0.000 000 198 616 390 889 583 271 928 131 107 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 198 616 390 889 583 271 928 131 107(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
0.000 000 198 616 390 889 583 271 928 131 107(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: 0.000 000 198 616 390 889 583 271 928 131 107.

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) 0.000 000 198 616 390 889 583 271 928 131 107 × 2 = 0 + 0.000 000 397 232 781 779 166 543 856 262 214;
  • 2) 0.000 000 397 232 781 779 166 543 856 262 214 × 2 = 0 + 0.000 000 794 465 563 558 333 087 712 524 428;
  • 3) 0.000 000 794 465 563 558 333 087 712 524 428 × 2 = 0 + 0.000 001 588 931 127 116 666 175 425 048 856;
  • 4) 0.000 001 588 931 127 116 666 175 425 048 856 × 2 = 0 + 0.000 003 177 862 254 233 332 350 850 097 712;
  • 5) 0.000 003 177 862 254 233 332 350 850 097 712 × 2 = 0 + 0.000 006 355 724 508 466 664 701 700 195 424;
  • 6) 0.000 006 355 724 508 466 664 701 700 195 424 × 2 = 0 + 0.000 012 711 449 016 933 329 403 400 390 848;
  • 7) 0.000 012 711 449 016 933 329 403 400 390 848 × 2 = 0 + 0.000 025 422 898 033 866 658 806 800 781 696;
  • 8) 0.000 025 422 898 033 866 658 806 800 781 696 × 2 = 0 + 0.000 050 845 796 067 733 317 613 601 563 392;
  • 9) 0.000 050 845 796 067 733 317 613 601 563 392 × 2 = 0 + 0.000 101 691 592 135 466 635 227 203 126 784;
  • 10) 0.000 101 691 592 135 466 635 227 203 126 784 × 2 = 0 + 0.000 203 383 184 270 933 270 454 406 253 568;
  • 11) 0.000 203 383 184 270 933 270 454 406 253 568 × 2 = 0 + 0.000 406 766 368 541 866 540 908 812 507 136;
  • 12) 0.000 406 766 368 541 866 540 908 812 507 136 × 2 = 0 + 0.000 813 532 737 083 733 081 817 625 014 272;
  • 13) 0.000 813 532 737 083 733 081 817 625 014 272 × 2 = 0 + 0.001 627 065 474 167 466 163 635 250 028 544;
  • 14) 0.001 627 065 474 167 466 163 635 250 028 544 × 2 = 0 + 0.003 254 130 948 334 932 327 270 500 057 088;
  • 15) 0.003 254 130 948 334 932 327 270 500 057 088 × 2 = 0 + 0.006 508 261 896 669 864 654 541 000 114 176;
  • 16) 0.006 508 261 896 669 864 654 541 000 114 176 × 2 = 0 + 0.013 016 523 793 339 729 309 082 000 228 352;
  • 17) 0.013 016 523 793 339 729 309 082 000 228 352 × 2 = 0 + 0.026 033 047 586 679 458 618 164 000 456 704;
  • 18) 0.026 033 047 586 679 458 618 164 000 456 704 × 2 = 0 + 0.052 066 095 173 358 917 236 328 000 913 408;
  • 19) 0.052 066 095 173 358 917 236 328 000 913 408 × 2 = 0 + 0.104 132 190 346 717 834 472 656 001 826 816;
  • 20) 0.104 132 190 346 717 834 472 656 001 826 816 × 2 = 0 + 0.208 264 380 693 435 668 945 312 003 653 632;
  • 21) 0.208 264 380 693 435 668 945 312 003 653 632 × 2 = 0 + 0.416 528 761 386 871 337 890 624 007 307 264;
  • 22) 0.416 528 761 386 871 337 890 624 007 307 264 × 2 = 0 + 0.833 057 522 773 742 675 781 248 014 614 528;
  • 23) 0.833 057 522 773 742 675 781 248 014 614 528 × 2 = 1 + 0.666 115 045 547 485 351 562 496 029 229 056;
  • 24) 0.666 115 045 547 485 351 562 496 029 229 056 × 2 = 1 + 0.332 230 091 094 970 703 124 992 058 458 112;
  • 25) 0.332 230 091 094 970 703 124 992 058 458 112 × 2 = 0 + 0.664 460 182 189 941 406 249 984 116 916 224;
  • 26) 0.664 460 182 189 941 406 249 984 116 916 224 × 2 = 1 + 0.328 920 364 379 882 812 499 968 233 832 448;
  • 27) 0.328 920 364 379 882 812 499 968 233 832 448 × 2 = 0 + 0.657 840 728 759 765 624 999 936 467 664 896;
  • 28) 0.657 840 728 759 765 624 999 936 467 664 896 × 2 = 1 + 0.315 681 457 519 531 249 999 872 935 329 792;
