square roots and squaring without multiplication and division

[Mikl__]

• Difficult to surprise anyone that √(10₂^2N)=10₂^N but √(10₂^2N+1)=10₂^N x √(2)
  √(2)=1.4142135623730950488016887242097...₁₀=1.6A09E66...₁₆
  for example integer square root 2⁶³ is √{2⁶³}=√{2} x 2³¹=3037000499=1.0110.1010.0000.1001.1110.0110.0110.011₂
• if quantity of 1-s is 2N then √(1111...111₂)= 111...1₂ quantity of 1-s is N in result
• if quantity of 1-s is 2N+1 then √(1111...111₂)=√(2)x 2^N
• if quantity of 1-s is N then 11..1₂²=111..1100..001 at the first location is quantity of 1-s equal (N-1), at the second location is quantity of 0-ss equal N, at the third location is 1

P.S. excuse me for my bad english, but I think you understood the idea

[Gunther]

Hi Mikl__,

your idea looks at a first glance not so bad. Could you provide a paper with the formulae, please? You could write it with Latex or an office writer with a formula editor.

Gunther

[Mikl__]

Hi, Gunther!
picture with formulas in attach
maybe I'll show specific examples?
√{2³⁷}=√137438953472=370727=1.0110.1010.0000.1001.11
√{2⁴⁹-1}=√562949953421311=23726566=1.0110.1010.0000.1001.1110.0110
√{2⁵⁰-1}=√1125899906842623=√11111111111111111111111111111111111111111111111111=33554431=1111111111111111111111111
1111111111111111111111111²=33554431²=1125899839733761=11111111111111111111111100000000000000000000000001
11111²=31²=961=1111000001₂

[dedndave]

can you cite the original article ?

[Mikl__]

I have not yet written an article about it :)

[qWord]

Exponentation: Identities and properties

e.g.
sqrt(2³⁷) = (2³² * 2⁴ * 2¹)^1/2 = (2³²)^1/2 (2⁴)^1/2 (2¹)^1/2 = 2¹⁶2²2^1/2

[Gunther]

Mikl__,

your image makes it clear. Thank you.

qWord,

:t

Gunther

[Mikl__]

Exponentation: Identities and properties

e.g.
sqrt(2³⁷) = (2³² * 2⁴ * 2¹)^1/2 = (2³²)^1/2 (2⁴)^1/2 (2¹)^1/2 = 2¹⁶2²2^1/2

Hi, qWord!
talked about another case
sqrt(2³⁷) =sqrt(2^2x18+1) =sqrt(2)x2¹⁸ in the square root is 18+1=19 binary digits, square root from 2 we know, it is 1,4142135623730950488016887242097 or 1.6A09E66...₁₆=1,0110.1010.0000.1001.1110.0110.0110...₂
therefore sqrt(2³⁷) =1.0110.1010.0000.1001.11=370727

[rrr314159]

Hi Miki,

It took some thought to see what you're doing here, pretty clever but I don't think it's worth a Fields medal yet.

1) ------------------------------------------------------------------------------------

First, let x = sqrt(2^(2n+1)) = sqrt(2^2n)*sqrt(2) = 2^n * sqrt(2) = sqrt(2) left-shifted by n.

So, if you express sqrt(2) in binary, = 1.0110101000001001 etc, you can immediately write down x by shifting the "decimal" (or, radix) point by n.

Thus, sqrt(2^7) = sqrt(2^(2n+1)) = 1011 (plus remainder of course) = 11, which is right - that's the truncated sqrt of 128. And, you give a good example in the last post using sqrt(2^37).

Now, this is generally true. Given any radix r, if we can express a number

x = sqrt(r^(2n+1)) = r^n*sqrt(r)

, and we know sqrt(r) in that radix, we can use the same trick. For example base 10:

x = sqrt(10^7) = 10^3 * sqrt(10) = 3.162278... shifted 3 places = 3,162 + remainder.

In fact it doesn't have to be a square root. If we know the number's representation in a radix, to find it times r^n just shift the radix point. For instance

pi*10^3 = 3.14159265 shifted 3 places (a.k.a. 3.14159265 e3) = 3141 (remainder .59265 etc)

Which of course is very well known. Another example, since 1.6875 = 1.1011 binary, times 8 = 1101.1 binary.

2) ------------------------------------------------------------------------------------

Now, the other main point you have is the fact that in binary, (111...(n times))^2 = 111(n-1 times)...000(n times)...1. An example will make it more clear,

In binary: 11111^2 = 1111000001. In decimal: 31^2 = 961 = 512+256+128+64+1

We can see why this is as follows. 111...(n times) = 2^(n+1) - 1. So

111...(n times)^2 = (2^(n+1) - 1)^2 = 2^(n+1)^2 - 2*2^(n+1) + 1 = 2^(2n+2) - 2^(n+2) + 1

Now in general, 2^m - 2^n (with m > n) will consist of (m-n) 1's followed by n zeroes. For example, 2^10 - 2^6 =

10000000000 -
       1000000 =
01111000000

You see, subtracting in the normal way, on the right it's all zeros till we get to "0-1" which gives a 1, carry the 1. then it continues being 1's until the end when the carry from the right makes the last digit 0. So in this case,

1024 - 64 = 960 = 512+256+128+64 (+ zero's)

Stick on the last 1, and we have the example given above, 31^2 = 961.

