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

• jackson6612
In summary, the statement that "Any decimal number in the range 0 to 2^(n-1) can be represented in binary form as an n-bit number" is not incorrect. While it is true that numbers outside of this range can also be represented in binary form as n-bit numbers, this does not make the statement false. In certain contexts, a smaller range may be sufficient to prove a point.
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]?

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

Thanks a lot, Mathman.

jackson6612 said:
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.

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.

## 1. How do you represent a decimal number in binary form?

In order to represent a decimal number in binary form, you need to convert it to its binary equivalent by dividing it by 2 and noting the remainders until the quotient becomes 0. Then, you write the remainders in reverse order to get the binary form.

## 2. What is the range of decimal numbers that can be represented in binary form?

The range of decimal numbers that can be represented in binary form is from 0 to 2^(n-1), where n is the number of bits used to represent the number.

## 3. Why is the range of decimal numbers limited to 2^(n-1) when representing in binary form?

The range is limited to 2^(n-1) because the first bit in a binary number represents the sign, with 0 being positive and 1 being negative. Therefore, the remaining n-1 bits can only represent numbers up to 2^(n-1)-1.

## 4. What if the decimal number falls outside the range of 0 to 2^(n-1)?

If the decimal number falls outside the range of 0 to 2^(n-1), it cannot be represented accurately in binary form and may result in errors or loss of information.

## 5. Can any decimal number be represented in binary form?

Yes, any decimal number can be represented in binary form as long as it falls within the range of 0 to 2^(n-1) and the proper conversion method is used. However, certain numbers with repeating decimals may result in an infinite binary representation.

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