0.007 812 501 848 078 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.007 812 501 848 078 4(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
0.007 812 501 848 078 4(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.007 812 501 848 078 4.

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) 0.007 812 501 848 078 4 × 2 = 0 + 0.015 625 003 696 156 8;
  • 2) 0.015 625 003 696 156 8 × 2 = 0 + 0.031 250 007 392 313 6;
  • 3) 0.031 250 007 392 313 6 × 2 = 0 + 0.062 500 014 784 627 2;
  • 4) 0.062 500 014 784 627 2 × 2 = 0 + 0.125 000 029 569 254 4;
  • 5) 0.125 000 029 569 254 4 × 2 = 0 + 0.250 000 059 138 508 8;
  • 6) 0.250 000 059 138 508 8 × 2 = 0 + 0.500 000 118 277 017 6;
  • 7) 0.500 000 118 277 017 6 × 2 = 1 + 0.000 000 236 554 035 2;
  • 8) 0.000 000 236 554 035 2 × 2 = 0 + 0.000 000 473 108 070 4;
  • 9) 0.000 000 473 108 070 4 × 2 = 0 + 0.000 000 946 216 140 8;
  • 10) 0.000 000 946 216 140 8 × 2 = 0 + 0.000 001 892 432 281 6;
  • 11) 0.000 001 892 432 281 6 × 2 = 0 + 0.000 003 784 864 563 2;
  • 12) 0.000 003 784 864 563 2 × 2 = 0 + 0.000 007 569 729 126 4;
  • 13) 0.000 007 569 729 126 4 × 2 = 0 + 0.000 015 139 458 252 8;
  • 14) 0.000 015 139 458 252 8 × 2 = 0 + 0.000 030 278 916 505 6;
  • 15) 0.000 030 278 916 505 6 × 2 = 0 + 0.000 060 557 833 011 2;
  • 16) 0.000 060 557 833 011 2 × 2 = 0 + 0.000 121 115 666 022 4;
  • 17) 0.000 121 115 666 022 4 × 2 = 0 + 0.000 242 231 332 044 8;
  • 18) 0.000 242 231 332 044 8 × 2 = 0 + 0.000 484 462 664 089 6;
  • 19) 0.000 484 462 664 089 6 × 2 = 0 + 0.000 968 925 328 179 2;
  • 20) 0.000 968 925 328 179 2 × 2 = 0 + 0.001 937 850 656 358 4;
  • 21) 0.001 937 850 656 358 4 × 2 = 0 + 0.003 875 701 312 716 8;
  • 22) 0.003 875 701 312 716 8 × 2 = 0 + 0.007 751 402 625 433 6;
  • 23) 0.007 751 402 625 433 6 × 2 = 0 + 0.015 502 805 250 867 2;
  • 24) 0.015 502 805 250 867 2 × 2 = 0 + 0.031 005 610 501 734 4;
  • 25) 0.031 005 610 501 734 4 × 2 = 0 + 0.062 011 221 003 468 8;
  • 26) 0.062 011 221 003 468 8 × 2 = 0 + 0.124 022 442 006 937 6;
  • 27) 0.124 022 442 006 937 6 × 2 = 0 + 0.248 044 884 013 875 2;
  • 28) 0.248 044 884 013 875 2 × 2 = 0 + 0.496 089 768 027 750 4;
  • 29) 0.496 089 768 027 750 4 × 2 = 0 + 0.992 179 536 055 500 8;
  • 30) 0.992 179 536 055 500 8 × 2 = 1 + 0.984 359 072 111 001 6;
  • 31) 0.984 359 072 111 001 6 × 2 = 1 + 0.968 718 144 222 003 2;
  • 32) 0.968 718 144 222 003 2 × 2 = 1 + 0.937 436 288 444 006 4;
  • 33) 0.937 436 288 444 006 4 × 2 = 1 + 0.874 872 576 888 012 8;
  • 34) 0.874 872 576 888 012 8 × 2 = 1 + 0.749 745 153 776 025 6;
  • 35) 0.749 745 153 776 025 6 × 2 = 1 + 0.499 490 307 552 051 2;
  • 36) 0.499 490 307 552 051 2 × 2 = 0 + 0.998 980 615 104 102 4;
  • 37) 0.998 980 615 104 102 4 × 2 = 1 + 0.997 961 230 208 204 8;
  • 38) 0.997 961 230 208 204 8 × 2 = 1 + 0.995 922 460 416 409 6;
  • 39) 0.995 922 460 416 409 6 × 2 = 1 + 0.991 844 920 832 819 2;
