-0.016 738 891 601 562 496 725 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 725(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
-0.016 738 891 601 562 496 725(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:

|-0.016 738 891 601 562 496 725| = 0.016 738 891 601 562 496 725


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: 0.016 738 891 601 562 496 725.

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) 0.016 738 891 601 562 496 725 × 2 = 0 + 0.033 477 783 203 124 993 45;
  • 2) 0.033 477 783 203 124 993 45 × 2 = 0 + 0.066 955 566 406 249 986 9;
  • 3) 0.066 955 566 406 249 986 9 × 2 = 0 + 0.133 911 132 812 499 973 8;
  • 4) 0.133 911 132 812 499 973 8 × 2 = 0 + 0.267 822 265 624 999 947 6;
  • 5) 0.267 822 265 624 999 947 6 × 2 = 0 + 0.535 644 531 249 999 895 2;
  • 6) 0.535 644 531 249 999 895 2 × 2 = 1 + 0.071 289 062 499 999 790 4;
  • 7) 0.071 289 062 499 999 790 4 × 2 = 0 + 0.142 578 124 999 999 580 8;
  • 8) 0.142 578 124 999 999 580 8 × 2 = 0 + 0.285 156 249 999 999 161 6;
  • 9) 0.285 156 249 999 999 161 6 × 2 = 0 + 0.570 312 499 999 998 323 2;
  • 10) 0.570 312 499 999 998 323 2 × 2 = 1 + 0.140 624 999 999 996 646 4;
  • 11) 0.140 624 999 999 996 646 4 × 2 = 0 + 0.281 249 999 999 993 292 8;
  • 12) 0.281 249 999 999 993 292 8 × 2 = 0 + 0.562 499 999 999 986 585 6;
  • 13) 0.562 499 999 999 986 585 6 × 2 = 1 + 0.124 999 999 999 973 171 2;
  • 14) 0.124 999 999 999 973 171 2 × 2 = 0 + 0.249 999 999 999 946 342 4;
  • 15) 0.249 999 999 999 946 342 4 × 2 = 0 + 0.499 999 999 999 892 684 8;
  • 16) 0.499 999 999 999 892 684 8 × 2 = 0 + 0.999 999 999 999 785 369 6;
  • 17) 0.999 999 999 999 785 369 6 × 2 = 1 + 0.999 999 999 999 570 739 2;
  • 18) 0.999 999 999 999 570 739 2 × 2 = 1 + 0.999 999 999 999 141 478 4;
  • 19) 0.999 999 999 999 141 478 4 × 2 = 1 + 0.999 999 999 998 282 956 8;
  • 20) 0.999 999 999 998 282 956 8 × 2 = 1 + 0.999 999 999 996 565 913 6;
  • 21) 0.999 999 999 996 565 913 6 × 2 = 1 + 0.999 999 999 993 131 827 2;
  • 22) 0.999 999 999 993 131 827 2 × 2 = 1 + 0.999 999 999 986 263 654 4;
  • 23) 0.999 999 999 986 263 654 4 × 2 = 1 + 0.999 999 999 972 527 308 8;
  • 24) 0.999 999 999 972 527 308 8 × 2 = 1 + 0.999 999 999 945 054 617 6;
  • 25) 0.999 999 999 945 054 617 6 × 2 = 1 + 0.999 999 999 890 109 235 2;
  • 26) 0.999 999 999 890 109 235 2 × 2 = 1 + 0.999 999 999 780 218 470 4;
  • 27) 0.999 999 999 780 218 470 4 × 2 = 1 + 0.999 999 999 560 436 940 8;
  • 28) 0.999 999 999 560 436 940 8 × 2 = 1 + 0.999 999 999 120 873 881 6;
  • 29) 0.999 999 999 120 873 881 6 × 2 = 1 + 0.999 999 998 241 747 763 2;
  • 30) 0.999 999 998 241 747 763 2 × 2 = 1 + 0.999 999 996 483 495 526 4;
  • 31) 0.999 999 996 483 495 526 4 × 2 = 1 + 0.999 999 992 966 991 052 8;
  • 32) 0.999 999 992 966 991 052 8 × 2 = 1 + 0.999 999 985 933 982 105 6;
  • 33) 0.999 999 985 933 982 105 6 × 2 = 1 + 0.999 999 971 867 964 211 2;
  • 34) 0.999 999 971 867 964 211 2 × 2 = 1 + 0.999 999 943 735 928 422 4;
  • 35) 0.999 999 943 735 928 422 4 × 2 = 1 + 0.999 999 887 471 856 844 8;
  • 36) 0.999 999 887 471 856 844 8 × 2 = 1 + 0.999 999 774 943 713 689 6;
  • 37) 0.999 999 774 943 713 689 6 × 2 = 1 + 0.999 999 549 887 427 379 2;
  • 38) 0.999 999 549 887 427 379 2 × 2 = 1 + 0.999 999 099 774 854 758 4;
  • 39) 0.999 999 099 774 854 758 4 × 2 = 1 + 0.999 998 199 549 709 516 8;
  • 40) 0.999 998 199 549 709 516 8 × 2 = 1 + 0.999 996 399 099 419 033 6;
  • 41) 0.999 996 399 099 419 033 6 × 2 = 1 + 0.999 992 798 198 838 067 2;
  • 42) 0.999 992 798 198 838 067 2 × 2 = 1 + 0.999 985 596 397 676 134 4;
  • 43) 0.999 985 596 397 676 134 4 × 2 = 1 + 0.999 971 192 795 352 268 8;
  • 44) 0.999 971 192 795 352 268 8 × 2 = 1 + 0.999 942 385 590 704 537 6;
  • 45) 0.999 942 385 590 704 537 6 × 2 = 1 + 0.999 884 771 181 409 075 2;
  • 46) 0.999 884 771 181 409 075 2 × 2 = 1 + 0.999 769 542 362 818 150 4;
  • 47) 0.999 769 542 362 818 150 4 × 2 = 1 + 0.999 539 084 725 636 300 8;
  • 48) 0.999 539 084 725 636 300 8 × 2 = 1 + 0.999 078 169 451 272 601 6;
  • 49) 0.999 078 169 451 272 601 6 × 2 = 1 + 0.998 156 338 902 545 203 2;
  • 50) 0.998 156 338 902 545 203 2 × 2 = 1 + 0.996 312 677 805 090 406 4;
  • 51) 0.996 312 677 805 090 406 4 × 2 = 1 + 0.992 625 355 610 180 812 8;
  • 52) 0.992 625 355 610 180 812 8 × 2 = 1 + 0.985 250 711 220 361 625 6;
  • 53) 0.985 250 711 220 361 625 6 × 2 = 1 + 0.970 501 422 440 723 251 2;
  • 54) 0.970 501 422 440 723 251 2 × 2 = 1 + 0.941 002 844 881 446 502 4;
  • 55) 0.941 002 844 881 446 502 4 × 2 = 1 + 0.882 005 689 762 893 004 8;
  • 56) 0.882 005 689 762 893 004 8 × 2 = 1 + 0.764 011 379 525 786 009 6;
  • 57) 0.764 011 379 525 786 009 6 × 2 = 1 + 0.528 022 759 051 572 019 2;
  • 58) 0.528 022 759 051 572 019 2 × 2 = 1 + 0.056 045 518 103 144 038 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:


0.016 738 891 601 562 496 725(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 725(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 725(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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;

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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 725 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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