-0.000 000 000 000 087 568 840 866 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 087 568 840 866(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
-0.000 000 000 000 087 568 840 866(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:

|-0.000 000 000 000 087 568 840 866| = 0.000 000 000 000 087 568 840 866


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: 0.000 000 000 000 087 568 840 866.

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) 0.000 000 000 000 087 568 840 866 × 2 = 0 + 0.000 000 000 000 175 137 681 732;
  • 2) 0.000 000 000 000 175 137 681 732 × 2 = 0 + 0.000 000 000 000 350 275 363 464;
  • 3) 0.000 000 000 000 350 275 363 464 × 2 = 0 + 0.000 000 000 000 700 550 726 928;
  • 4) 0.000 000 000 000 700 550 726 928 × 2 = 0 + 0.000 000 000 001 401 101 453 856;
  • 5) 0.000 000 000 001 401 101 453 856 × 2 = 0 + 0.000 000 000 002 802 202 907 712;
  • 6) 0.000 000 000 002 802 202 907 712 × 2 = 0 + 0.000 000 000 005 604 405 815 424;
  • 7) 0.000 000 000 005 604 405 815 424 × 2 = 0 + 0.000 000 000 011 208 811 630 848;
  • 8) 0.000 000 000 011 208 811 630 848 × 2 = 0 + 0.000 000 000 022 417 623 261 696;
  • 9) 0.000 000 000 022 417 623 261 696 × 2 = 0 + 0.000 000 000 044 835 246 523 392;
  • 10) 0.000 000 000 044 835 246 523 392 × 2 = 0 + 0.000 000 000 089 670 493 046 784;
  • 11) 0.000 000 000 089 670 493 046 784 × 2 = 0 + 0.000 000 000 179 340 986 093 568;
  • 12) 0.000 000 000 179 340 986 093 568 × 2 = 0 + 0.000 000 000 358 681 972 187 136;
  • 13) 0.000 000 000 358 681 972 187 136 × 2 = 0 + 0.000 000 000 717 363 944 374 272;
  • 14) 0.000 000 000 717 363 944 374 272 × 2 = 0 + 0.000 000 001 434 727 888 748 544;
  • 15) 0.000 000 001 434 727 888 748 544 × 2 = 0 + 0.000 000 002 869 455 777 497 088;
  • 16) 0.000 000 002 869 455 777 497 088 × 2 = 0 + 0.000 000 005 738 911 554 994 176;
  • 17) 0.000 000 005 738 911 554 994 176 × 2 = 0 + 0.000 000 011 477 823 109 988 352;
  • 18) 0.000 000 011 477 823 109 988 352 × 2 = 0 + 0.000 000 022 955 646 219 976 704;
  • 19) 0.000 000 022 955 646 219 976 704 × 2 = 0 + 0.000 000 045 911 292 439 953 408;
  • 20) 0.000 000 045 911 292 439 953 408 × 2 = 0 + 0.000 000 091 822 584 879 906 816;
  • 21) 0.000 000 091 822 584 879 906 816 × 2 = 0 + 0.000 000 183 645 169 759 813 632;
  • 22) 0.000 000 183 645 169 759 813 632 × 2 = 0 + 0.000 000 367 290 339 519 627 264;
  • 23) 0.000 000 367 290 339 519 627 264 × 2 = 0 + 0.000 000 734 580 679 039 254 528;
  • 24) 0.000 000 734 580 679 039 254 528 × 2 = 0 + 0.000 001 469 161 358 078 509 056;
  • 25) 0.000 001 469 161 358 078 509 056 × 2 = 0 + 0.000 002 938 322 716 157 018 112;
  • 26) 0.000 002 938 322 716 157 018 112 × 2 = 0 + 0.000 005 876 645 432 314 036 224;
  • 27) 0.000 005 876 645 432 314 036 224 × 2 = 0 + 0.000 011 753 290 864 628 072 448;
  • 28) 0.000 011 753 290 864 628 072 448 × 2 = 0 + 0.000 023 506 581 729 256 144 896;
  • 29) 0.000 023 506 581 729 256 144 896 × 2 = 0 + 0.000 047 013 163 458 512 289 792;
  • 30) 0.000 047 013 163 458 512 289 792 × 2 = 0 + 0.000 094 026 326 917 024 579 584;
  • 31) 0.000 094 026 326 917 024 579 584 × 2 = 0 + 0.000 188 052 653 834 049 159 168;
  • 32) 0.000 188 052 653 834 049 159 168 × 2 = 0 + 0.000 376 105 307 668 098 318 336;
  • 33) 0.000 376 105 307 668 098 318 336 × 2 = 0 + 0.000 752 210 615 336 196 636 672;
