32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 047 6 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 047 6(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 047 6| = 0.000 000 047 6

2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 047 6.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 047 6 × 2 = 0 + 0.000 000 095 2;
  • 2) 0.000 000 095 2 × 2 = 0 + 0.000 000 190 4;
  • 3) 0.000 000 190 4 × 2 = 0 + 0.000 000 380 8;
  • 4) 0.000 000 380 8 × 2 = 0 + 0.000 000 761 6;
  • 5) 0.000 000 761 6 × 2 = 0 + 0.000 001 523 2;
  • 6) 0.000 001 523 2 × 2 = 0 + 0.000 003 046 4;
  • 7) 0.000 003 046 4 × 2 = 0 + 0.000 006 092 8;
  • 8) 0.000 006 092 8 × 2 = 0 + 0.000 012 185 6;
  • 9) 0.000 012 185 6 × 2 = 0 + 0.000 024 371 2;
  • 10) 0.000 024 371 2 × 2 = 0 + 0.000 048 742 4;
  • 11) 0.000 048 742 4 × 2 = 0 + 0.000 097 484 8;
  • 12) 0.000 097 484 8 × 2 = 0 + 0.000 194 969 6;
  • 13) 0.000 194 969 6 × 2 = 0 + 0.000 389 939 2;
  • 14) 0.000 389 939 2 × 2 = 0 + 0.000 779 878 4;
  • 15) 0.000 779 878 4 × 2 = 0 + 0.001 559 756 8;
  • 16) 0.001 559 756 8 × 2 = 0 + 0.003 119 513 6;
  • 17) 0.003 119 513 6 × 2 = 0 + 0.006 239 027 2;
  • 18) 0.006 239 027 2 × 2 = 0 + 0.012 478 054 4;
  • 19) 0.012 478 054 4 × 2 = 0 + 0.024 956 108 8;
  • 20) 0.024 956 108 8 × 2 = 0 + 0.049 912 217 6;
  • 21) 0.049 912 217 6 × 2 = 0 + 0.099 824 435 2;
  • 22) 0.099 824 435 2 × 2 = 0 + 0.199 648 870 4;
  • 23) 0.199 648 870 4 × 2 = 0 + 0.399 297 740 8;
  • 24) 0.399 297 740 8 × 2 = 0 + 0.798 595 481 6;
  • 25) 0.798 595 481 6 × 2 = 1 + 0.597 190 963 2;
  • 26) 0.597 190 963 2 × 2 = 1 + 0.194 381 926 4;
  • 27) 0.194 381 926 4 × 2 = 0 + 0.388 763 852 8;
  • 28) 0.388 763 852 8 × 2 = 0 + 0.777 527 705 6;
  • 29) 0.777 527 705 6 × 2 = 1 + 0.555 055 411 2;
  • 30) 0.555 055 411 2 × 2 = 1 + 0.110 110 822 4;
  • 31) 0.110 110 822 4 × 2 = 0 + 0.220 221 644 8;
  • 32) 0.220 221 644 8 × 2 = 0 + 0.440 443 289 6;
  • 33) 0.440 443 289 6 × 2 = 0 + 0.880 886 579 2;
  • 34) 0.880 886 579 2 × 2 = 1 + 0.761 773 158 4;
  • 35) 0.761 773 158 4 × 2 = 1 + 0.523 546 316 8;
  • 36) 0.523 546 316 8 × 2 = 1 + 0.047 092 633 6;
  • 37) 0.047 092 633 6 × 2 = 0 + 0.094 185 267 2;
  • 38) 0.094 185 267 2 × 2 = 0 + 0.188 370 534 4;
  • 39) 0.188 370 534 4 × 2 = 0 + 0.376 741 068 8;
  • 40) 0.376 741 068 8 × 2 = 0 + 0.753 482 137 6;
  • 41) 0.753 482 137 6 × 2 = 1 + 0.506 964 275 2;
  • 42) 0.506 964 275 2 × 2 = 1 + 0.013 928 550 4;
  • 43) 0.013 928 550 4 × 2 = 0 + 0.027 857 100 8;
  • 44) 0.027 857 100 8 × 2 = 0 + 0.055 714 201 6;
  • 45) 0.055 714 201 6 × 2 = 0 + 0.111 428 403 2;
  • 46) 0.111 428 403 2 × 2 = 0 + 0.222 856 806 4;
  • 47) 0.222 856 806 4 × 2 = 0 + 0.445 713 612 8;
  • 48) 0.445 713 612 8 × 2 = 0 + 0.891 427 225 6;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 047 6(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 0111 0000 1100 0000(2)


6. Positive number before normalization:

0.000 000 047 6(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 0111 0000 1100 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 047 6(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 0111 0000 1100 0000(2) =

0.0000 0000 0000 0000 0000 0000 1100 1100 0111 0000 1100 0000(2) × 20 =

1.1001 1000 1110 0001 1000 000(2) × 2-25


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

Sign 1 (a negative number)

Exponent (unadjusted): -25

Mantissa (not normalized):
1.1001 1000 1110 0001 1000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

102(10) =

0110 0110(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1100 0111 0000 1100 0000 =

100 1100 0111 0000 1100 0000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0110 0110

Mantissa (23 bits) =
100 1100 0111 0000 1100 0000


The base ten decimal number -0.000 000 047 6 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0110 0110 - 100 1100 0111 0000 1100 0000

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number -0.000 000 047 7 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number -0.000 000 047 5 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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

Number 52.218 78 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number -0.014 459 890 236 287 751 815 702 126 261 5 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 0.545 454 545 459 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 0.000 007 629 394 536 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 0.000 000 000 009 543 491 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 20 711 126 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number -670.1 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number -72.082 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 0.000 000 000 107 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 UTC (GMT)
Number 1 019 740 148 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 05 15:35 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

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111