  • 40) 0.991 844 920 832 819 2 × 2 = 1 + 0.983 689 841 665 638 4;
  • 41) 0.983 689 841 665 638 4 × 2 = 1 + 0.967 379 683 331 276 8;
  • 42) 0.967 379 683 331 276 8 × 2 = 1 + 0.934 759 366 662 553 6;
  • 43) 0.934 759 366 662 553 6 × 2 = 1 + 0.869 518 733 325 107 2;
  • 44) 0.869 518 733 325 107 2 × 2 = 1 + 0.739 037 466 650 214 4;
  • 45) 0.739 037 466 650 214 4 × 2 = 1 + 0.478 074 933 300 428 8;
  • 46) 0.478 074 933 300 428 8 × 2 = 0 + 0.956 149 866 600 857 6;
  • 47) 0.956 149 866 600 857 6 × 2 = 1 + 0.912 299 733 201 715 2;
  • 48) 0.912 299 733 201 715 2 × 2 = 1 + 0.824 599 466 403 430 4;
  • 49) 0.824 599 466 403 430 4 × 2 = 1 + 0.649 198 932 806 860 8;
  • 50) 0.649 198 932 806 860 8 × 2 = 1 + 0.298 397 865 613 721 6;
  • 51) 0.298 397 865 613 721 6 × 2 = 0 + 0.596 795 731 227 443 2;
  • 52) 0.596 795 731 227 443 2 × 2 = 1 + 0.193 591 462 454 886 4;
  • 53) 0.193 591 462 454 886 4 × 2 = 0 + 0.387 182 924 909 772 8;
  • 54) 0.387 182 924 909 772 8 × 2 = 0 + 0.774 365 849 819 545 6;
  • 55) 0.774 365 849 819 545 6 × 2 = 1 + 0.548 731 699 639 091 2;
  • 56) 0.548 731 699 639 091 2 × 2 = 1 + 0.097 463 399 278 182 4;
  • 57) 0.097 463 399 278 182 4 × 2 = 0 + 0.194 926 798 556 364 8;
  • 58) 0.194 926 798 556 364 8 × 2 = 0 + 0.389 853 597 112 729 6;
  • 59) 0.389 853 597 112 729 6 × 2 = 0 + 0.779 707 194 225 459 2;

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).


4. 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:


0.007 812 501 848 078 4(10) =

0.0000 0010 0000 0000 0000 0000 0000 0111 1110 1111 1111 1011 1101 0011 000(2)

5. Positive number before normalization:

0.007 812 501 848 078 4(10) =

0.0000 0010 0000 0000 0000 0000 0000 0111 1110 1111 1111 1011 1101 0011 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.007 812 501 848 078 4(10) =

0.0000 0010 0000 0000 0000 0000 0000 0111 1110 1111 1111 1011 1101 0011 000(2) =

0.0000 0010 0000 0000 0000 0000 0000 0111 1110 1111 1111 1011 1101 0011 000(2) × 20 =

1.0000 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000(2) × 2-7


7. 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 0 (a positive number)

Exponent (unadjusted): -7

Mantissa (not normalized):
1.0000 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(10)


9. 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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1016(10) =

011 1111 1000(2)


11. 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 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000 =

0000 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1000

Mantissa (52 bits) =
0000 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000


Decimal number 0.007 812 501 848 078 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1000 - 0000 0000 0000 0000 0000 0011 1111 0111 1111 1101 1110 1001 1000

» Convert decimal 0.007 812 501 848 078 5 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