-0.000 007 820 329 949 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 329 949 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
-0.000 007 820 329 949 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:

|-0.000 007 820 329 949 3| = 0.000 007 820 329 949 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: 0.000 007 820 329 949 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) 0.000 007 820 329 949 3 × 2 = 0 + 0.000 015 640 659 898 6;
  • 2) 0.000 015 640 659 898 6 × 2 = 0 + 0.000 031 281 319 797 2;
  • 3) 0.000 031 281 319 797 2 × 2 = 0 + 0.000 062 562 639 594 4;
  • 4) 0.000 062 562 639 594 4 × 2 = 0 + 0.000 125 125 279 188 8;
  • 5) 0.000 125 125 279 188 8 × 2 = 0 + 0.000 250 250 558 377 6;
  • 6) 0.000 250 250 558 377 6 × 2 = 0 + 0.000 500 501 116 755 2;
  • 7) 0.000 500 501 116 755 2 × 2 = 0 + 0.001 001 002 233 510 4;
  • 8) 0.001 001 002 233 510 4 × 2 = 0 + 0.002 002 004 467 020 8;
  • 9) 0.002 002 004 467 020 8 × 2 = 0 + 0.004 004 008 934 041 6;
  • 10) 0.004 004 008 934 041 6 × 2 = 0 + 0.008 008 017 868 083 2;
  • 11) 0.008 008 017 868 083 2 × 2 = 0 + 0.016 016 035 736 166 4;
  • 12) 0.016 016 035 736 166 4 × 2 = 0 + 0.032 032 071 472 332 8;
  • 13) 0.032 032 071 472 332 8 × 2 = 0 + 0.064 064 142 944 665 6;
  • 14) 0.064 064 142 944 665 6 × 2 = 0 + 0.128 128 285 889 331 2;
  • 15) 0.128 128 285 889 331 2 × 2 = 0 + 0.256 256 571 778 662 4;
  • 16) 0.256 256 571 778 662 4 × 2 = 0 + 0.512 513 143 557 324 8;
  • 17) 0.512 513 143 557 324 8 × 2 = 1 + 0.025 026 287 114 649 6;
  • 18) 0.025 026 287 114 649 6 × 2 = 0 + 0.050 052 574 229 299 2;
  • 19) 0.050 052 574 229 299 2 × 2 = 0 + 0.100 105 148 458 598 4;
  • 20) 0.100 105 148 458 598 4 × 2 = 0 + 0.200 210 296 917 196 8;
  • 21) 0.200 210 296 917 196 8 × 2 = 0 + 0.400 420 593 834 393 6;
  • 22) 0.400 420 593 834 393 6 × 2 = 0 + 0.800 841 187 668 787 2;
  • 23) 0.800 841 187 668 787 2 × 2 = 1 + 0.601 682 375 337 574 4;
  • 24) 0.601 682 375 337 574 4 × 2 = 1 + 0.203 364 750 675 148 8;
  • 25) 0.203 364 750 675 148 8 × 2 = 0 + 0.406 729 501 350 297 6;
  • 26) 0.406 729 501 350 297 6 × 2 = 0 + 0.813 459 002 700 595 2;
  • 27) 0.813 459 002 700 595 2 × 2 = 1 + 0.626 918 005 401 190 4;
  • 28) 0.626 918 005 401 190 4 × 2 = 1 + 0.253 836 010 802 380 8;
  • 29) 0.253 836 010 802 380 8 × 2 = 0 + 0.507 672 021 604 761 6;
  • 30) 0.507 672 021 604 761 6 × 2 = 1 + 0.015 344 043 209 523 2;
  • 31) 0.015 344 043 209 523 2 × 2 = 0 + 0.030 688 086 419 046 4;
  • 32) 0.030 688 086 419 046 4 × 2 = 0 + 0.061 376 172 838 092 8;
  • 33) 0.061 376 172 838 092 8 × 2 = 0 + 0.122 752 345 676 185 6;
  • 34) 0.122 752 345 676 185 6 × 2 = 0 + 0.245 504 691 352 371 2;
  • 35) 0.245 504 691 352 371 2 × 2 = 0 + 0.491 009 382 704 742 4;
  • 36) 0.491 009 382 704 742 4 × 2 = 0 + 0.982 018 765 409 484 8;
  • 37) 0.982 018 765 409 484 8 × 2 = 1 + 0.964 037 530 818 969 6;
  • 38) 0.964 037 530 818 969 6 × 2 = 1 + 0.928 075 061 637 939 2;
  • 39) 0.928 075 061 637 939 2 × 2 = 1 + 0.856 150 123 275 878 4;
  • 40) 0.856 150 123 275 878 4 × 2 = 1 + 0.712 300 246 551 756 8;
