-0.016 738 891 601 562 496 536 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 536 2(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
-0.016 738 891 601 562 496 536 2(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:

|-0.016 738 891 601 562 496 536 2| = 0.016 738 891 601 562 496 536 2


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: 0.016 738 891 601 562 496 536 2.

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) 0.016 738 891 601 562 496 536 2 × 2 = 0 + 0.033 477 783 203 124 993 072 4;
  • 2) 0.033 477 783 203 124 993 072 4 × 2 = 0 + 0.066 955 566 406 249 986 144 8;
  • 3) 0.066 955 566 406 249 986 144 8 × 2 = 0 + 0.133 911 132 812 499 972 289 6;
  • 4) 0.133 911 132 812 499 972 289 6 × 2 = 0 + 0.267 822 265 624 999 944 579 2;
  • 5) 0.267 822 265 624 999 944 579 2 × 2 = 0 + 0.535 644 531 249 999 889 158 4;
  • 6) 0.535 644 531 249 999 889 158 4 × 2 = 1 + 0.071 289 062 499 999 778 316 8;
  • 7) 0.071 289 062 499 999 778 316 8 × 2 = 0 + 0.142 578 124 999 999 556 633 6;
  • 8) 0.142 578 124 999 999 556 633 6 × 2 = 0 + 0.285 156 249 999 999 113 267 2;
  • 9) 0.285 156 249 999 999 113 267 2 × 2 = 0 + 0.570 312 499 999 998 226 534 4;
  • 10) 0.570 312 499 999 998 226 534 4 × 2 = 1 + 0.140 624 999 999 996 453 068 8;
  • 11) 0.140 624 999 999 996 453 068 8 × 2 = 0 + 0.281 249 999 999 992 906 137 6;
  • 12) 0.281 249 999 999 992 906 137 6 × 2 = 0 + 0.562 499 999 999 985 812 275 2;
  • 13) 0.562 499 999 999 985 812 275 2 × 2 = 1 + 0.124 999 999 999 971 624 550 4;
  • 14) 0.124 999 999 999 971 624 550 4 × 2 = 0 + 0.249 999 999 999 943 249 100 8;
  • 15) 0.249 999 999 999 943 249 100 8 × 2 = 0 + 0.499 999 999 999 886 498 201 6;
  • 16) 0.499 999 999 999 886 498 201 6 × 2 = 0 + 0.999 999 999 999 772 996 403 2;
  • 17) 0.999 999 999 999 772 996 403 2 × 2 = 1 + 0.999 999 999 999 545 992 806 4;
  • 18) 0.999 999 999 999 545 992 806 4 × 2 = 1 + 0.999 999 999 999 091 985 612 8;
  • 19) 0.999 999 999 999 091 985 612 8 × 2 = 1 + 0.999 999 999 998 183 971 225 6;
  • 20) 0.999 999 999 998 183 971 225 6 × 2 = 1 + 0.999 999 999 996 367 942 451 2;
  • 21) 0.999 999 999 996 367 942 451 2 × 2 = 1 + 0.999 999 999 992 735 884 902 4;
  • 22) 0.999 999 999 992 735 884 902 4 × 2 = 1 + 0.999 999 999 985 471 769 804 8;
  • 23) 0.999 999 999 985 471 769 804 8 × 2 = 1 + 0.999 999 999 970 943 539 609 6;
  • 24) 0.999 999 999 970 943 539 609 6 × 2 = 1 + 0.999 999 999 941 887 079 219 2;
  • 25) 0.999 999 999 941 887 079 219 2 × 2 = 1 + 0.999 999 999 883 774 158 438 4;
  • 26) 0.999 999 999 883 774 158 438 4 × 2 = 1 + 0.999 999 999 767 548 316 876 8;
  • 27) 0.999 999 999 767 548 316 876 8 × 2 = 1 + 0.999 999 999 535 096 633 753 6;
  • 28) 0.999 999 999 535 096 633 753 6 × 2 = 1 + 0.999 999 999 070 193 267 507 2;
  • 29) 0.999 999 999 070 193 267 507 2 × 2 = 1 + 0.999 999 998 140 386 535 014 4;
