-0.000 000 005 917 031 260 61 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 917 031 260 61(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 61(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 61| = 0.000 000 005 917 031 260 61


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 61.

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 61 × 2 = 0 + 0.000 000 011 834 062 521 22;
  • 2) 0.000 000 011 834 062 521 22 × 2 = 0 + 0.000 000 023 668 125 042 44;
  • 3) 0.000 000 023 668 125 042 44 × 2 = 0 + 0.000 000 047 336 250 084 88;
  • 4) 0.000 000 047 336 250 084 88 × 2 = 0 + 0.000 000 094 672 500 169 76;
  • 5) 0.000 000 094 672 500 169 76 × 2 = 0 + 0.000 000 189 345 000 339 52;
  • 6) 0.000 000 189 345 000 339 52 × 2 = 0 + 0.000 000 378 690 000 679 04;
  • 7) 0.000 000 378 690 000 679 04 × 2 = 0 + 0.000 000 757 380 001 358 08;
  • 8) 0.000 000 757 380 001 358 08 × 2 = 0 + 0.000 001 514 760 002 716 16;
  • 9) 0.000 001 514 760 002 716 16 × 2 = 0 + 0.000 003 029 520 005 432 32;
  • 10) 0.000 003 029 520 005 432 32 × 2 = 0 + 0.000 006 059 040 010 864 64;
  • 11) 0.000 006 059 040 010 864 64 × 2 = 0 + 0.000 012 118 080 021 729 28;
  • 12) 0.000 012 118 080 021 729 28 × 2 = 0 + 0.000 024 236 160 043 458 56;
  • 13) 0.000 024 236 160 043 458 56 × 2 = 0 + 0.000 048 472 320 086 917 12;
  • 14) 0.000 048 472 320 086 917 12 × 2 = 0 + 0.000 096 944 640 173 834 24;
  • 15) 0.000 096 944 640 173 834 24 × 2 = 0 + 0.000 193 889 280 347 668 48;
  • 16) 0.000 193 889 280 347 668 48 × 2 = 0 + 0.000 387 778 560 695 336 96;
  • 17) 0.000 387 778 560 695 336 96 × 2 = 0 + 0.000 775 557 121 390 673 92;
  • 18) 0.000 775 557 121 390 673 92 × 2 = 0 + 0.001 551 114 242 781 347 84;
  • 19) 0.001 551 114 242 781 347 84 × 2 = 0 + 0.003 102 228 485 562 695 68;
  • 20) 0.003 102 228 485 562 695 68 × 2 = 0 + 0.006 204 456 971 125 391 36;
  • 21) 0.006 204 456 971 125 391 36 × 2 = 0 + 0.012 408 913 942 250 782 72;
  • 22) 0.012 408 913 942 250 782 72 × 2 = 0 + 0.024 817 827 884 501 565 44;
  • 23) 0.024 817 827 884 501 565 44 × 2 = 0 + 0.049 635 655 769 003 130 88;
  • 24) 0.049 635 655 769 003 130 88 × 2 = 0 + 0.099 271 311 538 006 261 76;
  • 25) 0.099 271 311 538 006 261 76 × 2 = 0 + 0.198 542 623 076 012 523 52;
  • 26) 0.198 542 623 076 012 523 52 × 2 = 0 + 0.397 085 246 152 025 047 04;
  • 27) 0.397 085 246 152 025 047 04 × 2 = 0 + 0.794 170 492 304 050 094 08;
  • 28) 0.794 170 492 304 050 094 08 × 2 = 1 + 0.588 340 984 608 100 188 16;
  • 29) 0.588 340 984 608 100 188 16 × 2 = 1 + 0.176 681 969 216 200 376 32;
  • 30) 0.176 681 969 216 200 376 32 × 2 = 0 + 0.353 363 938 432 400 752 64;
  • 31) 0.353 363 938 432 400 752 64 × 2 = 0 + 0.706 727 876 864 801 505 28;
  • 32) 0.706 727 876 864 801 505 28 × 2 = 1 + 0.413 455 753 729 603 010 56;
  • 33) 0.413 455 753 729 603 010 56 × 2 = 0 + 0.826 911 507 459 206 021 12;
  • 34) 0.826 911 507 459 206 021 12 × 2 = 1 + 0.653 823 014 918 412 042 24;
  • 35) 0.653 823 014 918 412 042 24 × 2 = 1 + 0.307 646 029 836 824 084 48;
  • 36) 0.307 646 029 836 824 084 48 × 2 = 0 + 0.615 292 059 673 648 168 96;
  • 37) 0.615 292 059 673 648 168 96 × 2 = 1 + 0.230 584 119 347 296 337 92;
  • 38) 0.230 584 119 347 296 337 92 × 2 = 0 + 0.461 168 238 694 592 675 84;
  • 39) 0.461 168 238 694 592 675 84 × 2 = 0 + 0.922 336 477 389 185 351 68;
  • 40) 0.922 336 477 389 185 351 68 × 2 = 1 + 0.844 672 954 778 370 703 36;
  • 41) 0.844 672 954 778 370 703 36 × 2 = 1 + 0.689 345 909 556 741 406 72;
  • 42) 0.689 345 909 556 741 406 72 × 2 = 1 + 0.378 691 819 113 482 813 44;
  • 43) 0.378 691 819 113 482 813 44 × 2 = 0 + 0.757 383 638 226 965 626 88;
  • 44) 0.757 383 638 226 965 626 88 × 2 = 1 + 0.514 767 276 453 931 253 76;
