-0.000 002 566 591 972 243 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 566 591 972 243 3(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
-0.000 002 566 591 972 243 3(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:

|-0.000 002 566 591 972 243 3| = 0.000 002 566 591 972 243 3


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: 0.000 002 566 591 972 243 3.

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) 0.000 002 566 591 972 243 3 × 2 = 0 + 0.000 005 133 183 944 486 6;
  • 2) 0.000 005 133 183 944 486 6 × 2 = 0 + 0.000 010 266 367 888 973 2;
  • 3) 0.000 010 266 367 888 973 2 × 2 = 0 + 0.000 020 532 735 777 946 4;
  • 4) 0.000 020 532 735 777 946 4 × 2 = 0 + 0.000 041 065 471 555 892 8;
  • 5) 0.000 041 065 471 555 892 8 × 2 = 0 + 0.000 082 130 943 111 785 6;
  • 6) 0.000 082 130 943 111 785 6 × 2 = 0 + 0.000 164 261 886 223 571 2;
  • 7) 0.000 164 261 886 223 571 2 × 2 = 0 + 0.000 328 523 772 447 142 4;
  • 8) 0.000 328 523 772 447 142 4 × 2 = 0 + 0.000 657 047 544 894 284 8;
  • 9) 0.000 657 047 544 894 284 8 × 2 = 0 + 0.001 314 095 089 788 569 6;
  • 10) 0.001 314 095 089 788 569 6 × 2 = 0 + 0.002 628 190 179 577 139 2;
  • 11) 0.002 628 190 179 577 139 2 × 2 = 0 + 0.005 256 380 359 154 278 4;
  • 12) 0.005 256 380 359 154 278 4 × 2 = 0 + 0.010 512 760 718 308 556 8;
  • 13) 0.010 512 760 718 308 556 8 × 2 = 0 + 0.021 025 521 436 617 113 6;
  • 14) 0.021 025 521 436 617 113 6 × 2 = 0 + 0.042 051 042 873 234 227 2;
  • 15) 0.042 051 042 873 234 227 2 × 2 = 0 + 0.084 102 085 746 468 454 4;
  • 16) 0.084 102 085 746 468 454 4 × 2 = 0 + 0.168 204 171 492 936 908 8;
  • 17) 0.168 204 171 492 936 908 8 × 2 = 0 + 0.336 408 342 985 873 817 6;
  • 18) 0.336 408 342 985 873 817 6 × 2 = 0 + 0.672 816 685 971 747 635 2;
  • 19) 0.672 816 685 971 747 635 2 × 2 = 1 + 0.345 633 371 943 495 270 4;
  • 20) 0.345 633 371 943 495 270 4 × 2 = 0 + 0.691 266 743 886 990 540 8;
  • 21) 0.691 266 743 886 990 540 8 × 2 = 1 + 0.382 533 487 773 981 081 6;
  • 22) 0.382 533 487 773 981 081 6 × 2 = 0 + 0.765 066 975 547 962 163 2;
  • 23) 0.765 066 975 547 962 163 2 × 2 = 1 + 0.530 133 951 095 924 326 4;
  • 24) 0.530 133 951 095 924 326 4 × 2 = 1 + 0.060 267 902 191 848 652 8;
  • 25) 0.060 267 902 191 848 652 8 × 2 = 0 + 0.120 535 804 383 697 305 6;
  • 26) 0.120 535 804 383 697 305 6 × 2 = 0 + 0.241 071 608 767 394 611 2;
  • 27) 0.241 071 608 767 394 611 2 × 2 = 0 + 0.482 143 217 534 789 222 4;
  • 28) 0.482 143 217 534 789 222 4 × 2 = 0 + 0.964 286 435 069 578 444 8;
  • 29) 0.964 286 435 069 578 444 8 × 2 = 1 + 0.928 572 870 139 156 889 6;
  • 30) 0.928 572 870 139 156 889 6 × 2 = 1 + 0.857 145 740 278 313 779 2;
  • 31) 0.857 145 740 278 313 779 2 × 2 = 1 + 0.714 291 480 556 627 558 4;
  • 32) 0.714 291 480 556 627 558 4 × 2 = 1 + 0.428 582 961 113 255 116 8;
  • 33) 0.428 582 961 113 255 116 8 × 2 = 0 + 0.857 165 922 226 510 233 6;
  • 34) 0.857 165 922 226 510 233 6 × 2 = 1 + 0.714 331 844 453 020 467 2;
  • 35) 0.714 331 844 453 020 467 2 × 2 = 1 + 0.428 663 688 906 040 934 4;
  • 36) 0.428 663 688 906 040 934 4 × 2 = 0 + 0.857 327 377 812 081 868 8;
  • 37) 0.857 327 377 812 081 868 8 × 2 = 1 + 0.714 654 755 624 163 737 6;
  • 38) 0.714 654 755 624 163 737 6 × 2 = 1 + 0.429 309 511 248 327 475 2;
  • 39) 0.429 309 511 248 327 475 2 × 2 = 0 + 0.858 619 022 496 654 950 4;
  • 40) 0.858 619 022 496 654 950 4 × 2 = 1 + 0.717 238 044 993 309 900 8;
  • 41) 0.717 238 044 993 309 900 8 × 2 = 1 + 0.434 476 089 986 619 801 6;
  • 42) 0.434 476 089 986 619 801 6 × 2 = 0 + 0.868 952 179 973 239 603 2;
