# Converting decimal number to binary IEEE 754 32-bit single-precision then hexadecimal

1. Jan 27, 2013

### s3a

1. The problem statement, all variables and given/known data
Convert the decimal number 2000.90210 to a single-precision IEEE 754 hexadecimal.

2. Relevant equations
(exponent field)_10 - 127_10 = exponent_10
The first bit is the sign bit.
The 2nd to 9th bits are the exponent bits.
The 10th to 32nd bits are the fraction field bits.

3. The attempt at a solution
My attempt at the solution is attached as MyWork.jpg.

I was planning to convert it to the binary IEEE 754 32-bit single-precision representation since that's what I am familiar with and then convert it to hexadecimal however, it seems that when I do this, my binary version is far off from the decimal value I'm initially given.

I'm not sure but, this might be related to the fact that the binary fraction field is (at least, seemingly) irrational.

Any help in solving this problem would be greatly appreciated!

#### Attached Files:

• ###### MyWork.jpg
File size:
32 KB
Views:
577
2. Jan 27, 2013

### Staff: Mentor

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

Rational <-> irrational is independent of the base you express a number in. There is no exact representation of the number with a finite number in base 2, but that does not matter. It just gives a small rounding error.

In which way?

3. Jan 27, 2013

### s3a

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

Hello.

Alright.

When I input my binary IEEE 754 answer to this ( http://www.h-schmidt.net/FloatConverter/IEEE754.html ) website, the decimal representation is around 4001.80. If it's not my work that's wrong, what's wrong with the Java applet software or with the way I am “communicating” with it?

4. Jan 27, 2013

### rcgldr

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

so this means
(exponent field)_10 = exponent_10 + 127_10

if exponent_10 is 127, then what should (exponent field)_10 be?

5. Jan 27, 2013

### s3a

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

The equation is (exponent field)_10 = exponent_10 – 127_10. (Notice the – instead of a +.) I did do that in my work, however. So, why is the Java applet software disagreeing strongly with me?

6. Jan 27, 2013

### rcgldr

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

(exponent field)_10 - 127_10 = exponent_10

then add 127_10 to both sides:

(exponent field)_10 = exponent_10 + 127_10

7. Jan 27, 2013

### Staff: Mentor

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

I would not call this "far off", it is exactly a factor of 2, indicating a problem with the exponent. rcgldr found the error.

8. Jan 27, 2013

### s3a

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

Good point, mfb. I just noticed that.

And, rcgldr, that was just what I needed to fix my entire answer to the problem.

Thank you both!

9. Jan 27, 2013

### s3a

Re: Converting decimal number to binary IEEE 754 32-bit single-precision then hexadec

Good point, mfb. I just noticed that.

And, rcgldr, that was just what I needed to fix my entire answer to the problem.

Thank you both!