  • 29) 0.315 681 457 519 531 249 999 872 935 329 792 × 2 = 0 + 0.631 362 915 039 062 499 999 745 870 659 584;
  • 30) 0.631 362 915 039 062 499 999 745 870 659 584 × 2 = 1 + 0.262 725 830 078 124 999 999 491 741 319 168;
  • 31) 0.262 725 830 078 124 999 999 491 741 319 168 × 2 = 0 + 0.525 451 660 156 249 999 998 983 482 638 336;
  • 32) 0.525 451 660 156 249 999 998 983 482 638 336 × 2 = 1 + 0.050 903 320 312 499 999 997 966 965 276 672;
  • 33) 0.050 903 320 312 499 999 997 966 965 276 672 × 2 = 0 + 0.101 806 640 624 999 999 995 933 930 553 344;
  • 34) 0.101 806 640 624 999 999 995 933 930 553 344 × 2 = 0 + 0.203 613 281 249 999 999 991 867 861 106 688;
  • 35) 0.203 613 281 249 999 999 991 867 861 106 688 × 2 = 0 + 0.407 226 562 499 999 999 983 735 722 213 376;
  • 36) 0.407 226 562 499 999 999 983 735 722 213 376 × 2 = 0 + 0.814 453 124 999 999 999 967 471 444 426 752;
  • 37) 0.814 453 124 999 999 999 967 471 444 426 752 × 2 = 1 + 0.628 906 249 999 999 999 934 942 888 853 504;
  • 38) 0.628 906 249 999 999 999 934 942 888 853 504 × 2 = 1 + 0.257 812 499 999 999 999 869 885 777 707 008;
  • 39) 0.257 812 499 999 999 999 869 885 777 707 008 × 2 = 0 + 0.515 624 999 999 999 999 739 771 555 414 016;
  • 40) 0.515 624 999 999 999 999 739 771 555 414 016 × 2 = 1 + 0.031 249 999 999 999 999 479 543 110 828 032;
  • 41) 0.031 249 999 999 999 999 479 543 110 828 032 × 2 = 0 + 0.062 499 999 999 999 998 959 086 221 656 064;
  • 42) 0.062 499 999 999 999 998 959 086 221 656 064 × 2 = 0 + 0.124 999 999 999 999 997 918 172 443 312 128;
  • 43) 0.124 999 999 999 999 997 918 172 443 312 128 × 2 = 0 + 0.249 999 999 999 999 995 836 344 886 624 256;
  • 44) 0.249 999 999 999 999 995 836 344 886 624 256 × 2 = 0 + 0.499 999 999 999 999 991 672 689 773 248 512;
  • 45) 0.499 999 999 999 999 991 672 689 773 248 512 × 2 = 0 + 0.999 999 999 999 999 983 345 379 546 497 024;
  • 46) 0.999 999 999 999 999 983 345 379 546 497 024 × 2 = 1 + 0.999 999 999 999 999 966 690 759 092 994 048;

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:


0.000 000 198 616 390 889 583 271 928 131 107(10) =

0.0000 0000 0000 0000 0000 0011 0101 0101 0000 1101 0000 01(2)

5. Positive number before normalization:

0.000 000 198 616 390 889 583 271 928 131 107(10) =

0.0000 0000 0000 0000 0000 0011 0101 0101 0000 1101 0000 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 198 616 390 889 583 271 928 131 107(10) =

0.0000 0000 0000 0000 0000 0011 0101 0101 0000 1101 0000 01(2) =

0.0000 0000 0000 0000 0000 0011 0101 0101 0000 1101 0000 01(2) × 20 =

1.1010 1010 1000 0110 1000 001(2) × 2-23


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): -23

Mantissa (not normalized):
1.1010 1010 1000 0110 1000 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

104(10) =

0110 1000(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 0101 0100 0011 0100 0001 =

101 0101 0100 0011 0100 0001


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) =
0110 1000

Mantissa (23 bits) =
101 0101 0100 0011 0100 0001


Decimal number 0.000 000 198 616 390 889 583 271 928 131 107 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 101 0101 0100 0011 0100 0001

Share

» Convert decimal 0.000 000 198 616 390 889 583 271 928 131 19 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 04, 2026 [April]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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