-0.000 001 324 478 226 373 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 324 478 226 373 63(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
-0.000 001 324 478 226 373 63(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:

|-0.000 001 324 478 226 373 63| = 0.000 001 324 478 226 373 63


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: 0.000 001 324 478 226 373 63.

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) 0.000 001 324 478 226 373 63 × 2 = 0 + 0.000 002 648 956 452 747 26;
  • 2) 0.000 002 648 956 452 747 26 × 2 = 0 + 0.000 005 297 912 905 494 52;
  • 3) 0.000 005 297 912 905 494 52 × 2 = 0 + 0.000 010 595 825 810 989 04;
  • 4) 0.000 010 595 825 810 989 04 × 2 = 0 + 0.000 021 191 651 621 978 08;
  • 5) 0.000 021 191 651 621 978 08 × 2 = 0 + 0.000 042 383 303 243 956 16;
  • 6) 0.000 042 383 303 243 956 16 × 2 = 0 + 0.000 084 766 606 487 912 32;
  • 7) 0.000 084 766 606 487 912 32 × 2 = 0 + 0.000 169 533 212 975 824 64;
  • 8) 0.000 169 533 212 975 824 64 × 2 = 0 + 0.000 339 066 425 951 649 28;
  • 9) 0.000 339 066 425 951 649 28 × 2 = 0 + 0.000 678 132 851 903 298 56;
  • 10) 0.000 678 132 851 903 298 56 × 2 = 0 + 0.001 356 265 703 806 597 12;
  • 11) 0.001 356 265 703 806 597 12 × 2 = 0 + 0.002 712 531 407 613 194 24;
  • 12) 0.002 712 531 407 613 194 24 × 2 = 0 + 0.005 425 062 815 226 388 48;
  • 13) 0.005 425 062 815 226 388 48 × 2 = 0 + 0.010 850 125 630 452 776 96;
  • 14) 0.010 850 125 630 452 776 96 × 2 = 0 + 0.021 700 251 260 905 553 92;
  • 15) 0.021 700 251 260 905 553 92 × 2 = 0 + 0.043 400 502 521 811 107 84;
  • 16) 0.043 400 502 521 811 107 84 × 2 = 0 + 0.086 801 005 043 622 215 68;
  • 17) 0.086 801 005 043 622 215 68 × 2 = 0 + 0.173 602 010 087 244 431 36;
  • 18) 0.173 602 010 087 244 431 36 × 2 = 0 + 0.347 204 020 174 488 862 72;
  • 19) 0.347 204 020 174 488 862 72 × 2 = 0 + 0.694 408 040 348 977 725 44;
  • 20) 0.694 408 040 348 977 725 44 × 2 = 1 + 0.388 816 080 697 955 450 88;
  • 21) 0.388 816 080 697 955 450 88 × 2 = 0 + 0.777 632 161 395 910 901 76;
  • 22) 0.777 632 161 395 910 901 76 × 2 = 1 + 0.555 264 322 791 821 803 52;
  • 23) 0.555 264 322 791 821 803 52 × 2 = 1 + 0.110 528 645 583 643 607 04;
  • 24) 0.110 528 645 583 643 607 04 × 2 = 0 + 0.221 057 291 167 287 214 08;
  • 25) 0.221 057 291 167 287 214 08 × 2 = 0 + 0.442 114 582 334 574 428 16;
  • 26) 0.442 114 582 334 574 428 16 × 2 = 0 + 0.884 229 164 669 148 856 32;
  • 27) 0.884 229 164 669 148 856 32 × 2 = 1 + 0.768 458 329 338 297 712 64;
  • 28) 0.768 458 329 338 297 712 64 × 2 = 1 + 0.536 916 658 676 595 425 28;
  • 29) 0.536 916 658 676 595 425 28 × 2 = 1 + 0.073 833 317 353 190 850 56;
  • 30) 0.073 833 317 353 190 850 56 × 2 = 0 + 0.147 666 634 706 381 701 12;
  • 31) 0.147 666 634 706 381 701 12 × 2 = 0 + 0.295 333 269 412 763 402 24;
  • 32) 0.295 333 269 412 763 402 24 × 2 = 0 + 0.590 666 538 825 526 804 48;
  • 33) 0.590 666 538 825 526 804 48 × 2 = 1 + 0.181 333 077 651 053 608 96;
  • 34) 0.181 333 077 651 053 608 96 × 2 = 0 + 0.362 666 155 302 107 217 92;
  • 35) 0.362 666 155 302 107 217 92 × 2 = 0 + 0.725 332 310 604 214 435 84;
  • 36) 0.725 332 310 604 214 435 84 × 2 = 1 + 0.450 664 621 208 428 871 68;
  • 37) 0.450 664 621 208 428 871 68 × 2 = 0 + 0.901 329 242 416 857 743 36;
  • 38) 0.901 329 242 416 857 743 36 × 2 = 1 + 0.802 658 484 833 715 486 72;
  • 39) 0.802 658 484 833 715 486 72 × 2 = 1 + 0.605 316 969 667 430 973 44;
  • 40) 0.605 316 969 667 430 973 44 × 2 = 1 + 0.210 633 939 334 861 946 88;
  • 41) 0.210 633 939 334 861 946 88 × 2 = 0 + 0.421 267 878 669 723 893 76;
  • 42) 0.421 267 878 669 723 893 76 × 2 = 0 + 0.842 535 757 339 447 787 52;
  • 43) 0.842 535 757 339 447 787 52 × 2 = 1 + 0.685 071 514 678 895 575 04;
