-0.000 002 916 619 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 916 619 8(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
-0.000 002 916 619 8(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:

|-0.000 002 916 619 8| = 0.000 002 916 619 8


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: 0.000 002 916 619 8.

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) 0.000 002 916 619 8 × 2 = 0 + 0.000 005 833 239 6;
  • 2) 0.000 005 833 239 6 × 2 = 0 + 0.000 011 666 479 2;
  • 3) 0.000 011 666 479 2 × 2 = 0 + 0.000 023 332 958 4;
  • 4) 0.000 023 332 958 4 × 2 = 0 + 0.000 046 665 916 8;
  • 5) 0.000 046 665 916 8 × 2 = 0 + 0.000 093 331 833 6;
  • 6) 0.000 093 331 833 6 × 2 = 0 + 0.000 186 663 667 2;
  • 7) 0.000 186 663 667 2 × 2 = 0 + 0.000 373 327 334 4;
  • 8) 0.000 373 327 334 4 × 2 = 0 + 0.000 746 654 668 8;
  • 9) 0.000 746 654 668 8 × 2 = 0 + 0.001 493 309 337 6;
  • 10) 0.001 493 309 337 6 × 2 = 0 + 0.002 986 618 675 2;
  • 11) 0.002 986 618 675 2 × 2 = 0 + 0.005 973 237 350 4;
  • 12) 0.005 973 237 350 4 × 2 = 0 + 0.011 946 474 700 8;
  • 13) 0.011 946 474 700 8 × 2 = 0 + 0.023 892 949 401 6;
  • 14) 0.023 892 949 401 6 × 2 = 0 + 0.047 785 898 803 2;
  • 15) 0.047 785 898 803 2 × 2 = 0 + 0.095 571 797 606 4;
  • 16) 0.095 571 797 606 4 × 2 = 0 + 0.191 143 595 212 8;
  • 17) 0.191 143 595 212 8 × 2 = 0 + 0.382 287 190 425 6;
  • 18) 0.382 287 190 425 6 × 2 = 0 + 0.764 574 380 851 2;
  • 19) 0.764 574 380 851 2 × 2 = 1 + 0.529 148 761 702 4;
  • 20) 0.529 148 761 702 4 × 2 = 1 + 0.058 297 523 404 8;
  • 21) 0.058 297 523 404 8 × 2 = 0 + 0.116 595 046 809 6;
  • 22) 0.116 595 046 809 6 × 2 = 0 + 0.233 190 093 619 2;
  • 23) 0.233 190 093 619 2 × 2 = 0 + 0.466 380 187 238 4;
  • 24) 0.466 380 187 238 4 × 2 = 0 + 0.932 760 374 476 8;
  • 25) 0.932 760 374 476 8 × 2 = 1 + 0.865 520 748 953 6;
  • 26) 0.865 520 748 953 6 × 2 = 1 + 0.731 041 497 907 2;
  • 27) 0.731 041 497 907 2 × 2 = 1 + 0.462 082 995 814 4;
  • 28) 0.462 082 995 814 4 × 2 = 0 + 0.924 165 991 628 8;
  • 29) 0.924 165 991 628 8 × 2 = 1 + 0.848 331 983 257 6;
  • 30) 0.848 331 983 257 6 × 2 = 1 + 0.696 663 966 515 2;
  • 31) 0.696 663 966 515 2 × 2 = 1 + 0.393 327 933 030 4;
  • 32) 0.393 327 933 030 4 × 2 = 0 + 0.786 655 866 060 8;
  • 33) 0.786 655 866 060 8 × 2 = 1 + 0.573 311 732 121 6;
  • 34) 0.573 311 732 121 6 × 2 = 1 + 0.146 623 464 243 2;
  • 35) 0.146 623 464 243 2 × 2 = 0 + 0.293 246 928 486 4;
  • 36) 0.293 246 928 486 4 × 2 = 0 + 0.586 493 856 972 8;
  • 37) 0.586 493 856 972 8 × 2 = 1 + 0.172 987 713 945 6;
  • 38) 0.172 987 713 945 6 × 2 = 0 + 0.345 975 427 891 2;
  • 39) 0.345 975 427 891 2 × 2 = 0 + 0.691 950 855 782 4;
  • 40) 0.691 950 855 782 4 × 2 = 1 + 0.383 901 711 564 8;
  • 41) 0.383 901 711 564 8 × 2 = 0 + 0.767 803 423 129 6;
  • 42) 0.767 803 423 129 6 × 2 = 1 + 0.535 606 846 259 2;
  • 43) 0.535 606 846 259 2 × 2 = 1 + 0.071 213 692 518 4;
  • 44) 0.071 213 692 518 4 × 2 = 0 + 0.142 427 385 036 8;
