Any decimal number in the range 0 to 2^(n-1)can be represented in binary form

[jackson6612]

I have read somewhere: Any decimal number in the range 0 to 2^(n-1) can be represented in binary form as an n-bit number.

I suspect it's wrong. Shouldn't it rather be 0 to [(2^n)-1]?

Please guide me. Thanks.

 

[mathman]

You're right. n=3 table
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111

7=2³ - 1

 

[jackson6612]

Thanks a lot, Mathman.

 

[AlephZero]

I have read somewhere: Any decimal number in the range 0 to 2^(n-1) can be represented in binary form as an n-bit number.

I suspect it's wrong. Shouldn't it rather be 0 to [(2^n)-1]?

It's not "wrong". For example if n = 3, then any decimal number in the range 0 to 2^2 = 4 CAN be expressed as an n-bit binary number.

Sure, there are some other numbers that can be expressed as well, like 5 6 and 7, but that doesn't make the statement false.

Often in math proofs, there is no value in stretching every condition to its ultimate limit just for the sake of it. Possibly, in the context where you read this, the smaller range was all that was relevant.

 

[CellCoree]

I can confirm that the statement is correct. Any decimal number in the range of 0 to 2^(n-1) can indeed be represented in binary form as an n-bit number. This is because the range of 0 to 2^(n-1) includes all possible combinations of n bits, from all 0's to all 1's. However, it is important to note that the highest possible decimal number that can be represented in n bits is actually (2^n)-1, not 2^(n-1). This is because the first bit in binary represents the number 0, not 1. Therefore, the correct range for representing a decimal number in binary form with n bits is 0 to (2^n)-1. I hope this clarifies any confusion.