-0.016 738 891 601 562 496 535 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 535 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
-0.016 738 891 601 562 496 535 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:

|-0.016 738 891 601 562 496 535 1| = 0.016 738 891 601 562 496 535 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: 0.016 738 891 601 562 496 535 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) 0.016 738 891 601 562 496 535 1 × 2 = 0 + 0.033 477 783 203 124 993 070 2;
  • 2) 0.033 477 783 203 124 993 070 2 × 2 = 0 + 0.066 955 566 406 249 986 140 4;
  • 3) 0.066 955 566 406 249 986 140 4 × 2 = 0 + 0.133 911 132 812 499 972 280 8;
  • 4) 0.133 911 132 812 499 972 280 8 × 2 = 0 + 0.267 822 265 624 999 944 561 6;
  • 5) 0.267 822 265 624 999 944 561 6 × 2 = 0 + 0.535 644 531 249 999 889 123 2;
  • 6) 0.535 644 531 249 999 889 123 2 × 2 = 1 + 0.071 289 062 499 999 778 246 4;
  • 7) 0.071 289 062 499 999 778 246 4 × 2 = 0 + 0.142 578 124 999 999 556 492 8;
  • 8) 0.142 578 124 999 999 556 492 8 × 2 = 0 + 0.285 156 249 999 999 112 985 6;
  • 9) 0.285 156 249 999 999 112 985 6 × 2 = 0 + 0.570 312 499 999 998 225 971 2;
  • 10) 0.570 312 499 999 998 225 971 2 × 2 = 1 + 0.140 624 999 999 996 451 942 4;
  • 11) 0.140 624 999 999 996 451 942 4 × 2 = 0 + 0.281 249 999 999 992 903 884 8;
  • 12) 0.281 249 999 999 992 903 884 8 × 2 = 0 + 0.562 499 999 999 985 807 769 6;
  • 13) 0.562 499 999 999 985 807 769 6 × 2 = 1 + 0.124 999 999 999 971 615 539 2;
  • 14) 0.124 999 999 999 971 615 539 2 × 2 = 0 + 0.249 999 999 999 943 231 078 4;
  • 15) 0.249 999 999 999 943 231 078 4 × 2 = 0 + 0.499 999 999 999 886 462 156 8;
  • 16) 0.499 999 999 999 886 462 156 8 × 2 = 0 + 0.999 999 999 999 772 924 313 6;
  • 17) 0.999 999 999 999 772 924 313 6 × 2 = 1 + 0.999 999 999 999 545 848 627 2;
  • 18) 0.999 999 999 999 545 848 627 2 × 2 = 1 + 0.999 999 999 999 091 697 254 4;
  • 19) 0.999 999 999 999 091 697 254 4 × 2 = 1 + 0.999 999 999 998 183 394 508 8;
  • 20) 0.999 999 999 998 183 394 508 8 × 2 = 1 + 0.999 999 999 996 366 789 017 6;
  • 21) 0.999 999 999 996 366 789 017 6 × 2 = 1 + 0.999 999 999 992 733 578 035 2;
  • 22) 0.999 999 999 992 733 578 035 2 × 2 = 1 + 0.999 999 999 985 467 156 070 4;
  • 23) 0.999 999 999 985 467 156 070 4 × 2 = 1 + 0.999 999 999 970 934 312 140 8;
  • 24) 0.999 999 999 970 934 312 140 8 × 2 = 1 + 0.999 999 999 941 868 624 281 6;
  • 25) 0.999 999 999 941 868 624 281 6 × 2 = 1 + 0.999 999 999 883 737 248 563 2;
  • 26) 0.999 999 999 883 737 248 563 2 × 2 = 1 + 0.999 999 999 767 474 497 126 4;
  • 27) 0.999 999 999 767 474 497 126 4 × 2 = 1 + 0.999 999 999 534 948 994 252 8;
  • 28) 0.999 999 999 534 948 994 252 8 × 2 = 1 + 0.999 999 999 069 897 988 505 6;
  • 29) 0.999 999 999 069 897 988 505 6 × 2 = 1 + 0.999 999 998 139 795 977 011 2;
