Question

Indices, Surd and Numeral Systems #2
Question 42
A binary number has n digits and all digits are '1's. Convert the binary number into a decimal number in terms of n.

Answer 1

lets try out a few small ones to see if theres a pattern

Answer 2

1 = 1 = 2 - 1
11 = 3 = 4 - 1
111 = 7 = 8 - 1

the value appears to be 2^n -1

Answer 3

Yes. But how to explain it?

Answer 4

if you add 1 to it you get:

1000....n which is 2^n

Answer 5

so subtract 1 from the 2^(n+1) spot

Answer 6

Sorry... but I don't understand that. I understand the part ' add 1', but not the 'subtract one'.

Answer 7

Why is it 2^(n+1), but not 2^n - 1?

Answer 8

if you have n 1s to begin, that is one less in value than the a string of digits that starts with a 1 with n zeros after it

Answer 9

maybe im mixing the n parts in my head ... tho

Answer 10

2^0 = 0
2^1 = 10
2^2 = 100
2^3 = 1000
2^4 = 10000

so there are n zeros behind the 1 digit

Answer 11

2^0 = 1 :)

Answer 12

so if we subtract 1 from those we get:

2^0 -1 = 0
2^1 - 1 = 1
2^2 - 1 = 11
2^3 - 1 = 111

yeah, the (n+1) was a misthought

Answer 13

correct

Answer 14

In this case, I should write
when 111.....1 + 1,
^ n '1's
It becomes, 100...0, that is 2^n
^ n '0's
So,
111...1 = 100...0 - 1 = 2^n -1
^n '1's ^n '0's
Is that correct?