Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.000 006 611 673 66 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 006 611 673 66(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 006 611 673 66| = 0.000 006 611 673 66


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: 0.000 006 611 673 66.

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) 0.000 006 611 673 66 × 2 = 0 + 0.000 013 223 347 32;
  • 2) 0.000 013 223 347 32 × 2 = 0 + 0.000 026 446 694 64;
  • 3) 0.000 026 446 694 64 × 2 = 0 + 0.000 052 893 389 28;
  • 4) 0.000 052 893 389 28 × 2 = 0 + 0.000 105 786 778 56;
  • 5) 0.000 105 786 778 56 × 2 = 0 + 0.000 211 573 557 12;
  • 6) 0.000 211 573 557 12 × 2 = 0 + 0.000 423 147 114 24;
  • 7) 0.000 423 147 114 24 × 2 = 0 + 0.000 846 294 228 48;
  • 8) 0.000 846 294 228 48 × 2 = 0 + 0.001 692 588 456 96;
  • 9) 0.001 692 588 456 96 × 2 = 0 + 0.003 385 176 913 92;
  • 10) 0.003 385 176 913 92 × 2 = 0 + 0.006 770 353 827 84;
  • 11) 0.006 770 353 827 84 × 2 = 0 + 0.013 540 707 655 68;
  • 12) 0.013 540 707 655 68 × 2 = 0 + 0.027 081 415 311 36;
  • 13) 0.027 081 415 311 36 × 2 = 0 + 0.054 162 830 622 72;
  • 14) 0.054 162 830 622 72 × 2 = 0 + 0.108 325 661 245 44;
  • 15) 0.108 325 661 245 44 × 2 = 0 + 0.216 651 322 490 88;
  • 16) 0.216 651 322 490 88 × 2 = 0 + 0.433 302 644 981 76;
  • 17) 0.433 302 644 981 76 × 2 = 0 + 0.866 605 289 963 52;
  • 18) 0.866 605 289 963 52 × 2 = 1 + 0.733 210 579 927 04;
  • 19) 0.733 210 579 927 04 × 2 = 1 + 0.466 421 159 854 08;
  • 20) 0.466 421 159 854 08 × 2 = 0 + 0.932 842 319 708 16;
  • 21) 0.932 842 319 708 16 × 2 = 1 + 0.865 684 639 416 32;
  • 22) 0.865 684 639 416 32 × 2 = 1 + 0.731 369 278 832 64;
  • 23) 0.731 369 278 832 64 × 2 = 1 + 0.462 738 557 665 28;
  • 24) 0.462 738 557 665 28 × 2 = 0 + 0.925 477 115 330 56;
  • 25) 0.925 477 115 330 56 × 2 = 1 + 0.850 954 230 661 12;
  • 26) 0.850 954 230 661 12 × 2 = 1 + 0.701 908 461 322 24;
  • 27) 0.701 908 461 322 24 × 2 = 1 + 0.403 816 922 644 48;
  • 28) 0.403 816 922 644 48 × 2 = 0 + 0.807 633 845 288 96;
  • 29) 0.807 633 845 288 96 × 2 = 1 + 0.615 267 690 577 92;
  • 30) 0.615 267 690 577 92 × 2 = 1 + 0.230 535 381 155 84;
  • 31) 0.230 535 381 155 84 × 2 = 0 + 0.461 070 762 311 68;
  • 32) 0.461 070 762 311 68 × 2 = 0 + 0.922 141 524 623 36;
  • 33) 0.922 141 524 623 36 × 2 = 1 + 0.844 283 049 246 72;
  • 34) 0.844 283 049 246 72 × 2 = 1 + 0.688 566 098 493 44;
  • 35) 0.688 566 098 493 44 × 2 = 1 + 0.377 132 196 986 88;
  • 36) 0.377 132 196 986 88 × 2 = 0 + 0.754 264 393 973 76;
  • 37) 0.754 264 393 973 76 × 2 = 1 + 0.508 528 787 947 52;
  • 38) 0.508 528 787 947 52 × 2 = 1 + 0.017 057 575 895 04;
  • 39) 0.017 057 575 895 04 × 2 = 0 + 0.034 115 151 790 08;
  • 40) 0.034 115 151 790 08 × 2 = 0 + 0.068 230 303 580 16;
  • 41) 0.068 230 303 580 16 × 2 = 0 + 0.136 460 607 160 32;
  • 42) 0.136 460 607 160 32 × 2 = 0 + 0.272 921 214 320 64;
  • 43) 0.272 921 214 320 64 × 2 = 0 + 0.545 842 428 641 28;
  • 44) 0.545 842 428 641 28 × 2 = 1 + 0.091 684 857 282 56;
  • 45) 0.091 684 857 282 56 × 2 = 0 + 0.183 369 714 565 12;
  • 46) 0.183 369 714 565 12 × 2 = 0 + 0.366 739 429 130 24;
  • 47) 0.366 739 429 130 24 × 2 = 0 + 0.733 478 858 260 48;
  • 48) 0.733 478 858 260 48 × 2 = 1 + 0.466 957 716 520 96;
  • 49) 0.466 957 716 520 96 × 2 = 0 + 0.933 915 433 041 92;
  • 50) 0.933 915 433 041 92 × 2 = 1 + 0.867 830 866 083 84;
  • 51) 0.867 830 866 083 84 × 2 = 1 + 0.735 661 732 167 68;
  • 52) 0.735 661 732 167 68 × 2 = 1 + 0.471 323 464 335 36;
  • 53) 0.471 323 464 335 36 × 2 = 0 + 0.942 646 928 670 72;
  • 54) 0.942 646 928 670 72 × 2 = 1 + 0.885 293 857 341 44;
  • 55) 0.885 293 857 341 44 × 2 = 1 + 0.770 587 714 682 88;
  • 56) 0.770 587 714 682 88 × 2 = 1 + 0.541 175 429 365 76;
  • 57) 0.541 175 429 365 76 × 2 = 1 + 0.082 350 858 731 52;
  • 58) 0.082 350 858 731 52 × 2 = 0 + 0.164 701 717 463 04;
  • 59) 0.164 701 717 463 04 × 2 = 0 + 0.329 403 434 926 08;
  • 60) 0.329 403 434 926 08 × 2 = 0 + 0.658 806 869 852 16;
  • 61) 0.658 806 869 852 16 × 2 = 1 + 0.317 613 739 704 32;
  • 62) 0.317 613 739 704 32 × 2 = 0 + 0.635 227 479 408 64;
  • 63) 0.635 227 479 408 64 × 2 = 1 + 0.270 454 958 817 28;
  • 64) 0.270 454 958 817 28 × 2 = 0 + 0.540 909 917 634 56;
  • 65) 0.540 909 917 634 56 × 2 = 1 + 0.081 819 835 269 12;
  • 66) 0.081 819 835 269 12 × 2 = 0 + 0.163 639 670 538 24;
  • 67) 0.163 639 670 538 24 × 2 = 0 + 0.327 279 341 076 48;
  • 68) 0.327 279 341 076 48 × 2 = 0 + 0.654 558 682 152 96;
  • 69) 0.654 558 682 152 96 × 2 = 1 + 0.309 117 364 305 92;
  • 70) 0.309 117 364 305 92 × 2 = 0 + 0.618 234 728 611 84;