  • 45) 0.514 767 276 453 931 253 76 × 2 = 1 + 0.029 534 552 907 862 507 52;
  • 46) 0.029 534 552 907 862 507 52 × 2 = 0 + 0.059 069 105 815 725 015 04;
  • 47) 0.059 069 105 815 725 015 04 × 2 = 0 + 0.118 138 211 631 450 030 08;
  • 48) 0.118 138 211 631 450 030 08 × 2 = 0 + 0.236 276 423 262 900 060 16;
  • 49) 0.236 276 423 262 900 060 16 × 2 = 0 + 0.472 552 846 525 800 120 32;
  • 50) 0.472 552 846 525 800 120 32 × 2 = 0 + 0.945 105 693 051 600 240 64;
  • 51) 0.945 105 693 051 600 240 64 × 2 = 1 + 0.890 211 386 103 200 481 28;
  • 52) 0.890 211 386 103 200 481 28 × 2 = 1 + 0.780 422 772 206 400 962 56;
  • 53) 0.780 422 772 206 400 962 56 × 2 = 1 + 0.560 845 544 412 801 925 12;
  • 54) 0.560 845 544 412 801 925 12 × 2 = 1 + 0.121 691 088 825 603 850 24;
  • 55) 0.121 691 088 825 603 850 24 × 2 = 0 + 0.243 382 177 651 207 700 48;
  • 56) 0.243 382 177 651 207 700 48 × 2 = 0 + 0.486 764 355 302 415 400 96;
  • 57) 0.486 764 355 302 415 400 96 × 2 = 0 + 0.973 528 710 604 830 801 92;
  • 58) 0.973 528 710 604 830 801 92 × 2 = 1 + 0.947 057 421 209 661 603 84;
  • 59) 0.947 057 421 209 661 603 84 × 2 = 1 + 0.894 114 842 419 323 207 68;
  • 60) 0.894 114 842 419 323 207 68 × 2 = 1 + 0.788 229 684 838 646 415 36;
  • 61) 0.788 229 684 838 646 415 36 × 2 = 1 + 0.576 459 369 677 292 830 72;
  • 62) 0.576 459 369 677 292 830 72 × 2 = 1 + 0.152 918 739 354 585 661 44;
  • 63) 0.152 918 739 354 585 661 44 × 2 = 0 + 0.305 837 478 709 171 322 88;
  • 64) 0.305 837 478 709 171 322 88 × 2 = 0 + 0.611 674 957 418 342 645 76;
  • 65) 0.611 674 957 418 342 645 76 × 2 = 1 + 0.223 349 914 836 685 291 52;
  • 66) 0.223 349 914 836 685 291 52 × 2 = 0 + 0.446 699 829 673 370 583 04;
  • 67) 0.446 699 829 673 370 583 04 × 2 = 0 + 0.893 399 659 346 741 166 08;
  • 68) 0.893 399 659 346 741 166 08 × 2 = 1 + 0.786 799 318 693 482 332 16;
  • 69) 0.786 799 318 693 482 332 16 × 2 = 1 + 0.573 598 637 386 964 664 32;
  • 70) 0.573 598 637 386 964 664 32 × 2 = 1 + 0.147 197 274 773 929 328 64;
  • 71) 0.147 197 274 773 929 328 64 × 2 = 0 + 0.294 394 549 547 858 657 28;
  • 72) 0.294 394 549 547 858 657 28 × 2 = 0 + 0.588 789 099 095 717 314 56;
  • 73) 0.588 789 099 095 717 314 56 × 2 = 1 + 0.177 578 198 191 434 629 12;
  • 74) 0.177 578 198 191 434 629 12 × 2 = 0 + 0.355 156 396 382 869 258 24;
  • 75) 0.355 156 396 382 869 258 24 × 2 = 0 + 0.710 312 792 765 738 516 48;
  • 76) 0.710 312 792 765 738 516 48 × 2 = 1 + 0.420 625 585 531 477 032 96;
  • 77) 0.420 625 585 531 477 032 96 × 2 = 0 + 0.841 251 171 062 954 065 92;
  • 78) 0.841 251 171 062 954 065 92 × 2 = 1 + 0.682 502 342 125 908 131 84;
  • 79) 0.682 502 342 125 908 131 84 × 2 = 1 + 0.365 004 684 251 816 263 68;
  • 80) 0.365 004 684 251 816 263 68 × 2 = 0 + 0.730 009 368 503 632 527 36;

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 61(10) =

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

6. Positive number before normalization:

0.000 000 005 917 031 260 61(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1100 1001 1100 1001 0110(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 61(10) =

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

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

1.1001 0110 1001 1101 1000 0011 1100 0111 1100 1001 1100 1001 0110(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 1100 1001 1100 1001 0110


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 1100 1001 1100 1001 0110 =

1001 0110 1001 1101 1000 0011 1100 0111 1100 1001 1100 1001 0110


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 1100 1001 1100 1001 0110


Decimal number -0.000 000 005 917 031 260 61 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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