  • 43) 0.868 952 179 973 239 603 2 × 2 = 1 + 0.737 904 359 946 479 206 4;
  • 44) 0.737 904 359 946 479 206 4 × 2 = 1 + 0.475 808 719 892 958 412 8;
  • 45) 0.475 808 719 892 958 412 8 × 2 = 0 + 0.951 617 439 785 916 825 6;
  • 46) 0.951 617 439 785 916 825 6 × 2 = 1 + 0.903 234 879 571 833 651 2;
  • 47) 0.903 234 879 571 833 651 2 × 2 = 1 + 0.806 469 759 143 667 302 4;
  • 48) 0.806 469 759 143 667 302 4 × 2 = 1 + 0.612 939 518 287 334 604 8;
  • 49) 0.612 939 518 287 334 604 8 × 2 = 1 + 0.225 879 036 574 669 209 6;
  • 50) 0.225 879 036 574 669 209 6 × 2 = 0 + 0.451 758 073 149 338 419 2;
  • 51) 0.451 758 073 149 338 419 2 × 2 = 0 + 0.903 516 146 298 676 838 4;
  • 52) 0.903 516 146 298 676 838 4 × 2 = 1 + 0.807 032 292 597 353 676 8;
  • 53) 0.807 032 292 597 353 676 8 × 2 = 1 + 0.614 064 585 194 707 353 6;
  • 54) 0.614 064 585 194 707 353 6 × 2 = 1 + 0.228 129 170 389 414 707 2;
  • 55) 0.228 129 170 389 414 707 2 × 2 = 0 + 0.456 258 340 778 829 414 4;
  • 56) 0.456 258 340 778 829 414 4 × 2 = 0 + 0.912 516 681 557 658 828 8;
  • 57) 0.912 516 681 557 658 828 8 × 2 = 1 + 0.825 033 363 115 317 657 6;
  • 58) 0.825 033 363 115 317 657 6 × 2 = 1 + 0.650 066 726 230 635 315 2;
  • 59) 0.650 066 726 230 635 315 2 × 2 = 1 + 0.300 133 452 461 270 630 4;
  • 60) 0.300 133 452 461 270 630 4 × 2 = 0 + 0.600 266 904 922 541 260 8;
  • 61) 0.600 266 904 922 541 260 8 × 2 = 1 + 0.200 533 809 845 082 521 6;
  • 62) 0.200 533 809 845 082 521 6 × 2 = 0 + 0.401 067 619 690 165 043 2;
  • 63) 0.401 067 619 690 165 043 2 × 2 = 0 + 0.802 135 239 380 330 086 4;
  • 64) 0.802 135 239 380 330 086 4 × 2 = 1 + 0.604 270 478 760 660 172 8;
  • 65) 0.604 270 478 760 660 172 8 × 2 = 1 + 0.208 540 957 521 320 345 6;
  • 66) 0.208 540 957 521 320 345 6 × 2 = 0 + 0.417 081 915 042 640 691 2;
  • 67) 0.417 081 915 042 640 691 2 × 2 = 0 + 0.834 163 830 085 281 382 4;
  • 68) 0.834 163 830 085 281 382 4 × 2 = 1 + 0.668 327 660 170 562 764 8;
  • 69) 0.668 327 660 170 562 764 8 × 2 = 1 + 0.336 655 320 341 125 529 6;
  • 70) 0.336 655 320 341 125 529 6 × 2 = 0 + 0.673 310 640 682 251 059 2;
  • 71) 0.673 310 640 682 251 059 2 × 2 = 1 + 0.346 621 281 364 502 118 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:


0.000 002 566 591 972 243 3(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1100 1110 1001 1001 101(2)

6. Positive number before normalization:

0.000 002 566 591 972 243 3(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1100 1110 1001 1001 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 002 566 591 972 243 3(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1100 1110 1001 1001 101(2) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1100 1110 1001 1001 101(2) × 20 =

1.0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101(2) × 2-19


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): -19

Mantissa (not normalized):
1.0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 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) =

1004(10) =

011 1110 1100(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. 0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101 =

0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101


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 1100

Mantissa (52 bits) =
0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101


Decimal number -0.000 002 566 591 972 243 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1100 - 0101 1000 0111 1011 0110 1101 1011 1100 1110 0111 0100 1100 1101

» Convert decimal -0.000 002 566 591 972 248 2 to 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