32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 000 011 100 111 8 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 000 011 100 111 8(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. 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: 0.000 000 000 011 100 111 8.

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) 0.000 000 000 011 100 111 8 × 2 = 0 + 0.000 000 000 022 200 223 6;
  • 2) 0.000 000 000 022 200 223 6 × 2 = 0 + 0.000 000 000 044 400 447 2;
  • 3) 0.000 000 000 044 400 447 2 × 2 = 0 + 0.000 000 000 088 800 894 4;
  • 4) 0.000 000 000 088 800 894 4 × 2 = 0 + 0.000 000 000 177 601 788 8;
  • 5) 0.000 000 000 177 601 788 8 × 2 = 0 + 0.000 000 000 355 203 577 6;
  • 6) 0.000 000 000 355 203 577 6 × 2 = 0 + 0.000 000 000 710 407 155 2;
  • 7) 0.000 000 000 710 407 155 2 × 2 = 0 + 0.000 000 001 420 814 310 4;
  • 8) 0.000 000 001 420 814 310 4 × 2 = 0 + 0.000 000 002 841 628 620 8;
  • 9) 0.000 000 002 841 628 620 8 × 2 = 0 + 0.000 000 005 683 257 241 6;
  • 10) 0.000 000 005 683 257 241 6 × 2 = 0 + 0.000 000 011 366 514 483 2;
  • 11) 0.000 000 011 366 514 483 2 × 2 = 0 + 0.000 000 022 733 028 966 4;
  • 12) 0.000 000 022 733 028 966 4 × 2 = 0 + 0.000 000 045 466 057 932 8;
  • 13) 0.000 000 045 466 057 932 8 × 2 = 0 + 0.000 000 090 932 115 865 6;
  • 14) 0.000 000 090 932 115 865 6 × 2 = 0 + 0.000 000 181 864 231 731 2;
  • 15) 0.000 000 181 864 231 731 2 × 2 = 0 + 0.000 000 363 728 463 462 4;
  • 16) 0.000 000 363 728 463 462 4 × 2 = 0 + 0.000 000 727 456 926 924 8;
  • 17) 0.000 000 727 456 926 924 8 × 2 = 0 + 0.000 001 454 913 853 849 6;
  • 18) 0.000 001 454 913 853 849 6 × 2 = 0 + 0.000 002 909 827 707 699 2;
  • 19) 0.000 002 909 827 707 699 2 × 2 = 0 + 0.000 005 819 655 415 398 4;
  • 20) 0.000 005 819 655 415 398 4 × 2 = 0 + 0.000 011 639 310 830 796 8;
  • 21) 0.000 011 639 310 830 796 8 × 2 = 0 + 0.000 023 278 621 661 593 6;
  • 22) 0.000 023 278 621 661 593 6 × 2 = 0 + 0.000 046 557 243 323 187 2;
  • 23) 0.000 046 557 243 323 187 2 × 2 = 0 + 0.000 093 114 486 646 374 4;
  • 24) 0.000 093 114 486 646 374 4 × 2 = 0 + 0.000 186 228 973 292 748 8;
  • 25) 0.000 186 228 973 292 748 8 × 2 = 0 + 0.000 372 457 946 585 497 6;
  • 26) 0.000 372 457 946 585 497 6 × 2 = 0 + 0.000 744 915 893 170 995 2;
  • 27) 0.000 744 915 893 170 995 2 × 2 = 0 + 0.001 489 831 786 341 990 4;
  • 28) 0.001 489 831 786 341 990 4 × 2 = 0 + 0.002 979 663 572 683 980 8;
  • 29) 0.002 979 663 572 683 980 8 × 2 = 0 + 0.005 959 327 145 367 961 6;
  • 30) 0.005 959 327 145 367 961 6 × 2 = 0 + 0.011 918 654 290 735 923 2;
  • 31) 0.011 918 654 290 735 923 2 × 2 = 0 + 0.023 837 308 581 471 846 4;
  • 32) 0.023 837 308 581 471 846 4 × 2 = 0 + 0.047 674 617 162 943 692 8;
  • 33) 0.047 674 617 162 943 692 8 × 2 = 0 + 0.095 349 234 325 887 385 6;
  • 34) 0.095 349 234 325 887 385 6 × 2 = 0 + 0.190 698 468 651 774 771 2;
  • 35) 0.190 698 468 651 774 771 2 × 2 = 0 + 0.381 396 937 303 549 542 4;
  • 36) 0.381 396 937 303 549 542 4 × 2 = 0 + 0.762 793 874 607 099 084 8;
  • 37) 0.762 793 874 607 099 084 8 × 2 = 1 + 0.525 587 749 214 198 169 6;
  • 38) 0.525 587 749 214 198 169 6 × 2 = 1 + 0.051 175 498 428 396 339 2;
