-0.000 000 005 917 031 260 546 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 917 031 260 546(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 005 917 031 260 546(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 005 917 031 260 546| = 0.000 000 005 917 031 260 546


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: 0.000 000 005 917 031 260 546.

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) 0.000 000 005 917 031 260 546 × 2 = 0 + 0.000 000 011 834 062 521 092;
  • 2) 0.000 000 011 834 062 521 092 × 2 = 0 + 0.000 000 023 668 125 042 184;
  • 3) 0.000 000 023 668 125 042 184 × 2 = 0 + 0.000 000 047 336 250 084 368;
  • 4) 0.000 000 047 336 250 084 368 × 2 = 0 + 0.000 000 094 672 500 168 736;
  • 5) 0.000 000 094 672 500 168 736 × 2 = 0 + 0.000 000 189 345 000 337 472;
  • 6) 0.000 000 189 345 000 337 472 × 2 = 0 + 0.000 000 378 690 000 674 944;
  • 7) 0.000 000 378 690 000 674 944 × 2 = 0 + 0.000 000 757 380 001 349 888;
  • 8) 0.000 000 757 380 001 349 888 × 2 = 0 + 0.000 001 514 760 002 699 776;
  • 9) 0.000 001 514 760 002 699 776 × 2 = 0 + 0.000 003 029 520 005 399 552;
  • 10) 0.000 003 029 520 005 399 552 × 2 = 0 + 0.000 006 059 040 010 799 104;
  • 11) 0.000 006 059 040 010 799 104 × 2 = 0 + 0.000 012 118 080 021 598 208;
  • 12) 0.000 012 118 080 021 598 208 × 2 = 0 + 0.000 024 236 160 043 196 416;
  • 13) 0.000 024 236 160 043 196 416 × 2 = 0 + 0.000 048 472 320 086 392 832;
  • 14) 0.000 048 472 320 086 392 832 × 2 = 0 + 0.000 096 944 640 172 785 664;
  • 15) 0.000 096 944 640 172 785 664 × 2 = 0 + 0.000 193 889 280 345 571 328;
  • 16) 0.000 193 889 280 345 571 328 × 2 = 0 + 0.000 387 778 560 691 142 656;
  • 17) 0.000 387 778 560 691 142 656 × 2 = 0 + 0.000 775 557 121 382 285 312;
  • 18) 0.000 775 557 121 382 285 312 × 2 = 0 + 0.001 551 114 242 764 570 624;
  • 19) 0.001 551 114 242 764 570 624 × 2 = 0 + 0.003 102 228 485 529 141 248;
  • 20) 0.003 102 228 485 529 141 248 × 2 = 0 + 0.006 204 456 971 058 282 496;
  • 21) 0.006 204 456 971 058 282 496 × 2 = 0 + 0.012 408 913 942 116 564 992;
  • 22) 0.012 408 913 942 116 564 992 × 2 = 0 + 0.024 817 827 884 233 129 984;
  • 23) 0.024 817 827 884 233 129 984 × 2 = 0 + 0.049 635 655 768 466 259 968;
  • 24) 0.049 635 655 768 466 259 968 × 2 = 0 + 0.099 271 311 536 932 519 936;
  • 25) 0.099 271 311 536 932 519 936 × 2 = 0 + 0.198 542 623 073 865 039 872;
  • 26) 0.198 542 623 073 865 039 872 × 2 = 0 + 0.397 085 246 147 730 079 744;
  • 27) 0.397 085 246 147 730 079 744 × 2 = 0 + 0.794 170 492 295 460 159 488;
  • 28) 0.794 170 492 295 460 159 488 × 2 = 1 + 0.588 340 984 590 920 318 976;
  • 29) 0.588 340 984 590 920 318 976 × 2 = 1 + 0.176 681 969 181 840 637 952;
  • 30) 0.176 681 969 181 840 637 952 × 2 = 0 + 0.353 363 938 363 681 275 904;
  • 31) 0.353 363 938 363 681 275 904 × 2 = 0 + 0.706 727 876 727 362 551 808;
  • 32) 0.706 727 876 727 362 551 808 × 2 = 1 + 0.413 455 753 454 725 103 616;
  • 33) 0.413 455 753 454 725 103 616 × 2 = 0 + 0.826 911 506 909 450 207 232;
  • 34) 0.826 911 506 909 450 207 232 × 2 = 1 + 0.653 823 013 818 900 414 464;
  • 35) 0.653 823 013 818 900 414 464 × 2 = 1 + 0.307 646 027 637 800 828 928;
  • 36) 0.307 646 027 637 800 828 928 × 2 = 0 + 0.615 292 055 275 601 657 856;
  • 37) 0.615 292 055 275 601 657 856 × 2 = 1 + 0.230 584 110 551 203 315 712;
  • 38) 0.230 584 110 551 203 315 712 × 2 = 0 + 0.461 168 221 102 406 631 424;
  • 39) 0.461 168 221 102 406 631 424 × 2 = 0 + 0.922 336 442 204 813 262 848;
  • 40) 0.922 336 442 204 813 262 848 × 2 = 1 + 0.844 672 884 409 626 525 696;
  • 41) 0.844 672 884 409 626 525 696 × 2 = 1 + 0.689 345 768 819 253 051 392;
  • 42) 0.689 345 768 819 253 051 392 × 2 = 1 + 0.378 691 537 638 506 102 784;
  • 43) 0.378 691 537 638 506 102 784 × 2 = 0 + 0.757 383 075 277 012 205 568;
  • 44) 0.757 383 075 277 012 205 568 × 2 = 1 + 0.514 766 150 554 024 411 136;
  • 45) 0.514 766 150 554 024 411 136 × 2 = 1 + 0.029 532 301 108 048 822 272;
  • 46) 0.029 532 301 108 048 822 272 × 2 = 0 + 0.059 064 602 216 097 644 544;
