-0.000 000 000 000 087 568 840 778 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 087 568 840 778(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
-0.000 000 000 000 087 568 840 778(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:

|-0.000 000 000 000 087 568 840 778| = 0.000 000 000 000 087 568 840 778


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: 0.000 000 000 000 087 568 840 778.

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) 0.000 000 000 000 087 568 840 778 × 2 = 0 + 0.000 000 000 000 175 137 681 556;
  • 2) 0.000 000 000 000 175 137 681 556 × 2 = 0 + 0.000 000 000 000 350 275 363 112;
  • 3) 0.000 000 000 000 350 275 363 112 × 2 = 0 + 0.000 000 000 000 700 550 726 224;
  • 4) 0.000 000 000 000 700 550 726 224 × 2 = 0 + 0.000 000 000 001 401 101 452 448;
  • 5) 0.000 000 000 001 401 101 452 448 × 2 = 0 + 0.000 000 000 002 802 202 904 896;
  • 6) 0.000 000 000 002 802 202 904 896 × 2 = 0 + 0.000 000 000 005 604 405 809 792;
  • 7) 0.000 000 000 005 604 405 809 792 × 2 = 0 + 0.000 000 000 011 208 811 619 584;
  • 8) 0.000 000 000 011 208 811 619 584 × 2 = 0 + 0.000 000 000 022 417 623 239 168;
  • 9) 0.000 000 000 022 417 623 239 168 × 2 = 0 + 0.000 000 000 044 835 246 478 336;
  • 10) 0.000 000 000 044 835 246 478 336 × 2 = 0 + 0.000 000 000 089 670 492 956 672;
  • 11) 0.000 000 000 089 670 492 956 672 × 2 = 0 + 0.000 000 000 179 340 985 913 344;
  • 12) 0.000 000 000 179 340 985 913 344 × 2 = 0 + 0.000 000 000 358 681 971 826 688;
  • 13) 0.000 000 000 358 681 971 826 688 × 2 = 0 + 0.000 000 000 717 363 943 653 376;
  • 14) 0.000 000 000 717 363 943 653 376 × 2 = 0 + 0.000 000 001 434 727 887 306 752;
  • 15) 0.000 000 001 434 727 887 306 752 × 2 = 0 + 0.000 000 002 869 455 774 613 504;
  • 16) 0.000 000 002 869 455 774 613 504 × 2 = 0 + 0.000 000 005 738 911 549 227 008;
  • 17) 0.000 000 005 738 911 549 227 008 × 2 = 0 + 0.000 000 011 477 823 098 454 016;
  • 18) 0.000 000 011 477 823 098 454 016 × 2 = 0 + 0.000 000 022 955 646 196 908 032;
  • 19) 0.000 000 022 955 646 196 908 032 × 2 = 0 + 0.000 000 045 911 292 393 816 064;
  • 20) 0.000 000 045 911 292 393 816 064 × 2 = 0 + 0.000 000 091 822 584 787 632 128;
  • 21) 0.000 000 091 822 584 787 632 128 × 2 = 0 + 0.000 000 183 645 169 575 264 256;
  • 22) 0.000 000 183 645 169 575 264 256 × 2 = 0 + 0.000 000 367 290 339 150 528 512;
  • 23) 0.000 000 367 290 339 150 528 512 × 2 = 0 + 0.000 000 734 580 678 301 057 024;
  • 24) 0.000 000 734 580 678 301 057 024 × 2 = 0 + 0.000 001 469 161 356 602 114 048;
  • 25) 0.000 001 469 161 356 602 114 048 × 2 = 0 + 0.000 002 938 322 713 204 228 096;
  • 26) 0.000 002 938 322 713 204 228 096 × 2 = 0 + 0.000 005 876 645 426 408 456 192;
  • 27) 0.000 005 876 645 426 408 456 192 × 2 = 0 + 0.000 011 753 290 852 816 912 384;
  • 28) 0.000 011 753 290 852 816 912 384 × 2 = 0 + 0.000 023 506 581 705 633 824 768;
  • 29) 0.000 023 506 581 705 633 824 768 × 2 = 0 + 0.000 047 013 163 411 267 649 536;
  • 30) 0.000 047 013 163 411 267 649 536 × 2 = 0 + 0.000 094 026 326 822 535 299 072;
  • 31) 0.000 094 026 326 822 535 299 072 × 2 = 0 + 0.000 188 052 653 645 070 598 144;
  • 32) 0.000 188 052 653 645 070 598 144 × 2 = 0 + 0.000 376 105 307 290 141 196 288;
  • 33) 0.000 376 105 307 290 141 196 288 × 2 = 0 + 0.000 752 210 614 580 282 392 576;
  • 34) 0.000 752 210 614 580 282 392 576 × 2 = 0 + 0.001 504 421 229 160 564 785 152;
  • 35) 0.001 504 421 229 160 564 785 152 × 2 = 0 + 0.003 008 842 458 321 129 570 304;
