0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99(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
0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99(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: 0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99.

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) 0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99 × 2 = 0 + 0.239 191 465 468 837 731 611 454 239 191 465 468 837 731 611 454 239 187 98;
  • 2) 0.239 191 465 468 837 731 611 454 239 191 465 468 837 731 611 454 239 187 98 × 2 = 0 + 0.478 382 930 937 675 463 222 908 478 382 930 937 675 463 222 908 478 375 96;
  • 3) 0.478 382 930 937 675 463 222 908 478 382 930 937 675 463 222 908 478 375 96 × 2 = 0 + 0.956 765 861 875 350 926 445 816 956 765 861 875 350 926 445 816 956 751 92;
  • 4) 0.956 765 861 875 350 926 445 816 956 765 861 875 350 926 445 816 956 751 92 × 2 = 1 + 0.913 531 723 750 701 852 891 633 913 531 723 750 701 852 891 633 913 503 84;
  • 5) 0.913 531 723 750 701 852 891 633 913 531 723 750 701 852 891 633 913 503 84 × 2 = 1 + 0.827 063 447 501 403 705 783 267 827 063 447 501 403 705 783 267 827 007 68;
  • 6) 0.827 063 447 501 403 705 783 267 827 063 447 501 403 705 783 267 827 007 68 × 2 = 1 + 0.654 126 895 002 807 411 566 535 654 126 895 002 807 411 566 535 654 015 36;
  • 7) 0.654 126 895 002 807 411 566 535 654 126 895 002 807 411 566 535 654 015 36 × 2 = 1 + 0.308 253 790 005 614 823 133 071 308 253 790 005 614 823 133 071 308 030 72;
  • 8) 0.308 253 790 005 614 823 133 071 308 253 790 005 614 823 133 071 308 030 72 × 2 = 0 + 0.616 507 580 011 229 646 266 142 616 507 580 011 229 646 266 142 616 061 44;
  • 9) 0.616 507 580 011 229 646 266 142 616 507 580 011 229 646 266 142 616 061 44 × 2 = 1 + 0.233 015 160 022 459 292 532 285 233 015 160 022 459 292 532 285 232 122 88;
  • 10) 0.233 015 160 022 459 292 532 285 233 015 160 022 459 292 532 285 232 122 88 × 2 = 0 + 0.466 030 320 044 918 585 064 570 466 030 320 044 918 585 064 570 464 245 76;
  • 11) 0.466 030 320 044 918 585 064 570 466 030 320 044 918 585 064 570 464 245 76 × 2 = 0 + 0.932 060 640 089 837 170 129 140 932 060 640 089 837 170 129 140 928 491 52;
  • 12) 0.932 060 640 089 837 170 129 140 932 060 640 089 837 170 129 140 928 491 52 × 2 = 1 + 0.864 121 280 179 674 340 258 281 864 121 280 179 674 340 258 281 856 983 04;
  • 13) 0.864 121 280 179 674 340 258 281 864 121 280 179 674 340 258 281 856 983 04 × 2 = 1 + 0.728 242 560 359 348 680 516 563 728 242 560 359 348 680 516 563 713 966 08;
  • 14) 0.728 242 560 359 348 680 516 563 728 242 560 359 348 680 516 563 713 966 08 × 2 = 1 + 0.456 485 120 718 697 361 033 127 456 485 120 718 697 361 033 127 427 932 16;
  • 15) 0.456 485 120 718 697 361 033 127 456 485 120 718 697 361 033 127 427 932 16 × 2 = 0 + 0.912 970 241 437 394 722 066 254 912 970 241 437 394 722 066 254 855 864 32;
  • 16) 0.912 970 241 437 394 722 066 254 912 970 241 437 394 722 066 254 855 864 32 × 2 = 1 + 0.825 940 482 874 789 444 132 509 825 940 482 874 789 444 132 509 711 728 64;
  • 17) 0.825 940 482 874 789 444 132 509 825 940 482 874 789 444 132 509 711 728 64 × 2 = 1 + 0.651 880 965 749 578 888 265 019 651 880 965 749 578 888 265 019 423 457 28;
  • 18) 0.651 880 965 749 578 888 265 019 651 880 965 749 578 888 265 019 423 457 28 × 2 = 1 + 0.303 761 931 499 157 776 530 039 303 761 931 499 157 776 530 038 846 914 56;
  • 19) 0.303 761 931 499 157 776 530 039 303 761 931 499 157 776 530 038 846 914 56 × 2 = 0 + 0.607 523 862 998 315 553 060 078 607 523 862 998 315 553 060 077 693 829 12;
  • 20) 0.607 523 862 998 315 553 060 078 607 523 862 998 315 553 060 077 693 829 12 × 2 = 1 + 0.215 047 725 996 631 106 120 157 215 047 725 996 631 106 120 155 387 658 24;
  • 21) 0.215 047 725 996 631 106 120 157 215 047 725 996 631 106 120 155 387 658 24 × 2 = 0 + 0.430 095 451 993 262 212 240 314 430 095 451 993 262 212 240 310 775 316 48;
  • 22) 0.430 095 451 993 262 212 240 314 430 095 451 993 262 212 240 310 775 316 48 × 2 = 0 + 0.860 190 903 986 524 424 480 628 860 190 903 986 524 424 480 621 550 632 96;
  • 23) 0.860 190 903 986 524 424 480 628 860 190 903 986 524 424 480 621 550 632 96 × 2 = 1 + 0.720 381 807 973 048 848 961 257 720 381 807 973 048 848 961 243 101 265 92;
  • 24) 0.720 381 807 973 048 848 961 257 720 381 807 973 048 848 961 243 101 265 92 × 2 = 1 + 0.440 763 615 946 097 697 922 515 440 763 615 946 097 697 922 486 202 531 84;
  • 25) 0.440 763 615 946 097 697 922 515 440 763 615 946 097 697 922 486 202 531 84 × 2 = 0 + 0.881 527 231 892 195 395 845 030 881 527 231 892 195 395 844 972 405 063 68;
  • 26) 0.881 527 231 892 195 395 845 030 881 527 231 892 195 395 844 972 405 063 68 × 2 = 1 + 0.763 054 463 784 390 791 690 061 763 054 463 784 390 791 689 944 810 127 36;
  • 27) 0.763 054 463 784 390 791 690 061 763 054 463 784 390 791 689 944 810 127 36 × 2 = 1 + 0.526 108 927 568 781 583 380 123 526 108 927 568 781 583 379 889 620 254 72;

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:


0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99(10) =

0.0001 1110 1001 1101 1101 0011 011(2)

5. Positive number before normalization:

0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99(10) =

0.0001 1110 1001 1101 1101 0011 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99(10) =

0.0001 1110 1001 1101 1101 0011 011(2) =

0.0001 1110 1001 1101 1101 0011 011(2) × 20 =

1.1110 1001 1101 1101 0011 011(2) × 2-4


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): -4

Mantissa (not normalized):
1.1110 1001 1101 1101 0011 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(8-1) - 1 =

(-4 + 127)(10) =

123(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;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

123(10) =

0111 1011(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. 111 0100 1110 1110 1001 1011 =

111 0100 1110 1110 1001 1011


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) =
0111 1011

Mantissa (23 bits) =
111 0100 1110 1110 1001 1011


Decimal number 0.119 595 732 734 418 865 805 727 119 595 732 734 418 865 805 727 119 593 99 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 111 0100 1110 1110 1001 1011

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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