
#1
May1413, 06:42 AM

P: 42

I wonder how do I find the root of 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, and so on?
And why the circle and the curve looks so smooth in the computer graphic software such as AutoCAD, Adobe Illustrator, etc., if the root is can not be found? It should be looks rough. Thank you 



#2
May1413, 11:12 AM

P: 428

What makes you think the root cannot be found? Are you worried because can't be written as a decimal or fraction to infinite precision? It can be approximated as close as you would like. 



#3
May1413, 11:38 AM

P: 42

If it be approximated, then the circle or curve in the computer graphic software should not looks smooth.
How do you approximate the root of 2,3,5,6,7,8,10, and so on? By the numerical method? 



#4
May1413, 12:04 PM

P: 771

What Is The Root of 5? Irrational 



#5
May1413, 01:09 PM

Mentor
P: 7,292

You have the equation:
x^{2}c=0 Where c is the number of your choice. There are lots of numerical methods which can be used on this. The most basic is the bisection method. Just for example let c=6 let x = 1 plug that into the above expression to get 5, now let x = 3, plug that into the equation to get 3, since the first try was negative and the second positive we must have a zero between 1 and 3, cut the interval in half and try x=1.5. What is the sign? It will be either + or , so you have narrowed the interval where the root lies. Continue this process until you achieve the accuracy you want. 



#6
May1413, 01:32 PM

Admin
P: 22,712

While the decimal expansion of the [itex]\sqrt 2[/itex] is infinite, we have a perfectly valid and exact representation of the number  [itex]\sqrt 2[/itex].
And lines on the screen look nice because of antialiasing, not for any other reason. Yes, Bresenham's line algorithm or midpoint circle algorithm produce jagged lines, but they are rarely used these days. 



#7
May1413, 02:03 PM

P: 235

In my experience a 32 sided polygon works well for small circles. As you draw bigger and bigger circles you just add more and more sides. 


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