# Base ten decimal number 247 079.925 converted to 32 bit single precision IEEE 754 binary floating point standard

## How to convert the decimal number 247 079.925(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: 247 079. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

• division = quotient + remainder;
• 247 079 ÷ 2 = 123 539 + 1;
• 123 539 ÷ 2 = 61 769 + 1;
• 61 769 ÷ 2 = 30 884 + 1;
• 30 884 ÷ 2 = 15 442 + 0;
• 15 442 ÷ 2 = 7 721 + 0;
• 7 721 ÷ 2 = 3 860 + 1;
• 3 860 ÷ 2 = 1 930 + 0;
• 1 930 ÷ 2 = 965 + 0;
• 965 ÷ 2 = 482 + 1;
• 482 ÷ 2 = 241 + 0;
• 241 ÷ 2 = 120 + 1;
• 120 ÷ 2 = 60 + 0;
• 60 ÷ 2 = 30 + 0;
• 30 ÷ 2 = 15 + 0;
• 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: 0.925. 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) 0.925 × 2 = 1 + 0.85;
• 2) 0.85 × 2 = 1 + 0.7;
• 3) 0.7 × 2 = 1 + 0.4;
• 4) 0.4 × 2 = 0 + 0.8;
• 5) 0.8 × 2 = 1 + 0.6;
• 6) 0.6 × 2 = 1 + 0.2;
• 7) 0.2 × 2 = 0 + 0.4;
• 8) 0.4 × 2 = 0 + 0.8;
• 9) 0.8 × 2 = 1 + 0.6;
• 10) 0.6 × 2 = 1 + 0.2;
• 11) 0.2 × 2 = 0 + 0.4;
• 12) 0.4 × 2 = 0 + 0.8;
• 13) 0.8 × 2 = 1 + 0.6;
• 14) 0.6 × 2 = 1 + 0.2;
• 15) 0.2 × 2 = 0 + 0.4;
• 16) 0.4 × 2 = 0 + 0.8;
• 17) 0.8 × 2 = 1 + 0.6;
• 18) 0.6 × 2 = 1 + 0.2;
• 19) 0.2 × 2 = 0 + 0.4;
• 20) 0.4 × 2 = 0 + 0.8;
• 21) 0.8 × 2 = 1 + 0.6;
• 22) 0.6 × 2 = 1 + 0.2;
• 23) 0.2 × 2 = 0 + 0.4;
• 24) 0.4 × 2 = 0 + 0.8;

### 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:

#### 144(10)

• division = quotient + remainder;
• 144 ÷ 2 = 72 + 0;
• 72 ÷ 2 = 36 + 0;
• 36 ÷ 2 = 18 + 0;
• 18 ÷ 2 = 9 + 0;
• 9 ÷ 2 = 4 + 1;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

## 0 - 1001 0000 - 111 0001 0100 1001 1111 1011

(32 bits IEEE 754)

• 0

31

• 1

30
• 0

29
• 0

28
• 1

27
• 0

26
• 0

25
• 0

24
• 0

23

• 1

22
• 1

21
• 1

20
• 0

19
• 0

18
• 0

17
• 1

16
• 0

15
• 1

14
• 0

13
• 0

12
• 1

11
• 0

10
• 0

9
• 1

8
• 1

7
• 1

6
• 1

5
• 1

4
• 1

3
• 0

2
• 1

1
• 1

0

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

 247 079.925 = 0 - 1001 0000 - 111 0001 0100 1001 1111 1011 Jun 19 19:37 UTC (GMT) 150.875 = 0 - 1000 0110 - 001 0110 1110 0000 0000 0000 Jun 19 19:37 UTC (GMT) -2.75 = 1 - 1000 0000 - 011 0000 0000 0000 0000 0000 Jun 19 19:36 UTC (GMT) 10.01 = 0 - 1000 0010 - 010 0000 0010 1000 1111 0101 Jun 19 19:36 UTC (GMT) 44 474 = 0 - 1000 1110 - 010 1101 1011 1010 0000 0000 Jun 19 19:36 UTC (GMT) 1 111 111 011 111 111 111 111 111 111 111 = 0 - 1110 0010 - 110 0000 0110 0011 0001 0111 Jun 19 19:35 UTC (GMT) 24 124.705 = 0 - 1000 1101 - 011 1100 0111 1001 0110 1000 Jun 19 19:35 UTC (GMT) 5.27 = 0 - 1000 0001 - 010 1000 1010 0011 1101 0111 Jun 19 19:35 UTC (GMT) 1 740 = 0 - 1000 1001 - 101 1001 1000 0000 0000 0000 Jun 19 19:34 UTC (GMT) 2 015.3 = 0 - 1000 1001 - 111 1011 1110 1001 1001 1001 Jun 19 19:33 UTC (GMT) 5 411 = 0 - 1000 1011 - 010 1001 0001 1000 0000 0000 Jun 19 19:33 UTC (GMT) 3 294 = 0 - 1000 1010 - 100 1101 1110 0000 0000 0000 Jun 19 19:33 UTC (GMT) 66.25 = 0 - 1000 0101 - 000 0100 1000 0000 0000 0000 Jun 19 19:33 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:
Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
• 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:

• 1. Start with the positive version of the number:

|-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)

Exponent (unadjusted): 4

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