  • 44) 0.685 071 514 678 895 575 04 × 2 = 1 + 0.370 143 029 357 791 150 08;
  • 45) 0.370 143 029 357 791 150 08 × 2 = 0 + 0.740 286 058 715 582 300 16;
  • 46) 0.740 286 058 715 582 300 16 × 2 = 1 + 0.480 572 117 431 164 600 32;
  • 47) 0.480 572 117 431 164 600 32 × 2 = 0 + 0.961 144 234 862 329 200 64;
  • 48) 0.961 144 234 862 329 200 64 × 2 = 1 + 0.922 288 469 724 658 401 28;
  • 49) 0.922 288 469 724 658 401 28 × 2 = 1 + 0.844 576 939 449 316 802 56;
  • 50) 0.844 576 939 449 316 802 56 × 2 = 1 + 0.689 153 878 898 633 605 12;
  • 51) 0.689 153 878 898 633 605 12 × 2 = 1 + 0.378 307 757 797 267 210 24;
  • 52) 0.378 307 757 797 267 210 24 × 2 = 0 + 0.756 615 515 594 534 420 48;
  • 53) 0.756 615 515 594 534 420 48 × 2 = 1 + 0.513 231 031 189 068 840 96;
  • 54) 0.513 231 031 189 068 840 96 × 2 = 1 + 0.026 462 062 378 137 681 92;
  • 55) 0.026 462 062 378 137 681 92 × 2 = 0 + 0.052 924 124 756 275 363 84;
  • 56) 0.052 924 124 756 275 363 84 × 2 = 0 + 0.105 848 249 512 550 727 68;
  • 57) 0.105 848 249 512 550 727 68 × 2 = 0 + 0.211 696 499 025 101 455 36;
  • 58) 0.211 696 499 025 101 455 36 × 2 = 0 + 0.423 392 998 050 202 910 72;
  • 59) 0.423 392 998 050 202 910 72 × 2 = 0 + 0.846 785 996 100 405 821 44;
  • 60) 0.846 785 996 100 405 821 44 × 2 = 1 + 0.693 571 992 200 811 642 88;
  • 61) 0.693 571 992 200 811 642 88 × 2 = 1 + 0.387 143 984 401 623 285 76;
  • 62) 0.387 143 984 401 623 285 76 × 2 = 0 + 0.774 287 968 803 246 571 52;
  • 63) 0.774 287 968 803 246 571 52 × 2 = 1 + 0.548 575 937 606 493 143 04;
  • 64) 0.548 575 937 606 493 143 04 × 2 = 1 + 0.097 151 875 212 986 286 08;
  • 65) 0.097 151 875 212 986 286 08 × 2 = 0 + 0.194 303 750 425 972 572 16;
  • 66) 0.194 303 750 425 972 572 16 × 2 = 0 + 0.388 607 500 851 945 144 32;
  • 67) 0.388 607 500 851 945 144 32 × 2 = 0 + 0.777 215 001 703 890 288 64;
  • 68) 0.777 215 001 703 890 288 64 × 2 = 1 + 0.554 430 003 407 780 577 28;
  • 69) 0.554 430 003 407 780 577 28 × 2 = 1 + 0.108 860 006 815 561 154 56;
  • 70) 0.108 860 006 815 561 154 56 × 2 = 0 + 0.217 720 013 631 122 309 12;
  • 71) 0.217 720 013 631 122 309 12 × 2 = 0 + 0.435 440 027 262 244 618 24;
  • 72) 0.435 440 027 262 244 618 24 × 2 = 0 + 0.870 880 054 524 489 236 48;

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:


0.000 001 324 478 226 373 63(10) =

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

6. Positive number before normalization:

0.000 001 324 478 226 373 63(10) =

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

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 324 478 226 373 63(10) =

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

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

1.0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000(2) × 2-20


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): -20

Mantissa (not normalized):
1.0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 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) =

1003(10) =

011 1110 1011(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. 0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000 =

0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000


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 1011

Mantissa (52 bits) =
0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000


Decimal number -0.000 001 324 478 226 373 63 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0011 1000 1001 0111 0011 0101 1110 1100 0001 1011 0001 1000

Share

» Convert decimal -0.000 001 324 478 226 373 67 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 04, 2026 [April]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 minutes

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