-0.000 000 000 000 087 568 840 819 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 087 568 840 819(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
-0.000 000 000 000 087 568 840 819(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. Start with the positive version of the number:

|-0.000 000 000 000 087 568 840 819| = 0.000 000 000 000 087 568 840 819


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: 0.000 000 000 000 087 568 840 819.

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) 0.000 000 000 000 087 568 840 819 × 2 = 0 + 0.000 000 000 000 175 137 681 638;
  • 2) 0.000 000 000 000 175 137 681 638 × 2 = 0 + 0.000 000 000 000 350 275 363 276;
  • 3) 0.000 000 000 000 350 275 363 276 × 2 = 0 + 0.000 000 000 000 700 550 726 552;
  • 4) 0.000 000 000 000 700 550 726 552 × 2 = 0 + 0.000 000 000 001 401 101 453 104;
  • 5) 0.000 000 000 001 401 101 453 104 × 2 = 0 + 0.000 000 000 002 802 202 906 208;
  • 6) 0.000 000 000 002 802 202 906 208 × 2 = 0 + 0.000 000 000 005 604 405 812 416;
  • 7) 0.000 000 000 005 604 405 812 416 × 2 = 0 + 0.000 000 000 011 208 811 624 832;
  • 8) 0.000 000 000 011 208 811 624 832 × 2 = 0 + 0.000 000 000 022 417 623 249 664;
  • 9) 0.000 000 000 022 417 623 249 664 × 2 = 0 + 0.000 000 000 044 835 246 499 328;
  • 10) 0.000 000 000 044 835 246 499 328 × 2 = 0 + 0.000 000 000 089 670 492 998 656;
  • 11) 0.000 000 000 089 670 492 998 656 × 2 = 0 + 0.000 000 000 179 340 985 997 312;
  • 12) 0.000 000 000 179 340 985 997 312 × 2 = 0 + 0.000 000 000 358 681 971 994 624;
  • 13) 0.000 000 000 358 681 971 994 624 × 2 = 0 + 0.000 000 000 717 363 943 989 248;
  • 14) 0.000 000 000 717 363 943 989 248 × 2 = 0 + 0.000 000 001 434 727 887 978 496;
  • 15) 0.000 000 001 434 727 887 978 496 × 2 = 0 + 0.000 000 002 869 455 775 956 992;
  • 16) 0.000 000 002 869 455 775 956 992 × 2 = 0 + 0.000 000 005 738 911 551 913 984;
  • 17) 0.000 000 005 738 911 551 913 984 × 2 = 0 + 0.000 000 011 477 823 103 827 968;
  • 18) 0.000 000 011 477 823 103 827 968 × 2 = 0 + 0.000 000 022 955 646 207 655 936;
  • 19) 0.000 000 022 955 646 207 655 936 × 2 = 0 + 0.000 000 045 911 292 415 311 872;
  • 20) 0.000 000 045 911 292 415 311 872 × 2 = 0 + 0.000 000 091 822 584 830 623 744;
  • 21) 0.000 000 091 822 584 830 623 744 × 2 = 0 + 0.000 000 183 645 169 661 247 488;
  • 22) 0.000 000 183 645 169 661 247 488 × 2 = 0 + 0.000 000 367 290 339 322 494 976;
  • 23) 0.000 000 367 290 339 322 494 976 × 2 = 0 + 0.000 000 734 580 678 644 989 952;
  • 24) 0.000 000 734 580 678 644 989 952 × 2 = 0 + 0.000 001 469 161 357 289 979 904;
  • 25) 0.000 001 469 161 357 289 979 904 × 2 = 0 + 0.000 002 938 322 714 579 959 808;
  • 26) 0.000 002 938 322 714 579 959 808 × 2 = 0 + 0.000 005 876 645 429 159 919 616;
  • 27) 0.000 005 876 645 429 159 919 616 × 2 = 0 + 0.000 011 753 290 858 319 839 232;
  • 28) 0.000 011 753 290 858 319 839 232 × 2 = 0 + 0.000 023 506 581 716 639 678 464;
  • 29) 0.000 023 506 581 716 639 678 464 × 2 = 0 + 0.000 047 013 163 433 279 356 928;
  • 30) 0.000 047 013 163 433 279 356 928 × 2 = 0 + 0.000 094 026 326 866 558 713 856;
  • 31) 0.000 094 026 326 866 558 713 856 × 2 = 0 + 0.000 188 052 653 733 117 427 712;
  • 32) 0.000 188 052 653 733 117 427 712 × 2 = 0 + 0.000 376 105 307 466 234 855 424;
  • 33) 0.000 376 105 307 466 234 855 424 × 2 = 0 + 0.000 752 210 614 932 469 710 848;
  • 34) 0.000 752 210 614 932 469 710 848 × 2 = 0 + 0.001 504 421 229 864 939 421 696;
  • 35) 0.001 504 421 229 864 939 421 696 × 2 = 0 + 0.003 008 842 459 729 878 843 392;
  • 36) 0.003 008 842 459 729 878 843 392 × 2 = 0 + 0.006 017 684 919 459 757 686 784;
  • 37) 0.006 017 684 919 459 757 686 784 × 2 = 0 + 0.012 035 369 838 919 515 373 568;
  • 38) 0.012 035 369 838 919 515 373 568 × 2 = 0 + 0.024 070 739 677 839 030 747 136;
  • 39) 0.024 070 739 677 839 030 747 136 × 2 = 0 + 0.048 141 479 355 678 061 494 272;
