0.000 000 000 050 403 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 050 403 4(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 050 403 4(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 050 403 4.

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) 0.000 000 000 050 403 4 × 2 = 0 + 0.000 000 000 100 806 8;
  • 2) 0.000 000 000 100 806 8 × 2 = 0 + 0.000 000 000 201 613 6;
  • 3) 0.000 000 000 201 613 6 × 2 = 0 + 0.000 000 000 403 227 2;
  • 4) 0.000 000 000 403 227 2 × 2 = 0 + 0.000 000 000 806 454 4;
  • 5) 0.000 000 000 806 454 4 × 2 = 0 + 0.000 000 001 612 908 8;
  • 6) 0.000 000 001 612 908 8 × 2 = 0 + 0.000 000 003 225 817 6;
  • 7) 0.000 000 003 225 817 6 × 2 = 0 + 0.000 000 006 451 635 2;
  • 8) 0.000 000 006 451 635 2 × 2 = 0 + 0.000 000 012 903 270 4;
  • 9) 0.000 000 012 903 270 4 × 2 = 0 + 0.000 000 025 806 540 8;
  • 10) 0.000 000 025 806 540 8 × 2 = 0 + 0.000 000 051 613 081 6;
  • 11) 0.000 000 051 613 081 6 × 2 = 0 + 0.000 000 103 226 163 2;
  • 12) 0.000 000 103 226 163 2 × 2 = 0 + 0.000 000 206 452 326 4;
  • 13) 0.000 000 206 452 326 4 × 2 = 0 + 0.000 000 412 904 652 8;
  • 14) 0.000 000 412 904 652 8 × 2 = 0 + 0.000 000 825 809 305 6;
  • 15) 0.000 000 825 809 305 6 × 2 = 0 + 0.000 001 651 618 611 2;
  • 16) 0.000 001 651 618 611 2 × 2 = 0 + 0.000 003 303 237 222 4;
  • 17) 0.000 003 303 237 222 4 × 2 = 0 + 0.000 006 606 474 444 8;
  • 18) 0.000 006 606 474 444 8 × 2 = 0 + 0.000 013 212 948 889 6;
  • 19) 0.000 013 212 948 889 6 × 2 = 0 + 0.000 026 425 897 779 2;
  • 20) 0.000 026 425 897 779 2 × 2 = 0 + 0.000 052 851 795 558 4;
  • 21) 0.000 052 851 795 558 4 × 2 = 0 + 0.000 105 703 591 116 8;
  • 22) 0.000 105 703 591 116 8 × 2 = 0 + 0.000 211 407 182 233 6;
  • 23) 0.000 211 407 182 233 6 × 2 = 0 + 0.000 422 814 364 467 2;
  • 24) 0.000 422 814 364 467 2 × 2 = 0 + 0.000 845 628 728 934 4;
  • 25) 0.000 845 628 728 934 4 × 2 = 0 + 0.001 691 257 457 868 8;
  • 26) 0.001 691 257 457 868 8 × 2 = 0 + 0.003 382 514 915 737 6;
  • 27) 0.003 382 514 915 737 6 × 2 = 0 + 0.006 765 029 831 475 2;
  • 28) 0.006 765 029 831 475 2 × 2 = 0 + 0.013 530 059 662 950 4;
  • 29) 0.013 530 059 662 950 4 × 2 = 0 + 0.027 060 119 325 900 8;
  • 30) 0.027 060 119 325 900 8 × 2 = 0 + 0.054 120 238 651 801 6;
  • 31) 0.054 120 238 651 801 6 × 2 = 0 + 0.108 240 477 303 603 2;
  • 32) 0.108 240 477 303 603 2 × 2 = 0 + 0.216 480 954 607 206 4;
  • 33) 0.216 480 954 607 206 4 × 2 = 0 + 0.432 961 909 214 412 8;
  • 34) 0.432 961 909 214 412 8 × 2 = 0 + 0.865 923 818 428 825 6;
  • 35) 0.865 923 818 428 825 6 × 2 = 1 + 0.731 847 636 857 651 2;
  • 36) 0.731 847 636 857 651 2 × 2 = 1 + 0.463 695 273 715 302 4;
  • 37) 0.463 695 273 715 302 4 × 2 = 0 + 0.927 390 547 430 604 8;
  • 38) 0.927 390 547 430 604 8 × 2 = 1 + 0.854 781 094 861 209 6;
  • 39) 0.854 781 094 861 209 6 × 2 = 1 + 0.709 562 189 722 419 2;
  • 40) 0.709 562 189 722 419 2 × 2 = 1 + 0.419 124 379 444 838 4;
  • 41) 0.419 124 379 444 838 4 × 2 = 0 + 0.838 248 758 889 676 8;
  • 42) 0.838 248 758 889 676 8 × 2 = 1 + 0.676 497 517 779 353 6;
  • 43) 0.676 497 517 779 353 6 × 2 = 1 + 0.352 995 035 558 707 2;
  • 44) 0.352 995 035 558 707 2 × 2 = 0 + 0.705 990 071 117 414 4;
  • 45) 0.705 990 071 117 414 4 × 2 = 1 + 0.411 980 142 234 828 8;
  • 46) 0.411 980 142 234 828 8 × 2 = 0 + 0.823 960 284 469 657 6;
  • 47) 0.823 960 284 469 657 6 × 2 = 1 + 0.647 920 568 939 315 2;
  • 48) 0.647 920 568 939 315 2 × 2 = 1 + 0.295 841 137 878 630 4;
  • 49) 0.295 841 137 878 630 4 × 2 = 0 + 0.591 682 275 757 260 8;
  • 50) 0.591 682 275 757 260 8 × 2 = 1 + 0.183 364 551 514 521 6;
  • 51) 0.183 364 551 514 521 6 × 2 = 0 + 0.366 729 103 029 043 2;
  • 52) 0.366 729 103 029 043 2 × 2 = 0 + 0.733 458 206 058 086 4;
  • 53) 0.733 458 206 058 086 4 × 2 = 1 + 0.466 916 412 116 172 8;
  • 54) 0.466 916 412 116 172 8 × 2 = 0 + 0.933 832 824 232 345 6;
  • 55) 0.933 832 824 232 345 6 × 2 = 1 + 0.867 665 648 464 691 2;
  • 56) 0.867 665 648 464 691 2 × 2 = 1 + 0.735 331 296 929 382 4;
  • 57) 0.735 331 296 929 382 4 × 2 = 1 + 0.470 662 593 858 764 8;
  • 58) 0.470 662 593 858 764 8 × 2 = 0 + 0.941 325 187 717 529 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 - the converted number we get in the end will be just a very good approximation of the initial one).


4. 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:


0.000 000 000 050 403 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0100 1011 10(2)

5. Positive number before normalization:

0.000 000 000 050 403 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0100 1011 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 35 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 050 403 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0100 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0100 1011 10(2) × 20 =

1.1011 1011 0101 1010 0101 110(2) × 2-35


7. 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 0 (a positive number)

Exponent (unadjusted): -35

Mantissa (not normalized):
1.1011 1011 0101 1010 0101 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-35 + 2(8-1) - 1 =

(-35 + 127)(10) =

92(10)


9. 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;
  • 92 ÷ 2 = 46 + 0;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

92(10) =

0101 1100(2)


11. 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. 101 1101 1010 1101 0010 1110 =

101 1101 1010 1101 0010 1110


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0101 1100

Mantissa (23 bits) =
101 1101 1010 1101 0010 1110


Decimal number 0.000 000 000 050 403 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1100 - 101 1101 1010 1101 0010 1110

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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