# Base ten decimal number 1 000.708 converted to 32 bit single precision IEEE 754 binary floating point standard

## How to convert the decimal number 1 000.708(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

### 1. First, convert to binary (base 2) the integer part: 1 000. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

• division = quotient + remainder;
• 1 000 ÷ 2 = 500 + 0;
• 500 ÷ 2 = 250 + 0;
• 250 ÷ 2 = 125 + 0;
• 125 ÷ 2 = 62 + 1;
• 62 ÷ 2 = 31 + 0;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

### 3. Convert to binary (base 2) the fractional part: 0.708. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

• #) multiplying = integer + fractional part;
• 1) 0.708 × 2 = 1 + 0.416;
• 2) 0.416 × 2 = 0 + 0.832;
• 3) 0.832 × 2 = 1 + 0.664;
• 4) 0.664 × 2 = 1 + 0.328;
• 5) 0.328 × 2 = 0 + 0.656;
• 6) 0.656 × 2 = 1 + 0.312;
• 7) 0.312 × 2 = 0 + 0.624;
• 8) 0.624 × 2 = 1 + 0.248;
• 9) 0.248 × 2 = 0 + 0.496;
• 10) 0.496 × 2 = 0 + 0.992;
• 11) 0.992 × 2 = 1 + 0.984;
• 12) 0.984 × 2 = 1 + 0.968;
• 13) 0.968 × 2 = 1 + 0.936;
• 14) 0.936 × 2 = 1 + 0.872;
• 15) 0.872 × 2 = 1 + 0.744;
• 16) 0.744 × 2 = 1 + 0.488;
• 17) 0.488 × 2 = 0 + 0.976;
• 18) 0.976 × 2 = 1 + 0.952;
• 19) 0.952 × 2 = 1 + 0.904;
• 20) 0.904 × 2 = 1 + 0.808;
• 21) 0.808 × 2 = 1 + 0.616;
• 22) 0.616 × 2 = 1 + 0.232;
• 23) 0.232 × 2 = 0 + 0.464;
• 24) 0.464 × 2 = 0 + 0.928;

### 6. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2:

#### 136(10)

• division = quotient + remainder;
• 136 ÷ 2 = 68 + 0;
• 68 ÷ 2 = 34 + 0;
• 34 ÷ 2 = 17 + 0;
• 17 ÷ 2 = 8 + 1;
• 8 ÷ 2 = 4 + 0;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

## 0 - 1000 1000 - 111 1010 0010 1101 0100 1111

(32 bits IEEE 754)

• 0

31

• 1

30
• 0

29
• 0

28
• 0

27
• 1

26
• 0

25
• 0

24
• 0

23

• 1

22
• 1

21
• 1

20
• 1

19
• 0

18
• 1

17
• 0

16
• 0

15
• 0

14
• 1

13
• 0

12
• 1

11
• 1

10
• 0

9
• 1

8
• 0

7
• 1

6
• 0

5
• 0

4
• 1

3
• 1

2
• 1

1
• 1

0

## Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

 1 000.708 = 0 - 1000 1000 - 111 1010 0010 1101 0100 1111 Aug 25 04:52 UTC (GMT) -6.375 = 1 - 1000 0001 - 100 1100 0000 0000 0000 0000 Aug 25 04:52 UTC (GMT) 51.523 = 0 - 1000 0100 - 100 1110 0001 0111 1000 1101 Aug 25 04:52 UTC (GMT) 32 100.121 3 = 0 - 1000 1101 - 111 1010 1100 1000 0011 1110 Aug 25 04:51 UTC (GMT) 50.665 283 203 125 = 0 - 1000 0100 - 100 1010 1010 1001 0100 0000 Aug 25 04:51 UTC (GMT) -16.531 25 = 1 - 1000 0011 - 000 0100 0100 0000 0000 0000 Aug 25 04:50 UTC (GMT) -174 = 1 - 1000 0110 - 010 1110 0000 0000 0000 0000 Aug 25 04:49 UTC (GMT) 0.35 = 0 - 0111 1101 - 011 0011 0011 0011 0011 0011 Aug 25 04:49 UTC (GMT) 2 147 483 904 = 0 - 1001 1110 - 000 0000 0000 0000 0000 0001 Aug 25 04:49 UTC (GMT) 94.573 602 236 1 = 0 - 1000 0101 - 011 1101 0010 0101 1010 1111 Aug 25 04:49 UTC (GMT) 0.000 000 000 000 000 000 000 021 32 = 0 - 0011 0011 - 100 1110 0011 0001 1100 0011 Aug 25 04:48 UTC (GMT) 4.67 = 0 - 1000 0001 - 001 0101 0111 0000 1010 0011 Aug 25 04:48 UTC (GMT) 100.25 = 0 - 1000 0101 - 100 1000 1000 0000 0000 0000 Aug 25 04:47 UTC (GMT) All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

## How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

### Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

• 1. If the number to be converted is negative, start with its the positive version.
• 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
• 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
• 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
• 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
• 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
• 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
• 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
• 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

### Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

|-25.347| = 25.347

• 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
• division = quotient + remainder;
• 25 ÷ 2 = 12 + 1;
• 12 ÷ 2 = 6 + 0;
• 6 ÷ 2 = 3 + 0;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;
• We have encountered a quotient that is ZERO => FULL STOP
• 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

25(10) = 1 1001(2)

• 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
• #) multiplying = integer + fractional part;
• 1) 0.347 × 2 = 0 + 0.694;
• 2) 0.694 × 2 = 1 + 0.388;
• 3) 0.388 × 2 = 0 + 0.776;
• 4) 0.776 × 2 = 1 + 0.552;
• 5) 0.552 × 2 = 1 + 0.104;
• 6) 0.104 × 2 = 0 + 0.208;
• 7) 0.208 × 2 = 0 + 0.416;
• 8) 0.416 × 2 = 0 + 0.832;
• 9) 0.832 × 2 = 1 + 0.664;
• 10) 0.664 × 2 = 1 + 0.328;
• 11) 0.328 × 2 = 0 + 0.656;
• 12) 0.656 × 2 = 1 + 0.312;
• 13) 0.312 × 2 = 0 + 0.624;
• 14) 0.624 × 2 = 1 + 0.248;
• 15) 0.248 × 2 = 0 + 0.496;
• 16) 0.496 × 2 = 0 + 0.992;
• 17) 0.992 × 2 = 1 + 0.984;
• 18) 0.984 × 2 = 1 + 0.968;
• 19) 0.968 × 2 = 1 + 0.936;
• 20) 0.936 × 2 = 1 + 0.872;
• 21) 0.872 × 2 = 1 + 0.744;
• 22) 0.744 × 2 = 1 + 0.488;
• 23) 0.488 × 2 = 0 + 0.976;
• 24) 0.976 × 2 = 1 + 0.952;
• We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

• 6. Summarizing - the positive number before normalization:

25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

• 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

25.347(10) =
1 1001.0101 1000 1101 0100 1111 1101(2) =
1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
1.1001 0101 1000 1101 0100 1111 1101(2) × 24

• 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

Sign: 1 (a negative number)

Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

• 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
1000 0011(2)

• 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

Mantissa (normalized): 100 1010 1100 0110 1010 0111

• Conclusion:

Sign (1 bit) = 1 (a negative number)

Exponent (8 bits) = 1000 0011

Mantissa (23 bits) = 100 1010 1100 0110 1010 0111