  • 30) 0.999 999 998 139 795 977 011 2 × 2 = 1 + 0.999 999 996 279 591 954 022 4;
  • 31) 0.999 999 996 279 591 954 022 4 × 2 = 1 + 0.999 999 992 559 183 908 044 8;
  • 32) 0.999 999 992 559 183 908 044 8 × 2 = 1 + 0.999 999 985 118 367 816 089 6;
  • 33) 0.999 999 985 118 367 816 089 6 × 2 = 1 + 0.999 999 970 236 735 632 179 2;
  • 34) 0.999 999 970 236 735 632 179 2 × 2 = 1 + 0.999 999 940 473 471 264 358 4;
  • 35) 0.999 999 940 473 471 264 358 4 × 2 = 1 + 0.999 999 880 946 942 528 716 8;
  • 36) 0.999 999 880 946 942 528 716 8 × 2 = 1 + 0.999 999 761 893 885 057 433 6;
  • 37) 0.999 999 761 893 885 057 433 6 × 2 = 1 + 0.999 999 523 787 770 114 867 2;
  • 38) 0.999 999 523 787 770 114 867 2 × 2 = 1 + 0.999 999 047 575 540 229 734 4;
  • 39) 0.999 999 047 575 540 229 734 4 × 2 = 1 + 0.999 998 095 151 080 459 468 8;
  • 40) 0.999 998 095 151 080 459 468 8 × 2 = 1 + 0.999 996 190 302 160 918 937 6;
  • 41) 0.999 996 190 302 160 918 937 6 × 2 = 1 + 0.999 992 380 604 321 837 875 2;
  • 42) 0.999 992 380 604 321 837 875 2 × 2 = 1 + 0.999 984 761 208 643 675 750 4;
  • 43) 0.999 984 761 208 643 675 750 4 × 2 = 1 + 0.999 969 522 417 287 351 500 8;
  • 44) 0.999 969 522 417 287 351 500 8 × 2 = 1 + 0.999 939 044 834 574 703 001 6;
  • 45) 0.999 939 044 834 574 703 001 6 × 2 = 1 + 0.999 878 089 669 149 406 003 2;
  • 46) 0.999 878 089 669 149 406 003 2 × 2 = 1 + 0.999 756 179 338 298 812 006 4;
  • 47) 0.999 756 179 338 298 812 006 4 × 2 = 1 + 0.999 512 358 676 597 624 012 8;
  • 48) 0.999 512 358 676 597 624 012 8 × 2 = 1 + 0.999 024 717 353 195 248 025 6;
  • 49) 0.999 024 717 353 195 248 025 6 × 2 = 1 + 0.998 049 434 706 390 496 051 2;
  • 50) 0.998 049 434 706 390 496 051 2 × 2 = 1 + 0.996 098 869 412 780 992 102 4;
  • 51) 0.996 098 869 412 780 992 102 4 × 2 = 1 + 0.992 197 738 825 561 984 204 8;
  • 52) 0.992 197 738 825 561 984 204 8 × 2 = 1 + 0.984 395 477 651 123 968 409 6;
  • 53) 0.984 395 477 651 123 968 409 6 × 2 = 1 + 0.968 790 955 302 247 936 819 2;
  • 54) 0.968 790 955 302 247 936 819 2 × 2 = 1 + 0.937 581 910 604 495 873 638 4;
  • 55) 0.937 581 910 604 495 873 638 4 × 2 = 1 + 0.875 163 821 208 991 747 276 8;
  • 56) 0.875 163 821 208 991 747 276 8 × 2 = 1 + 0.750 327 642 417 983 494 553 6;
  • 57) 0.750 327 642 417 983 494 553 6 × 2 = 1 + 0.500 655 284 835 966 989 107 2;
  • 58) 0.500 655 284 835 966 989 107 2 × 2 = 1 + 0.001 310 569 671 933 978 214 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:


0.016 738 891 601 562 496 535 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 535 1(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:


0.016 738 891 601 562 496 535 1(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 -0.016 738 891 601 562 496 535 1 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