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Is it possible to find √n using only the

*times table*and the 4 operations?You are using an out of date browser. It may not display this or other websites correctly.

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Is it possible to find √n using only the* times table* and the 4 operations?

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Office_Shredder

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It is possible:Using only finitely many such operations, you cannot....

if n = x² ,

[likewise, you can find in

if n = x² , digits > 6, : you can still guess three digits and get x with a few operations,

with a pocket calculator you can get any root in a few seconds. Can you believe that?

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- #4

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[tex]

x_{n + 1} = \frac{1}{2} \, \left( x_n+ \frac{a}{x_n} \right)

[/tex]

However, as this is a limiting process, you can only get an exact result in the infinite limit [itex]n \rightarrow \infty[/itex]. The convergence is fast though, and you get a good approximate result in just several iterations. Actually, you will exhaust the number of digits on your calculator pretty fast.

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That is the Babylonian method, according to wiki. Is there a general algorithm that does not use powers of the same order (k) of n = x^k ?

Cam you find[itex]\sqrt{403225}[/itex] using only logics, knowing that n= x²?

Cam you find[itex]\sqrt{403225}[/itex] using only logics, knowing that n= x²?

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(1+x)^n = 1 + n x + n(n-1) x^2/2! + n(n-1)(n-2) x^3/3! + ...

Substitute n with 1/2 and there you got the square root. Keep adding terms till you get the precision that satisfied you.

That's actually how computers calculate square-roots.

http://en.wikipedia.org/wiki/Taylor_series

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Hurkyl

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It depends on what you mean by "find".Is it possible to find √n using only thetimes tableand the 4 operations?

For example, for many purposes, an algorithm to produce arbitrarily good approximations (e.g. the method in this post) counts as "finding" the number.

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Hurkyl

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Reference? I find this claim rather unlikely....That's actually how computers calculate square-roots.

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Office_Shredder

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That's actually how computers calculate square-roots.

Are you sure about that? Taylor series are typically not a very fast way of calculating the function they are approximating. And your Taylor series only converges for |x|<1 which is pretty limited (the wikipedia article on 'calculating square roots' mentions how to adapt for this but also says the method is fairly slow compared to the others mentioned)

logics, if you just want the square root of something you know is an integer you can just use a binary search to find the square root without doing anything but multiplication a lot of times. For example to find the square root of 403225:

500

750

625

687

We need a number between 687 and 625

650

We need a number between 625 and 650:

637

We need a number between 625 and 635

etc, we find the square root is 635.

This technique can also be used to find the square root of any number to whatever precision you want, for example if you want the square root of 403226, we know that it is larger than 635 and smaller than 636. So we try 635.5, and if the square is too large, 635.25, and if the square is too large 635.125, and eventually when the number squared is smaller than 403226 we have a bound for the square root (and at that point we can keep dividing the interval in half in order to get more precision)

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Hurkyl

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Is it cheating to use the operation "<"?

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Is it cheating to use the operation "<"?

What operation is that? What is the output of such an operation?

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Office_Shredder

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your example is the midpoint method for solving the equation [itex]x^2 - a = 0[/itex]. The Babylonian method is nothing but the Newton-Raphson method for solving the same equation.

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When I say can you find x =[itex]\sqrt{635²}[/itex] using only logics I mean using only your mind, knowing by hearlogics, if you just want the square root of something you know is an integer you can just use a binary search.....

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Mentallic

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When I say can you find x =[itex]\sqrt{635²}[/itex] using only logics I mean using only your mind, knowing by heart the times table, using no calculator, no pencil and paper.

I guess...

[itex]635^2=403,225[/itex] just for reference.

So we're trying to find the answer of [itex]\sqrt{403,225}[/itex]. We know that since [itex]100^2<403,225<1000^2[/itex] then the answer will be a 3-digit number that we will denote as ABC.

To find the hundreds value, A, just realize that the first 2 digits of 403,225 are 40, and that is between 6

For C, the last 2 digits are 25, so I'd guess that it would be 5.

To find B - arguably the trickiest of all - I guess I'd flex my brain muscles by not pulling out a pen and paper and expanding (600+10B+5)

So [tex](600+10B+5)^2=(600+(10B+5))^2=600^2+1,200(10B+5)+(10B+5)^2[/tex]

Clearly the influence would be that middle term [itex]1,200(10B+5)[/itex]

So now let's do some simplifications. [itex]1,200(10B+5)>12,000B[/itex] and [itex]403,225-360,000 \approx 40,000[/itex] and then [tex]3<\frac{40,000}{12,000}<4[/tex] and so I would make the logical guess that B=3.

Thus, ABC = 635

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Well done, Mentallic, if you can manage all that in your mind!So [tex](600+10B+5)^2=(600+(10B+5))^2=600^2+1,200(10B+5)+(10B+5)^2[/tex]

Clearly the influence would be that middle term [itex]1,200(10B+5)[/itex]......Thus, ABC = 635

But you must find something simpler if you want to solve the challenge in the twin-thread about the cube root... (try to find x = [itex]\sqrt[3]{723³}[/itex]).

If you do not wish to move on to that thread, here is a slightly tougher [but still easy] challenge: find x = [itex]\sqrt{277729}[/itex]

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- #17

Mentallic

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Well done, Mentallic, if you can manage all that in your mind!

But you must find something simpler if you want to solve the challenge in the twin-thread about the cube root... (try to find x = [itex]\sqrt[3]{723³}[/itex]).

If you do not wish to move on to that thread, here is a slightly tougher [but still easy] challenge: find x = [itex]\sqrt{277729}[/itex]

With a bit of practice, I could answer the square root problems more efficiently, as long as they're either 4 or 6 digits, which is pretty narrowed down and a useless skill to have if you ask me.

And obviously I'd have little luck extending it to the cube roots.

- #18

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Do you "find" sqrt(2) by the operation sqrt?

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Practice [or luck] is not needed, only an efficient method, algorithm i.e.: logics.With a bit ofpractice, I could answer the square root problems ...a uselessskill

And obviously I'd have littleluckextending it to thecube roots.

We have, so far, 3 or 4 methods:

Taylor is probably the worst, Newton [(x * x) -a] : 2 * x requires 4 operations per round (so does O.Shredder), Babylon (x + a : x) : 2 requires only 3;

can you find an algorithm that requires only 2 operations per round?

If a = x²,x³... (x need not be an integer) you can find x easily bearing in mind a few rules, or performing only a couple of simple (2-3 digit) subtractions or divisions (perhaps with a pencil, if one wishes not to strain one's brain).

You have been very clever working out A, where http://wiki, as a rough extimate for [itex]\sqrt{125348}[/itex], instead of 3[00] gives a poor 6[00]. You had more luck than wiki? You have more practice? Surely not, they have a poor algorithm!

If you work out the right pattern, you'll discover that [itex]\sqrt[3]{x³}[/itex] is easier and that [itex]\sqrt[5]{x^5}[/itex] is even easier: you can easily find x even when many digits are missing.

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That is a good method, Dickfore, do you know the Bakhshali method? Does it converge faster than Babylon, as wiki hints?You can find the square root [tex]x_{n + 1} = \frac{1}{2} \, \left( x_n+ \frac{a}{x_n} \right)[/tex] The convergence is fast though,.

Can anyone work out how many operations are required to find:

x = 7123456789² ?

Starting with x

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