paste this number it shows the total amount of decimals which is 3.1415926535898 if I subtract 3.141592 and multiply with 10^6 then it does not recover more decimal points, so I guess the copy paste displays the maximum amount of decimals.
  8. Jun 6, 2015 #7


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    Alright so assuming your calculator actually calculates to 13 decimal places (dp) for all values and isn't just using a stored value of pi and a other hard-coded constants, I searched for an irrational that has 0-4 on the 14th dp such that a 13dp display would round down instead. Ironically, I actually had to search through all of the square root irrationals until ##\sqrt{14}## finally had the property I was looking for. Try that and see what your calculator displays for this one. For reference:

    $$\sqrt{14} = 3.74165738677394...$$

    The 14th dp is the 4 at the end.
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