  • 45) 0.142 427 385 036 8 × 2 = 0 + 0.284 854 770 073 6;
  • 46) 0.284 854 770 073 6 × 2 = 0 + 0.569 709 540 147 2;
  • 47) 0.569 709 540 147 2 × 2 = 1 + 0.139 419 080 294 4;
  • 48) 0.139 419 080 294 4 × 2 = 0 + 0.278 838 160 588 8;
  • 49) 0.278 838 160 588 8 × 2 = 0 + 0.557 676 321 177 6;
  • 50) 0.557 676 321 177 6 × 2 = 1 + 0.115 352 642 355 2;
  • 51) 0.115 352 642 355 2 × 2 = 0 + 0.230 705 284 710 4;
  • 52) 0.230 705 284 710 4 × 2 = 0 + 0.461 410 569 420 8;
  • 53) 0.461 410 569 420 8 × 2 = 0 + 0.922 821 138 841 6;
  • 54) 0.922 821 138 841 6 × 2 = 1 + 0.845 642 277 683 2;
  • 55) 0.845 642 277 683 2 × 2 = 1 + 0.691 284 555 366 4;
  • 56) 0.691 284 555 366 4 × 2 = 1 + 0.382 569 110 732 8;
  • 57) 0.382 569 110 732 8 × 2 = 0 + 0.765 138 221 465 6;
  • 58) 0.765 138 221 465 6 × 2 = 1 + 0.530 276 442 931 2;
  • 59) 0.530 276 442 931 2 × 2 = 1 + 0.060 552 885 862 4;
  • 60) 0.060 552 885 862 4 × 2 = 0 + 0.121 105 771 724 8;
  • 61) 0.121 105 771 724 8 × 2 = 0 + 0.242 211 543 449 6;
  • 62) 0.242 211 543 449 6 × 2 = 0 + 0.484 423 086 899 2;
  • 63) 0.484 423 086 899 2 × 2 = 0 + 0.968 846 173 798 4;
  • 64) 0.968 846 173 798 4 × 2 = 1 + 0.937 692 347 596 8;
  • 65) 0.937 692 347 596 8 × 2 = 1 + 0.875 384 695 193 6;
  • 66) 0.875 384 695 193 6 × 2 = 1 + 0.750 769 390 387 2;
  • 67) 0.750 769 390 387 2 × 2 = 1 + 0.501 538 780 774 4;
  • 68) 0.501 538 780 774 4 × 2 = 1 + 0.003 077 561 548 8;
  • 69) 0.003 077 561 548 8 × 2 = 0 + 0.006 155 123 097 6;
  • 70) 0.006 155 123 097 6 × 2 = 0 + 0.012 310 246 195 2;
  • 71) 0.012 310 246 195 2 × 2 = 0 + 0.024 620 492 390 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:


0.000 002 916 619 8(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1001 0110 0010 0100 0111 0110 0001 1111 000(2)

6. Positive number before normalization:

0.000 002 916 619 8(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1001 0110 0010 0100 0111 0110 0001 1111 000(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:


0.000 002 916 619 8(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1001 0110 0010 0100 0111 0110 0001 1111 000(2) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1001 0110 0010 0100 0111 0110 0001 1111 000(2) × 20 =

1.1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 1000(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):
1.1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 1000


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. 1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 1000 =

1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 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 1100

Mantissa (52 bits) =
1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 1000


Decimal number -0.000 002 916 619 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1100 - 1000 0111 0111 0110 0100 1011 0001 0010 0011 1011 0000 1111 1000

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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