-0.000 002 916 622 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 916 622 1(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
-0.000 002 916 622 1(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:

|-0.000 002 916 622 1| = 0.000 002 916 622 1


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: 0.000 002 916 622 1.

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) 0.000 002 916 622 1 × 2 = 0 + 0.000 005 833 244 2;
  • 2) 0.000 005 833 244 2 × 2 = 0 + 0.000 011 666 488 4;
  • 3) 0.000 011 666 488 4 × 2 = 0 + 0.000 023 332 976 8;
  • 4) 0.000 023 332 976 8 × 2 = 0 + 0.000 046 665 953 6;
  • 5) 0.000 046 665 953 6 × 2 = 0 + 0.000 093 331 907 2;
  • 6) 0.000 093 331 907 2 × 2 = 0 + 0.000 186 663 814 4;
  • 7) 0.000 186 663 814 4 × 2 = 0 + 0.000 373 327 628 8;
  • 8) 0.000 373 327 628 8 × 2 = 0 + 0.000 746 655 257 6;
  • 9) 0.000 746 655 257 6 × 2 = 0 + 0.001 493 310 515 2;
  • 10) 0.001 493 310 515 2 × 2 = 0 + 0.002 986 621 030 4;
  • 11) 0.002 986 621 030 4 × 2 = 0 + 0.005 973 242 060 8;
  • 12) 0.005 973 242 060 8 × 2 = 0 + 0.011 946 484 121 6;
  • 13) 0.011 946 484 121 6 × 2 = 0 + 0.023 892 968 243 2;
  • 14) 0.023 892 968 243 2 × 2 = 0 + 0.047 785 936 486 4;
  • 15) 0.047 785 936 486 4 × 2 = 0 + 0.095 571 872 972 8;
  • 16) 0.095 571 872 972 8 × 2 = 0 + 0.191 143 745 945 6;
  • 17) 0.191 143 745 945 6 × 2 = 0 + 0.382 287 491 891 2;
  • 18) 0.382 287 491 891 2 × 2 = 0 + 0.764 574 983 782 4;
  • 19) 0.764 574 983 782 4 × 2 = 1 + 0.529 149 967 564 8;
  • 20) 0.529 149 967 564 8 × 2 = 1 + 0.058 299 935 129 6;
  • 21) 0.058 299 935 129 6 × 2 = 0 + 0.116 599 870 259 2;
  • 22) 0.116 599 870 259 2 × 2 = 0 + 0.233 199 740 518 4;
  • 23) 0.233 199 740 518 4 × 2 = 0 + 0.466 399 481 036 8;
  • 24) 0.466 399 481 036 8 × 2 = 0 + 0.932 798 962 073 6;
  • 25) 0.932 798 962 073 6 × 2 = 1 + 0.865 597 924 147 2;
  • 26) 0.865 597 924 147 2 × 2 = 1 + 0.731 195 848 294 4;
  • 27) 0.731 195 848 294 4 × 2 = 1 + 0.462 391 696 588 8;
  • 28) 0.462 391 696 588 8 × 2 = 0 + 0.924 783 393 177 6;
  • 29) 0.924 783 393 177 6 × 2 = 1 + 0.849 566 786 355 2;
  • 30) 0.849 566 786 355 2 × 2 = 1 + 0.699 133 572 710 4;
  • 31) 0.699 133 572 710 4 × 2 = 1 + 0.398 267 145 420 8;
  • 32) 0.398 267 145 420 8 × 2 = 0 + 0.796 534 290 841 6;
  • 33) 0.796 534 290 841 6 × 2 = 1 + 0.593 068 581 683 2;
  • 34) 0.593 068 581 683 2 × 2 = 1 + 0.186 137 163 366 4;
  • 35) 0.186 137 163 366 4 × 2 = 0 + 0.372 274 326 732 8;
  • 36) 0.372 274 326 732 8 × 2 = 0 + 0.744 548 653 465 6;
  • 37) 0.744 548 653 465 6 × 2 = 1 + 0.489 097 306 931 2;
  • 38) 0.489 097 306 931 2 × 2 = 0 + 0.978 194 613 862 4;
  • 39) 0.978 194 613 862 4 × 2 = 1 + 0.956 389 227 724 8;
  • 40) 0.956 389 227 724 8 × 2 = 1 + 0.912 778 455 449 6;
  • 41) 0.912 778 455 449 6 × 2 = 1 + 0.825 556 910 899 2;
  • 42) 0.825 556 910 899 2 × 2 = 1 + 0.651 113 821 798 4;
  • 43) 0.651 113 821 798 4 × 2 = 1 + 0.302 227 643 596 8;
  • 44) 0.302 227 643 596 8 × 2 = 0 + 0.604 455 287 193 6;