  • 30) 0.999 999 998 140 386 535 014 4 × 2 = 1 + 0.999 999 996 280 773 070 028 8;
  • 31) 0.999 999 996 280 773 070 028 8 × 2 = 1 + 0.999 999 992 561 546 140 057 6;
  • 32) 0.999 999 992 561 546 140 057 6 × 2 = 1 + 0.999 999 985 123 092 280 115 2;
  • 33) 0.999 999 985 123 092 280 115 2 × 2 = 1 + 0.999 999 970 246 184 560 230 4;
  • 34) 0.999 999 970 246 184 560 230 4 × 2 = 1 + 0.999 999 940 492 369 120 460 8;
  • 35) 0.999 999 940 492 369 120 460 8 × 2 = 1 + 0.999 999 880 984 738 240 921 6;
  • 36) 0.999 999 880 984 738 240 921 6 × 2 = 1 + 0.999 999 761 969 476 481 843 2;
  • 37) 0.999 999 761 969 476 481 843 2 × 2 = 1 + 0.999 999 523 938 952 963 686 4;
  • 38) 0.999 999 523 938 952 963 686 4 × 2 = 1 + 0.999 999 047 877 905 927 372 8;
  • 39) 0.999 999 047 877 905 927 372 8 × 2 = 1 + 0.999 998 095 755 811 854 745 6;
  • 40) 0.999 998 095 755 811 854 745 6 × 2 = 1 + 0.999 996 191 511 623 709 491 2;
  • 41) 0.999 996 191 511 623 709 491 2 × 2 = 1 + 0.999 992 383 023 247 418 982 4;
  • 42) 0.999 992 383 023 247 418 982 4 × 2 = 1 + 0.999 984 766 046 494 837 964 8;
  • 43) 0.999 984 766 046 494 837 964 8 × 2 = 1 + 0.999 969 532 092 989 675 929 6;
  • 44) 0.999 969 532 092 989 675 929 6 × 2 = 1 + 0.999 939 064 185 979 351 859 2;
  • 45) 0.999 939 064 185 979 351 859 2 × 2 = 1 + 0.999 878 128 371 958 703 718 4;
  • 46) 0.999 878 128 371 958 703 718 4 × 2 = 1 + 0.999 756 256 743 917 407 436 8;
  • 47) 0.999 756 256 743 917 407 436 8 × 2 = 1 + 0.999 512 513 487 834 814 873 6;
  • 48) 0.999 512 513 487 834 814 873 6 × 2 = 1 + 0.999 025 026 975 669 629 747 2;
  • 49) 0.999 025 026 975 669 629 747 2 × 2 = 1 + 0.998 050 053 951 339 259 494 4;
  • 50) 0.998 050 053 951 339 259 494 4 × 2 = 1 + 0.996 100 107 902 678 518 988 8;
  • 51) 0.996 100 107 902 678 518 988 8 × 2 = 1 + 0.992 200 215 805 357 037 977 6;
  • 52) 0.992 200 215 805 357 037 977 6 × 2 = 1 + 0.984 400 431 610 714 075 955 2;
  • 53) 0.984 400 431 610 714 075 955 2 × 2 = 1 + 0.968 800 863 221 428 151 910 4;
  • 54) 0.968 800 863 221 428 151 910 4 × 2 = 1 + 0.937 601 726 442 856 303 820 8;
  • 55) 0.937 601 726 442 856 303 820 8 × 2 = 1 + 0.875 203 452 885 712 607 641 6;
  • 56) 0.875 203 452 885 712 607 641 6 × 2 = 1 + 0.750 406 905 771 425 215 283 2;
  • 57) 0.750 406 905 771 425 215 283 2 × 2 = 1 + 0.500 813 811 542 850 430 566 4;
  • 58) 0.500 813 811 542 850 430 566 4 × 2 = 1 + 0.001 627 623 085 700 861 132 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:


0.016 738 891 601 562 496 536 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 536 2(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:


0.016 738 891 601 562 496 536 2(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 -0.016 738 891 601 562 496 536 2 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