  • 47) 0.059 064 602 216 097 644 544 × 2 = 0 + 0.118 129 204 432 195 289 088;
  • 48) 0.118 129 204 432 195 289 088 × 2 = 0 + 0.236 258 408 864 390 578 176;
  • 49) 0.236 258 408 864 390 578 176 × 2 = 0 + 0.472 516 817 728 781 156 352;
  • 50) 0.472 516 817 728 781 156 352 × 2 = 0 + 0.945 033 635 457 562 312 704;
  • 51) 0.945 033 635 457 562 312 704 × 2 = 1 + 0.890 067 270 915 124 625 408;
  • 52) 0.890 067 270 915 124 625 408 × 2 = 1 + 0.780 134 541 830 249 250 816;
  • 53) 0.780 134 541 830 249 250 816 × 2 = 1 + 0.560 269 083 660 498 501 632;
  • 54) 0.560 269 083 660 498 501 632 × 2 = 1 + 0.120 538 167 320 997 003 264;
  • 55) 0.120 538 167 320 997 003 264 × 2 = 0 + 0.241 076 334 641 994 006 528;
  • 56) 0.241 076 334 641 994 006 528 × 2 = 0 + 0.482 152 669 283 988 013 056;
  • 57) 0.482 152 669 283 988 013 056 × 2 = 0 + 0.964 305 338 567 976 026 112;
  • 58) 0.964 305 338 567 976 026 112 × 2 = 1 + 0.928 610 677 135 952 052 224;
  • 59) 0.928 610 677 135 952 052 224 × 2 = 1 + 0.857 221 354 271 904 104 448;
  • 60) 0.857 221 354 271 904 104 448 × 2 = 1 + 0.714 442 708 543 808 208 896;
  • 61) 0.714 442 708 543 808 208 896 × 2 = 1 + 0.428 885 417 087 616 417 792;
  • 62) 0.428 885 417 087 616 417 792 × 2 = 0 + 0.857 770 834 175 232 835 584;
  • 63) 0.857 770 834 175 232 835 584 × 2 = 1 + 0.715 541 668 350 465 671 168;
  • 64) 0.715 541 668 350 465 671 168 × 2 = 1 + 0.431 083 336 700 931 342 336;
  • 65) 0.431 083 336 700 931 342 336 × 2 = 0 + 0.862 166 673 401 862 684 672;
  • 66) 0.862 166 673 401 862 684 672 × 2 = 1 + 0.724 333 346 803 725 369 344;
  • 67) 0.724 333 346 803 725 369 344 × 2 = 1 + 0.448 666 693 607 450 738 688;
  • 68) 0.448 666 693 607 450 738 688 × 2 = 0 + 0.897 333 387 214 901 477 376;
  • 69) 0.897 333 387 214 901 477 376 × 2 = 1 + 0.794 666 774 429 802 954 752;
  • 70) 0.794 666 774 429 802 954 752 × 2 = 1 + 0.589 333 548 859 605 909 504;
  • 71) 0.589 333 548 859 605 909 504 × 2 = 1 + 0.178 667 097 719 211 819 008;
  • 72) 0.178 667 097 719 211 819 008 × 2 = 0 + 0.357 334 195 438 423 638 016;
  • 73) 0.357 334 195 438 423 638 016 × 2 = 0 + 0.714 668 390 876 847 276 032;
  • 74) 0.714 668 390 876 847 276 032 × 2 = 1 + 0.429 336 781 753 694 552 064;
  • 75) 0.429 336 781 753 694 552 064 × 2 = 0 + 0.858 673 563 507 389 104 128;
  • 76) 0.858 673 563 507 389 104 128 × 2 = 1 + 0.717 347 127 014 778 208 256;
  • 77) 0.717 347 127 014 778 208 256 × 2 = 1 + 0.434 694 254 029 556 416 512;
  • 78) 0.434 694 254 029 556 416 512 × 2 = 0 + 0.869 388 508 059 112 833 024;
  • 79) 0.869 388 508 059 112 833 024 × 2 = 1 + 0.738 777 016 118 225 666 048;
  • 80) 0.738 777 016 118 225 666 048 × 2 = 1 + 0.477 554 032 236 451 332 096;

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:


0.000 000 005 917 031 260 546(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011(2)

6. Positive number before normalization:

0.000 000 005 917 031 260 546(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 917 031 260 546(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011(2) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011(2) × 20 =

1.1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011(2) × 2-28


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -28

Mantissa (not normalized):
1.1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-28 + 2(11-1) - 1 =

(-28 + 1 023)(10) =

995(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 995 ÷ 2 = 497 + 1;
  • 497 ÷ 2 = 248 + 1;
  • 248 ÷ 2 = 124 + 0;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

995(10) =

011 1110 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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011 =

1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1110 0011

Mantissa (52 bits) =
1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011


Decimal number -0.000 000 005 917 031 260 546 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0011 - 1001 0110 1001 1101 1000 0011 1100 0111 1011 0110 1110 0101 1011

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100