  • 40) 0.048 141 479 355 678 061 494 272 × 2 = 0 + 0.096 282 958 711 356 122 988 544;
  • 41) 0.096 282 958 711 356 122 988 544 × 2 = 0 + 0.192 565 917 422 712 245 977 088;
  • 42) 0.192 565 917 422 712 245 977 088 × 2 = 0 + 0.385 131 834 845 424 491 954 176;
  • 43) 0.385 131 834 845 424 491 954 176 × 2 = 0 + 0.770 263 669 690 848 983 908 352;
  • 44) 0.770 263 669 690 848 983 908 352 × 2 = 1 + 0.540 527 339 381 697 967 816 704;
  • 45) 0.540 527 339 381 697 967 816 704 × 2 = 1 + 0.081 054 678 763 395 935 633 408;
  • 46) 0.081 054 678 763 395 935 633 408 × 2 = 0 + 0.162 109 357 526 791 871 266 816;
  • 47) 0.162 109 357 526 791 871 266 816 × 2 = 0 + 0.324 218 715 053 583 742 533 632;
  • 48) 0.324 218 715 053 583 742 533 632 × 2 = 0 + 0.648 437 430 107 167 485 067 264;
  • 49) 0.648 437 430 107 167 485 067 264 × 2 = 1 + 0.296 874 860 214 334 970 134 528;
  • 50) 0.296 874 860 214 334 970 134 528 × 2 = 0 + 0.593 749 720 428 669 940 269 056;
  • 51) 0.593 749 720 428 669 940 269 056 × 2 = 1 + 0.187 499 440 857 339 880 538 112;
  • 52) 0.187 499 440 857 339 880 538 112 × 2 = 0 + 0.374 998 881 714 679 761 076 224;
  • 53) 0.374 998 881 714 679 761 076 224 × 2 = 0 + 0.749 997 763 429 359 522 152 448;
  • 54) 0.749 997 763 429 359 522 152 448 × 2 = 1 + 0.499 995 526 858 719 044 304 896;
  • 55) 0.499 995 526 858 719 044 304 896 × 2 = 0 + 0.999 991 053 717 438 088 609 792;
  • 56) 0.999 991 053 717 438 088 609 792 × 2 = 1 + 0.999 982 107 434 876 177 219 584;
  • 57) 0.999 982 107 434 876 177 219 584 × 2 = 1 + 0.999 964 214 869 752 354 439 168;
  • 58) 0.999 964 214 869 752 354 439 168 × 2 = 1 + 0.999 928 429 739 504 708 878 336;
  • 59) 0.999 928 429 739 504 708 878 336 × 2 = 1 + 0.999 856 859 479 009 417 756 672;
  • 60) 0.999 856 859 479 009 417 756 672 × 2 = 1 + 0.999 713 718 958 018 835 513 344;
  • 61) 0.999 713 718 958 018 835 513 344 × 2 = 1 + 0.999 427 437 916 037 671 026 688;
  • 62) 0.999 427 437 916 037 671 026 688 × 2 = 1 + 0.998 854 875 832 075 342 053 376;
  • 63) 0.998 854 875 832 075 342 053 376 × 2 = 1 + 0.997 709 751 664 150 684 106 752;
  • 64) 0.997 709 751 664 150 684 106 752 × 2 = 1 + 0.995 419 503 328 301 368 213 504;
  • 65) 0.995 419 503 328 301 368 213 504 × 2 = 1 + 0.990 839 006 656 602 736 427 008;
  • 66) 0.990 839 006 656 602 736 427 008 × 2 = 1 + 0.981 678 013 313 205 472 854 016;
  • 67) 0.981 678 013 313 205 472 854 016 × 2 = 1 + 0.963 356 026 626 410 945 708 032;

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).


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:


0.000 000 000 000 087 568 840 819(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0101 1111 1111 111(2)

6. Positive number before normalization:

0.000 000 000 000 087 568 840 819(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0101 1111 1111 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 44 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 087 568 840 819(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0101 1111 1111 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0101 1111 1111 111(2) × 20 =

1.1000 1010 0101 1111 1111 111(2) × 2-44


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): -44

Mantissa (not normalized):
1.1000 1010 0101 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-44 + 2(8-1) - 1 =

(-44 + 127)(10) =

83(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;
  • 83 ÷ 2 = 41 + 1;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

83(10) =

0101 0011(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 0101 0010 1111 1111 1111 =

100 0101 0010 1111 1111 1111


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) =
0101 0011

Mantissa (23 bits) =
100 0101 0010 1111 1111 1111


Decimal number -0.000 000 000 000 087 568 840 819 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 0011 - 100 0101 0010 1111 1111 1111

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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