Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.000 000 005 917 031 260 9 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 000 005 917 031 260 9(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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 9| = 0.000 000 005 917 031 260 9


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 9.

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 9 × 2 = 0 + 0.000 000 011 834 062 521 8;
  • 2) 0.000 000 011 834 062 521 8 × 2 = 0 + 0.000 000 023 668 125 043 6;
  • 3) 0.000 000 023 668 125 043 6 × 2 = 0 + 0.000 000 047 336 250 087 2;
  • 4) 0.000 000 047 336 250 087 2 × 2 = 0 + 0.000 000 094 672 500 174 4;
  • 5) 0.000 000 094 672 500 174 4 × 2 = 0 + 0.000 000 189 345 000 348 8;
  • 6) 0.000 000 189 345 000 348 8 × 2 = 0 + 0.000 000 378 690 000 697 6;
  • 7) 0.000 000 378 690 000 697 6 × 2 = 0 + 0.000 000 757 380 001 395 2;
  • 8) 0.000 000 757 380 001 395 2 × 2 = 0 + 0.000 001 514 760 002 790 4;
  • 9) 0.000 001 514 760 002 790 4 × 2 = 0 + 0.000 003 029 520 005 580 8;
  • 10) 0.000 003 029 520 005 580 8 × 2 = 0 + 0.000 006 059 040 011 161 6;
  • 11) 0.000 006 059 040 011 161 6 × 2 = 0 + 0.000 012 118 080 022 323 2;
  • 12) 0.000 012 118 080 022 323 2 × 2 = 0 + 0.000 024 236 160 044 646 4;
  • 13) 0.000 024 236 160 044 646 4 × 2 = 0 + 0.000 048 472 320 089 292 8;
  • 14) 0.000 048 472 320 089 292 8 × 2 = 0 + 0.000 096 944 640 178 585 6;
  • 15) 0.000 096 944 640 178 585 6 × 2 = 0 + 0.000 193 889 280 357 171 2;
  • 16) 0.000 193 889 280 357 171 2 × 2 = 0 + 0.000 387 778 560 714 342 4;
  • 17) 0.000 387 778 560 714 342 4 × 2 = 0 + 0.000 775 557 121 428 684 8;
  • 18) 0.000 775 557 121 428 684 8 × 2 = 0 + 0.001 551 114 242 857 369 6;
  • 19) 0.001 551 114 242 857 369 6 × 2 = 0 + 0.003 102 228 485 714 739 2;
  • 20) 0.003 102 228 485 714 739 2 × 2 = 0 + 0.006 204 456 971 429 478 4;
  • 21) 0.006 204 456 971 429 478 4 × 2 = 0 + 0.012 408 913 942 858 956 8;
  • 22) 0.012 408 913 942 858 956 8 × 2 = 0 + 0.024 817 827 885 717 913 6;
  • 23) 0.024 817 827 885 717 913 6 × 2 = 0 + 0.049 635 655 771 435 827 2;
  • 24) 0.049 635 655 771 435 827 2 × 2 = 0 + 0.099 271 311 542 871 654 4;
  • 25) 0.099 271 311 542 871 654 4 × 2 = 0 + 0.198 542 623 085 743 308 8;
  • 26) 0.198 542 623 085 743 308 8 × 2 = 0 + 0.397 085 246 171 486 617 6;
  • 27) 0.397 085 246 171 486 617 6 × 2 = 0 + 0.794 170 492 342 973 235 2;
  • 28) 0.794 170 492 342 973 235 2 × 2 = 1 + 0.588 340 984 685 946 470 4;
  • 29) 0.588 340 984 685 946 470 4 × 2 = 1 + 0.176 681 969 371 892 940 8;
  • 30) 0.176 681 969 371 892 940 8 × 2 = 0 + 0.353 363 938 743 785 881 6;
  • 31) 0.353 363 938 743 785 881 6 × 2 = 0 + 0.706 727 877 487 571 763 2;
  • 32) 0.706 727 877 487 571 763 2 × 2 = 1 + 0.413 455 754 975 143 526 4;
  • 33) 0.413 455 754 975 143 526 4 × 2 = 0 + 0.826 911 509 950 287 052 8;
  • 34) 0.826 911 509 950 287 052 8 × 2 = 1 + 0.653 823 019 900 574 105 6;
  • 35) 0.653 823 019 900 574 105 6 × 2 = 1 + 0.307 646 039 801 148 211 2;
  • 36) 0.307 646 039 801 148 211 2 × 2 = 0 + 0.615 292 079 602 296 422 4;
  • 37) 0.615 292 079 602 296 422 4 × 2 = 1 + 0.230 584 159 204 592 844 8;
  • 38) 0.230 584 159 204 592 844 8 × 2 = 0 + 0.461 168 318 409 185 689 6;
  • 39) 0.461 168 318 409 185 689 6 × 2 = 0 + 0.922 336 636 818 371 379 2;
  • 40) 0.922 336 636 818 371 379 2 × 2 = 1 + 0.844 673 273 636 742 758 4;
  • 41) 0.844 673 273 636 742 758 4 × 2 = 1 + 0.689 346 547 273 485 516 8;
  • 42) 0.689 346 547 273 485 516 8 × 2 = 1 + 0.378 693 094 546 971 033 6;
  • 43) 0.378 693 094 546 971 033 6 × 2 = 0 + 0.757 386 189 093 942 067 2;
  • 44) 0.757 386 189 093 942 067 2 × 2 = 1 + 0.514 772 378 187 884 134 4;
  • 45) 0.514 772 378 187 884 134 4 × 2 = 1 + 0.029 544 756 375 768 268 8;
