How to explain this binary number problem.

Question

If I have a number made of 128 1 and 128 0 (256 bits) and I convert it to 10 base integer I get:

115792089237316195423570985008687907852929702298719625575994209400481361428480

and if I square the number 2 with 256 I get:

115792089237316195423570985008687907853269984665640564039457584007913129639936

But taking a close look, the second number which should represent maximal 256 bits number is smaller than the first one which has 256 bits. How can this be explained?

What does it mean "if I squre number 2 with 256"? – user Jun 22 '19 at 16:59 — user, Jun 22 '19 at 16:59

score 3 · Accepted Answer · answered Jun 22 '19 at 16:19

3

Actually, the second number is larger. I indicated the first digit where they change:

$11579208923731619542357098500868790785\color{red}2929702298719625575994209400481361428480$

$11579208923731619542357098500868790785\color{red}3269984665640564039457584007913129639936$

answered Jun 22 '19 at 16:19

Bram28

100,612
6
70
118

No, they are equal! https://www.google.com/search?q=115792089237316195423570985008687907852929702298719625575994209400481361428480-115792089237316195423570985008687907853269984665640564039457584007913129639936 – dcolazin Jun 22 '19 at 16:20
1

And Google never lies! OMG – Bram28 Jun 22 '19 at 16:21

Somos · Answer 2 · 2019-06-22T20:11:25.483

2

Given $n=128,$ then the first number is $4^n-2^n$ while the second is $4^n$ which is obviously larger. If you look closely, the two number have the same digits up to halfway and differ after that. By the way, the number $4^n-1$ is the maximum number with $256$ bits and $4^n$ is the first number that requires $257$ bits.

edited Jun 22 '19 at 20:11

answered Jun 22 '19 at 17:44

Somos

35,251
3
30
76

How to explain this binary number problem.

2 Answers2