  • 39) 0.051 175 498 428 396 339 2 × 2 = 0 + 0.102 350 996 856 792 678 4;
  • 40) 0.102 350 996 856 792 678 4 × 2 = 0 + 0.204 701 993 713 585 356 8;
  • 41) 0.204 701 993 713 585 356 8 × 2 = 0 + 0.409 403 987 427 170 713 6;
  • 42) 0.409 403 987 427 170 713 6 × 2 = 0 + 0.818 807 974 854 341 427 2;
  • 43) 0.818 807 974 854 341 427 2 × 2 = 1 + 0.637 615 949 708 682 854 4;
  • 44) 0.637 615 949 708 682 854 4 × 2 = 1 + 0.275 231 899 417 365 708 8;
  • 45) 0.275 231 899 417 365 708 8 × 2 = 0 + 0.550 463 798 834 731 417 6;
  • 46) 0.550 463 798 834 731 417 6 × 2 = 1 + 0.100 927 597 669 462 835 2;
  • 47) 0.100 927 597 669 462 835 2 × 2 = 0 + 0.201 855 195 338 925 670 4;
  • 48) 0.201 855 195 338 925 670 4 × 2 = 0 + 0.403 710 390 677 851 340 8;
  • 49) 0.403 710 390 677 851 340 8 × 2 = 0 + 0.807 420 781 355 702 681 6;
  • 50) 0.807 420 781 355 702 681 6 × 2 = 1 + 0.614 841 562 711 405 363 2;
  • 51) 0.614 841 562 711 405 363 2 × 2 = 1 + 0.229 683 125 422 810 726 4;
  • 52) 0.229 683 125 422 810 726 4 × 2 = 0 + 0.459 366 250 845 621 452 8;
  • 53) 0.459 366 250 845 621 452 8 × 2 = 0 + 0.918 732 501 691 242 905 6;
  • 54) 0.918 732 501 691 242 905 6 × 2 = 1 + 0.837 465 003 382 485 811 2;
  • 55) 0.837 465 003 382 485 811 2 × 2 = 1 + 0.674 930 006 764 971 622 4;
  • 56) 0.674 930 006 764 971 622 4 × 2 = 1 + 0.349 860 013 529 943 244 8;
  • 57) 0.349 860 013 529 943 244 8 × 2 = 0 + 0.699 720 027 059 886 489 6;
  • 58) 0.699 720 027 059 886 489 6 × 2 = 1 + 0.399 440 054 119 772 979 2;
  • 59) 0.399 440 054 119 772 979 2 × 2 = 0 + 0.798 880 108 239 545 958 4;
  • 60) 0.798 880 108 239 545 958 4 × 2 = 1 + 0.597 760 216 479 091 916 8;

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...)


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:


0.000 000 000 011 100 111 8(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0011 0100 0110 0111 0101(2)


5. Positive number before normalization:

0.000 000 000 011 100 111 8(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0011 0100 0110 0111 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 011 100 111 8(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0011 0100 0110 0111 0101(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0011 0100 0110 0111 0101(2) × 20 =


1.1000 0110 1000 1100 1110 101(2) × 2-37


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): -37


Mantissa (not normalized):
1.1000 0110 1000 1100 1110 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


-37 + 2(8-1) - 1 =


(-37 + 127)(10) =


90(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;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 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) =


90(10) =


0101 1010(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. 100 0011 0100 0110 0111 0101 =


100 0011 0100 0110 0111 0101


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 1010


Mantissa (23 bits) =
100 0011 0100 0110 0111 0101


The base ten decimal number 0.000 000 000 011 100 111 8 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0101 1010 - 100 0011 0100 0110 0111 0101

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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