- #1

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**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.

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- #1

- 334

- 1

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

Please guide me. Thanks.

- #2

Science Advisor

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- 571

You're right. n=3 table

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

7=2^{3} - 1

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

7=2

- #3

- 334

- 1

Thanks a lot, Mathman.

- #4

Science Advisor

Homework Helper

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I have read somewhere: Any decimal number in the range0 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.

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