  • 46) 0.029 544 756 375 768 268 8 × 2 = 0 + 0.059 089 512 751 536 537 6;
  • 47) 0.059 089 512 751 536 537 6 × 2 = 0 + 0.118 179 025 503 073 075 2;
  • 48) 0.118 179 025 503 073 075 2 × 2 = 0 + 0.236 358 051 006 146 150 4;
  • 49) 0.236 358 051 006 146 150 4 × 2 = 0 + 0.472 716 102 012 292 300 8;
  • 50) 0.472 716 102 012 292 300 8 × 2 = 0 + 0.945 432 204 024 584 601 6;
  • 51) 0.945 432 204 024 584 601 6 × 2 = 1 + 0.890 864 408 049 169 203 2;
  • 52) 0.890 864 408 049 169 203 2 × 2 = 1 + 0.781 728 816 098 338 406 4;
  • 53) 0.781 728 816 098 338 406 4 × 2 = 1 + 0.563 457 632 196 676 812 8;
  • 54) 0.563 457 632 196 676 812 8 × 2 = 1 + 0.126 915 264 393 353 625 6;
  • 55) 0.126 915 264 393 353 625 6 × 2 = 0 + 0.253 830 528 786 707 251 2;
  • 56) 0.253 830 528 786 707 251 2 × 2 = 0 + 0.507 661 057 573 414 502 4;
  • 57) 0.507 661 057 573 414 502 4 × 2 = 1 + 0.015 322 115 146 829 004 8;
  • 58) 0.015 322 115 146 829 004 8 × 2 = 0 + 0.030 644 230 293 658 009 6;
  • 59) 0.030 644 230 293 658 009 6 × 2 = 0 + 0.061 288 460 587 316 019 2;
  • 60) 0.061 288 460 587 316 019 2 × 2 = 0 + 0.122 576 921 174 632 038 4;
  • 61) 0.122 576 921 174 632 038 4 × 2 = 0 + 0.245 153 842 349 264 076 8;
  • 62) 0.245 153 842 349 264 076 8 × 2 = 0 + 0.490 307 684 698 528 153 6;
  • 63) 0.490 307 684 698 528 153 6 × 2 = 0 + 0.980 615 369 397 056 307 2;
  • 64) 0.980 615 369 397 056 307 2 × 2 = 1 + 0.961 230 738 794 112 614 4;
  • 65) 0.961 230 738 794 112 614 4 × 2 = 1 + 0.922 461 477 588 225 228 8;
  • 66) 0.922 461 477 588 225 228 8 × 2 = 1 + 0.844 922 955 176 450 457 6;
  • 67) 0.844 922 955 176 450 457 6 × 2 = 1 + 0.689 845 910 352 900 915 2;
  • 68) 0.689 845 910 352 900 915 2 × 2 = 1 + 0.379 691 820 705 801 830 4;
  • 69) 0.379 691 820 705 801 830 4 × 2 = 0 + 0.759 383 641 411 603 660 8;
  • 70) 0.759 383 641 411 603 660 8 × 2 = 1 + 0.518 767 282 823 207 321 6;
  • 71) 0.518 767 282 823 207 321 6 × 2 = 1 + 0.037 534 565 646 414 643 2;
  • 72) 0.037 534 565 646 414 643 2 × 2 = 0 + 0.075 069 131 292 829 286 4;
  • 73) 0.075 069 131 292 829 286 4 × 2 = 0 + 0.150 138 262 585 658 572 8;
  • 74) 0.150 138 262 585 658 572 8 × 2 = 0 + 0.300 276 525 171 317 145 6;
  • 75) 0.300 276 525 171 317 145 6 × 2 = 0 + 0.600 553 050 342 634 291 2;
  • 76) 0.600 553 050 342 634 291 2 × 2 = 1 + 0.201 106 100 685 268 582 4;
  • 77) 0.201 106 100 685 268 582 4 × 2 = 0 + 0.402 212 201 370 537 164 8;
  • 78) 0.402 212 201 370 537 164 8 × 2 = 0 + 0.804 424 402 741 074 329 6;
  • 79) 0.804 424 402 741 074 329 6 × 2 = 1 + 0.608 848 805 482 148 659 2;
  • 80) 0.608 848 805 482 148 659 2 × 2 = 1 + 0.217 697 610 964 297 318 4;

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 9(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011(2)

6. Positive number before normalization:

0.000 000 005 917 031 260 9(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011(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 9(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011(2) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011(2) × 20 =

1.1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011(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 1000 0001 1111 0110 0001 0011


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 1000 0001 1111 0110 0001 0011 =

1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011


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 1000 0001 1111 0110 0001 0011


The base ten decimal number -0.000 000 005 917 031 260 9 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0011 - 1001 0110 1001 1101 1000 0011 1100 1000 0001 1111 0110 0001 0011

» Decimal number in base ten -0.000 000 005 917 031 256 6 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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