  • 36) 0.003 008 842 458 321 129 570 304 × 2 = 0 + 0.006 017 684 916 642 259 140 608;
  • 37) 0.006 017 684 916 642 259 140 608 × 2 = 0 + 0.012 035 369 833 284 518 281 216;
  • 38) 0.012 035 369 833 284 518 281 216 × 2 = 0 + 0.024 070 739 666 569 036 562 432;
  • 39) 0.024 070 739 666 569 036 562 432 × 2 = 0 + 0.048 141 479 333 138 073 124 864;
  • 40) 0.048 141 479 333 138 073 124 864 × 2 = 0 + 0.096 282 958 666 276 146 249 728;
  • 41) 0.096 282 958 666 276 146 249 728 × 2 = 0 + 0.192 565 917 332 552 292 499 456;
  • 42) 0.192 565 917 332 552 292 499 456 × 2 = 0 + 0.385 131 834 665 104 584 998 912;
  • 43) 0.385 131 834 665 104 584 998 912 × 2 = 0 + 0.770 263 669 330 209 169 997 824;
  • 44) 0.770 263 669 330 209 169 997 824 × 2 = 1 + 0.540 527 338 660 418 339 995 648;
  • 45) 0.540 527 338 660 418 339 995 648 × 2 = 1 + 0.081 054 677 320 836 679 991 296;
  • 46) 0.081 054 677 320 836 679 991 296 × 2 = 0 + 0.162 109 354 641 673 359 982 592;
  • 47) 0.162 109 354 641 673 359 982 592 × 2 = 0 + 0.324 218 709 283 346 719 965 184;
  • 48) 0.324 218 709 283 346 719 965 184 × 2 = 0 + 0.648 437 418 566 693 439 930 368;
  • 49) 0.648 437 418 566 693 439 930 368 × 2 = 1 + 0.296 874 837 133 386 879 860 736;
  • 50) 0.296 874 837 133 386 879 860 736 × 2 = 0 + 0.593 749 674 266 773 759 721 472;
  • 51) 0.593 749 674 266 773 759 721 472 × 2 = 1 + 0.187 499 348 533 547 519 442 944;
  • 52) 0.187 499 348 533 547 519 442 944 × 2 = 0 + 0.374 998 697 067 095 038 885 888;
  • 53) 0.374 998 697 067 095 038 885 888 × 2 = 0 + 0.749 997 394 134 190 077 771 776;
  • 54) 0.749 997 394 134 190 077 771 776 × 2 = 1 + 0.499 994 788 268 380 155 543 552;
  • 55) 0.499 994 788 268 380 155 543 552 × 2 = 0 + 0.999 989 576 536 760 311 087 104;
  • 56) 0.999 989 576 536 760 311 087 104 × 2 = 1 + 0.999 979 153 073 520 622 174 208;
  • 57) 0.999 979 153 073 520 622 174 208 × 2 = 1 + 0.999 958 306 147 041 244 348 416;
  • 58) 0.999 958 306 147 041 244 348 416 × 2 = 1 + 0.999 916 612 294 082 488 696 832;
  • 59) 0.999 916 612 294 082 488 696 832 × 2 = 1 + 0.999 833 224 588 164 977 393 664;
  • 60) 0.999 833 224 588 164 977 393 664 × 2 = 1 + 0.999 666 449 176 329 954 787 328;
  • 61) 0.999 666 449 176 329 954 787 328 × 2 = 1 + 0.999 332 898 352 659 909 574 656;
  • 62) 0.999 332 898 352 659 909 574 656 × 2 = 1 + 0.998 665 796 705 319 819 149 312;
  • 63) 0.998 665 796 705 319 819 149 312 × 2 = 1 + 0.997 331 593 410 639 638 298 624;
  • 64) 0.997 331 593 410 639 638 298 624 × 2 = 1 + 0.994 663 186 821 279 276 597 248;
  • 65) 0.994 663 186 821 279 276 597 248 × 2 = 1 + 0.989 326 373 642 558 553 194 496;
  • 66) 0.989 326 373 642 558 553 194 496 × 2 = 1 + 0.978 652 747 285 117 106 388 992;
  • 67) 0.978 652 747 285 117 106 388 992 × 2 = 1 + 0.957 305 494 570 234 212 777 984;

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:


0.000 000 000 000 087 568 840 778(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:

0.000 000 000 000 087 568 840 778(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:


0.000 000 000 000 087 568 840 778(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 -0.000 000 000 000 087 568 840 778 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 0011 - 100 0101 0010 1111 1111 1111

Share

» Convert decimal -0.000 000 000 000 087 568 840 822 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 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. Gemini Notebook YouTube Video of 7.50 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