  • 34) 0.000 752 210 615 336 196 636 672 × 2 = 0 + 0.001 504 421 230 672 393 273 344;
  • 35) 0.001 504 421 230 672 393 273 344 × 2 = 0 + 0.003 008 842 461 344 786 546 688;
  • 36) 0.003 008 842 461 344 786 546 688 × 2 = 0 + 0.006 017 684 922 689 573 093 376;
  • 37) 0.006 017 684 922 689 573 093 376 × 2 = 0 + 0.012 035 369 845 379 146 186 752;
  • 38) 0.012 035 369 845 379 146 186 752 × 2 = 0 + 0.024 070 739 690 758 292 373 504;
  • 39) 0.024 070 739 690 758 292 373 504 × 2 = 0 + 0.048 141 479 381 516 584 747 008;
  • 40) 0.048 141 479 381 516 584 747 008 × 2 = 0 + 0.096 282 958 763 033 169 494 016;
  • 41) 0.096 282 958 763 033 169 494 016 × 2 = 0 + 0.192 565 917 526 066 338 988 032;
  • 42) 0.192 565 917 526 066 338 988 032 × 2 = 0 + 0.385 131 835 052 132 677 976 064;
  • 43) 0.385 131 835 052 132 677 976 064 × 2 = 0 + 0.770 263 670 104 265 355 952 128;
  • 44) 0.770 263 670 104 265 355 952 128 × 2 = 1 + 0.540 527 340 208 530 711 904 256;
  • 45) 0.540 527 340 208 530 711 904 256 × 2 = 1 + 0.081 054 680 417 061 423 808 512;
  • 46) 0.081 054 680 417 061 423 808 512 × 2 = 0 + 0.162 109 360 834 122 847 617 024;
  • 47) 0.162 109 360 834 122 847 617 024 × 2 = 0 + 0.324 218 721 668 245 695 234 048;
  • 48) 0.324 218 721 668 245 695 234 048 × 2 = 0 + 0.648 437 443 336 491 390 468 096;
  • 49) 0.648 437 443 336 491 390 468 096 × 2 = 1 + 0.296 874 886 672 982 780 936 192;
  • 50) 0.296 874 886 672 982 780 936 192 × 2 = 0 + 0.593 749 773 345 965 561 872 384;
  • 51) 0.593 749 773 345 965 561 872 384 × 2 = 1 + 0.187 499 546 691 931 123 744 768;
  • 52) 0.187 499 546 691 931 123 744 768 × 2 = 0 + 0.374 999 093 383 862 247 489 536;
  • 53) 0.374 999 093 383 862 247 489 536 × 2 = 0 + 0.749 998 186 767 724 494 979 072;
  • 54) 0.749 998 186 767 724 494 979 072 × 2 = 1 + 0.499 996 373 535 448 989 958 144;
  • 55) 0.499 996 373 535 448 989 958 144 × 2 = 0 + 0.999 992 747 070 897 979 916 288;
  • 56) 0.999 992 747 070 897 979 916 288 × 2 = 1 + 0.999 985 494 141 795 959 832 576;
  • 57) 0.999 985 494 141 795 959 832 576 × 2 = 1 + 0.999 970 988 283 591 919 665 152;
  • 58) 0.999 970 988 283 591 919 665 152 × 2 = 1 + 0.999 941 976 567 183 839 330 304;
  • 59) 0.999 941 976 567 183 839 330 304 × 2 = 1 + 0.999 883 953 134 367 678 660 608;
  • 60) 0.999 883 953 134 367 678 660 608 × 2 = 1 + 0.999 767 906 268 735 357 321 216;
  • 61) 0.999 767 906 268 735 357 321 216 × 2 = 1 + 0.999 535 812 537 470 714 642 432;
  • 62) 0.999 535 812 537 470 714 642 432 × 2 = 1 + 0.999 071 625 074 941 429 284 864;
  • 63) 0.999 071 625 074 941 429 284 864 × 2 = 1 + 0.998 143 250 149 882 858 569 728;
  • 64) 0.998 143 250 149 882 858 569 728 × 2 = 1 + 0.996 286 500 299 765 717 139 456;
  • 65) 0.996 286 500 299 765 717 139 456 × 2 = 1 + 0.992 573 000 599 531 434 278 912;
  • 66) 0.992 573 000 599 531 434 278 912 × 2 = 1 + 0.985 146 001 199 062 868 557 824;
  • 67) 0.985 146 001 199 062 868 557 824 × 2 = 1 + 0.970 292 002 398 125 737 115 648;

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:


0.000 000 000 000 087 568 840 866(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:

0.000 000 000 000 087 568 840 866(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:


0.000 000 000 000 087 568 840 866(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 -0.000 000 000 000 087 568 840 866 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