  • 41) 0.712 300 246 551 756 8 × 2 = 1 + 0.424 600 493 103 513 6;
  • 42) 0.424 600 493 103 513 6 × 2 = 0 + 0.849 200 986 207 027 2;
  • 43) 0.849 200 986 207 027 2 × 2 = 1 + 0.698 401 972 414 054 4;
  • 44) 0.698 401 972 414 054 4 × 2 = 1 + 0.396 803 944 828 108 8;
  • 45) 0.396 803 944 828 108 8 × 2 = 0 + 0.793 607 889 656 217 6;
  • 46) 0.793 607 889 656 217 6 × 2 = 1 + 0.587 215 779 312 435 2;
  • 47) 0.587 215 779 312 435 2 × 2 = 1 + 0.174 431 558 624 870 4;
  • 48) 0.174 431 558 624 870 4 × 2 = 0 + 0.348 863 117 249 740 8;
  • 49) 0.348 863 117 249 740 8 × 2 = 0 + 0.697 726 234 499 481 6;
  • 50) 0.697 726 234 499 481 6 × 2 = 1 + 0.395 452 468 998 963 2;
  • 51) 0.395 452 468 998 963 2 × 2 = 0 + 0.790 904 937 997 926 4;
  • 52) 0.790 904 937 997 926 4 × 2 = 1 + 0.581 809 875 995 852 8;
  • 53) 0.581 809 875 995 852 8 × 2 = 1 + 0.163 619 751 991 705 6;
  • 54) 0.163 619 751 991 705 6 × 2 = 0 + 0.327 239 503 983 411 2;
  • 55) 0.327 239 503 983 411 2 × 2 = 0 + 0.654 479 007 966 822 4;
  • 56) 0.654 479 007 966 822 4 × 2 = 1 + 0.308 958 015 933 644 8;
  • 57) 0.308 958 015 933 644 8 × 2 = 0 + 0.617 916 031 867 289 6;
  • 58) 0.617 916 031 867 289 6 × 2 = 1 + 0.235 832 063 734 579 2;
  • 59) 0.235 832 063 734 579 2 × 2 = 0 + 0.471 664 127 469 158 4;
  • 60) 0.471 664 127 469 158 4 × 2 = 0 + 0.943 328 254 938 316 8;
  • 61) 0.943 328 254 938 316 8 × 2 = 1 + 0.886 656 509 876 633 6;
  • 62) 0.886 656 509 876 633 6 × 2 = 1 + 0.773 313 019 753 267 2;
  • 63) 0.773 313 019 753 267 2 × 2 = 1 + 0.546 626 039 506 534 4;
  • 64) 0.546 626 039 506 534 4 × 2 = 1 + 0.093 252 079 013 068 8;
  • 65) 0.093 252 079 013 068 8 × 2 = 0 + 0.186 504 158 026 137 6;
  • 66) 0.186 504 158 026 137 6 × 2 = 0 + 0.373 008 316 052 275 2;
  • 67) 0.373 008 316 052 275 2 × 2 = 0 + 0.746 016 632 104 550 4;
  • 68) 0.746 016 632 104 550 4 × 2 = 1 + 0.492 033 264 209 100 8;
  • 69) 0.492 033 264 209 100 8 × 2 = 0 + 0.984 066 528 418 201 6;

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:


0.000 007 820 329 949 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0000 1111 1011 0110 0101 1001 0100 1111 0001 0(2)

6. Positive number before normalization:

0.000 007 820 329 949 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0000 1111 1011 0110 0101 1001 0100 1111 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 329 949 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0000 1111 1011 0110 0101 1001 0100 1111 0001 0(2) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0000 1111 1011 0110 0101 1001 0100 1111 0001 0(2) × 20 =

1.0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010 =

0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010


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 1110

Mantissa (52 bits) =
0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010


Decimal number -0.000 007 820 329 949 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1000 0001 1111 0110 1100 1011 0010 1001 1110 0010

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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