And so, in binary given n-1 1's followed by n 0's and a 1, we can immediately write the sqrt:

sqrt(1) = 1, sqrt(1001) = 11, sqrt(110001) = 111, sqrt(11100001) = 1111, sqrt(11111100000001) = 1111111 etc

i.e. sqrt(9) = 3, sqrt(32+16+1) = sqrt(49) = 7, sqrt(128+64+32+1) = sqrt(225) = 15, sqrt(16129) = 127 etc

In base 10 there's a similar trick, but instead of a 1 we must use 5, namely, half the radix.

99995^2 = (2^5 - 5)^2 = 9999000025

Because the same subtraction as above gives a very analogous result, for the same reason (consider the "carry"):

10000000000 -
       1000000 =
09999000000

So given n 9's followed by n zero's then 25, we immediately know the square root is n 9's and a 5, without having to compute it. For instance,

sqrt(9999999999000000000025) = 99999999995. Or sqrt(9025) = 95, sqrt(990025) = 995, etc.

3) ------------------------------------------------------------------------------------

Ok, apart from those rather clever points, you have a couple mistakes. You seem to have confused 2^n with n 1's. Thus you say,

sqrt(111...(2n 1's)) = 111...(n 1's)

, in the first post and also in the picture, but no that's not true. For instance sqrt(15) does not equal 3. What is true, sqrt(2^2n) = 2^n, thus sqrt(16) does equal 4. And, the same confusion appears in couple other places. Actually a string of 1's, in radix 2, can never be a perfect square.

4) ------------------------------------------------------------------------------------

Off the top of my head I thought you could never have such a form in any radix. Thus in base 10 it can easily be shown a string of 1's can't be a perfect square: 11, 111, 1111, 11111 ... never square, forget it. Well, of course, "1" is square in any radix, and "11" in any radix 1 less than a square: thus 11 (radix 3) = 4, 11 (radix 63) = 64 etc. But 3 or more, seems impossible at first.

But if you think about it, it's interesting to note there are two cases (at least).

radix 3: sqrt(11111) = (decimal) sqrt(81+27+9+3+1) = sqrt(121) = sqrt(11) or, back to radix 3, = 102

and,

radix 7: sqrt(1111) = (decimal) sqrt(343+49+7+1) = sqrt(400) = sqrt(20) or, back to radix 7, = 26

I doubt there are any others?

------------------------------------------------------------------------------------

Miki, keep at it, a Fields medal is only $15,000 but a millenium prize is a million bucks...! I expect a cut, for helping you on the road to success ;)

[Mikl__]

Hi, rrr314159!
many thanks for the tips

Ok, apart from those rather clever points, you have a couple mistakes. You seem to have confused 2^n with n 1's. Thus you say,

sqrt(111...(2n 1's)) = 111...(n 1's)

, in the first post and also in the picture, but no that's not true. For instance sqrt(15) does not equal 3. What is true, sqrt(2^2n) = 2^n, thus sqrt(16) does equal 4. And, the same confusion appears in couple other places. Actually a string of 1's, in radix 2, can never be a perfect square.

I wrote about the integer part of the square root. It's easy enough to check:
integer(√1111)=integer(√15)=integer(3,8729833462074168851792653997824)=11₂
integer(√11111111)=integer(√255)=integer(15,968719422671311999070245176981)=1111₂
integer(√1111111111111111)=integer(√65535)=integer(255,99804686754936255911520419041)=11111111₂
P.S. I do not fall under the criteria for awarding the Fields Medal, since I'm not a mathematician and I am more than 40

[rrr314159]

Of course, it's correct with the understanding that you're talking about the integer part. I don't see it in the picture you attached, but that's alright.

ps, the millennium prize is still open to old people, and since Perlman didn't take it, I think they've got plenty of cash on hand

[Mikl__]

rrr314159,

the millennium prize is still open to old people

After these words, I feel decrepit invalid. I wanted to write that "The Fields Medal is a prize awarded to two, three, or four mathematicians under 40 years of age at the International Congress of the International Mathematical Union (IMU), a meeting that takes place every four years." and I am 42

[Gunther]

... and I am 42

That's a little bit to old, my friend. Bad luck.

Gunther

[Mikl__]

That's a little bit to old

Gunther,

Bad luck.

Why?

[Gunther]

Why?

It has to do with the age limit for the Fields Medal.  :lol: But there are other open questions, for example the Collatz conjecture. By solving it, you can achieve a lot of fame. It is perhaps partly solved, but I'm not sure. You can check some interesting details here.

Gunther