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:


0.000 006 611 673 66(10) =

0.0000 0000 0000 0000 0110 1110 1110 1100 1110 1100 0001 0001 0111 0111 1000 1010 1000 10(2)

6. Positive number before normalization:

0.000 006 611 673 66(10) =

0.0000 0000 0000 0000 0110 1110 1110 1100 1110 1100 0001 0001 0111 0111 1000 1010 1000 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 611 673 66(10) =

0.0000 0000 0000 0000 0110 1110 1110 1100 1110 1100 0001 0001 0111 0111 1000 1010 1000 10(2) =

0.0000 0000 0000 0000 0110 1110 1110 1100 1110 1100 0001 0001 0111 0111 1000 1010 1000 10(2) × 20 =

1.1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 0010(2) × 2-18


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): -18

Mantissa (not normalized):
1.1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(11-1) - 1 =

(-18 + 1 023)(10) =

1 005(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;
  • 1 005 ÷ 2 = 502 + 1;
  • 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) =

1005(10) =

011 1110 1101(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. 1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 0010 =

1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 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 1101

Mantissa (52 bits) =
1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 0010


The base ten decimal number -0.000 006 611 673 66 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1101 - 1011 1011 1011 0011 1011 0000 0100 0101 1101 1110 0010 1010 0010

» Decimal number in base ten -0.000 006 611 673 19 converted and written as 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