  • 45) 0.604 455 287 193 6 × 2 = 1 + 0.208 910 574 387 2;
  • 46) 0.208 910 574 387 2 × 2 = 0 + 0.417 821 148 774 4;
  • 47) 0.417 821 148 774 4 × 2 = 0 + 0.835 642 297 548 8;
  • 48) 0.835 642 297 548 8 × 2 = 1 + 0.671 284 595 097 6;
  • 49) 0.671 284 595 097 6 × 2 = 1 + 0.342 569 190 195 2;
  • 50) 0.342 569 190 195 2 × 2 = 0 + 0.685 138 380 390 4;
  • 51) 0.685 138 380 390 4 × 2 = 1 + 0.370 276 760 780 8;
  • 52) 0.370 276 760 780 8 × 2 = 0 + 0.740 553 521 561 6;
  • 53) 0.740 553 521 561 6 × 2 = 1 + 0.481 107 043 123 2;
  • 54) 0.481 107 043 123 2 × 2 = 0 + 0.962 214 086 246 4;
  • 55) 0.962 214 086 246 4 × 2 = 1 + 0.924 428 172 492 8;
  • 56) 0.924 428 172 492 8 × 2 = 1 + 0.848 856 344 985 6;
  • 57) 0.848 856 344 985 6 × 2 = 1 + 0.697 712 689 971 2;
  • 58) 0.697 712 689 971 2 × 2 = 1 + 0.395 425 379 942 4;
  • 59) 0.395 425 379 942 4 × 2 = 0 + 0.790 850 759 884 8;
  • 60) 0.790 850 759 884 8 × 2 = 1 + 0.581 701 519 769 6;
  • 61) 0.581 701 519 769 6 × 2 = 1 + 0.163 403 039 539 2;
  • 62) 0.163 403 039 539 2 × 2 = 0 + 0.326 806 079 078 4;
  • 63) 0.326 806 079 078 4 × 2 = 0 + 0.653 612 158 156 8;
  • 64) 0.653 612 158 156 8 × 2 = 1 + 0.307 224 316 313 6;
  • 65) 0.307 224 316 313 6 × 2 = 0 + 0.614 448 632 627 2;
  • 66) 0.614 448 632 627 2 × 2 = 1 + 0.228 897 265 254 4;
  • 67) 0.228 897 265 254 4 × 2 = 0 + 0.457 794 530 508 8;
  • 68) 0.457 794 530 508 8 × 2 = 0 + 0.915 589 061 017 6;
  • 69) 0.915 589 061 017 6 × 2 = 1 + 0.831 178 122 035 2;
  • 70) 0.831 178 122 035 2 × 2 = 1 + 0.662 356 244 070 4;
  • 71) 0.662 356 244 070 4 × 2 = 1 + 0.324 712 488 140 8;

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:


0.000 002 916 622 1(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1011 1110 1001 1010 1011 1101 1001 0100 111(2)

6. Positive number before normalization:

0.000 002 916 622 1(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1011 1110 1001 1010 1011 1101 1001 0100 111(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:


0.000 002 916 622 1(10) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1011 1110 1001 1010 1011 1101 1001 0100 111(2) =

0.0000 0000 0000 0000 0011 0000 1110 1110 1100 1011 1110 1001 1010 1011 1101 1001 0100 111(2) × 20 =

1.1000 0111 0111 0110 0101 1111 0100 1101 0101 1110 1100 1010 0111(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 0101 1111 0100 1101 0101 1110 1100 1010 0111


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 0101 1111 0100 1101 0101 1110 1100 1010 0111 =

1000 0111 0111 0110 0101 1111 0100 1101 0101 1110 1100 1010 0111


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 0101 1111 0100 1101 0101 1110 1100 1010 0111


Decimal number -0.000 002 916 622 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1100 - 1000 0111 0111 0110 0101 1111 0100 1101 0101 1110 1100 1010 0111

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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