-0.016 738 891 601 562 496 552 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 552 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
-0.016 738 891 601 562 496 552 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:

|-0.016 738 891 601 562 496 552 2| = 0.016 738 891 601 562 496 552 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: 0.016 738 891 601 562 496 552 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) 0.016 738 891 601 562 496 552 2 × 2 = 0 + 0.033 477 783 203 124 993 104 4;
  • 2) 0.033 477 783 203 124 993 104 4 × 2 = 0 + 0.066 955 566 406 249 986 208 8;
  • 3) 0.066 955 566 406 249 986 208 8 × 2 = 0 + 0.133 911 132 812 499 972 417 6;
  • 4) 0.133 911 132 812 499 972 417 6 × 2 = 0 + 0.267 822 265 624 999 944 835 2;
  • 5) 0.267 822 265 624 999 944 835 2 × 2 = 0 + 0.535 644 531 249 999 889 670 4;
  • 6) 0.535 644 531 249 999 889 670 4 × 2 = 1 + 0.071 289 062 499 999 779 340 8;
  • 7) 0.071 289 062 499 999 779 340 8 × 2 = 0 + 0.142 578 124 999 999 558 681 6;
  • 8) 0.142 578 124 999 999 558 681 6 × 2 = 0 + 0.285 156 249 999 999 117 363 2;
  • 9) 0.285 156 249 999 999 117 363 2 × 2 = 0 + 0.570 312 499 999 998 234 726 4;
  • 10) 0.570 312 499 999 998 234 726 4 × 2 = 1 + 0.140 624 999 999 996 469 452 8;
  • 11) 0.140 624 999 999 996 469 452 8 × 2 = 0 + 0.281 249 999 999 992 938 905 6;
  • 12) 0.281 249 999 999 992 938 905 6 × 2 = 0 + 0.562 499 999 999 985 877 811 2;
  • 13) 0.562 499 999 999 985 877 811 2 × 2 = 1 + 0.124 999 999 999 971 755 622 4;
  • 14) 0.124 999 999 999 971 755 622 4 × 2 = 0 + 0.249 999 999 999 943 511 244 8;
  • 15) 0.249 999 999 999 943 511 244 8 × 2 = 0 + 0.499 999 999 999 887 022 489 6;
  • 16) 0.499 999 999 999 887 022 489 6 × 2 = 0 + 0.999 999 999 999 774 044 979 2;
  • 17) 0.999 999 999 999 774 044 979 2 × 2 = 1 + 0.999 999 999 999 548 089 958 4;
  • 18) 0.999 999 999 999 548 089 958 4 × 2 = 1 + 0.999 999 999 999 096 179 916 8;
  • 19) 0.999 999 999 999 096 179 916 8 × 2 = 1 + 0.999 999 999 998 192 359 833 6;
  • 20) 0.999 999 999 998 192 359 833 6 × 2 = 1 + 0.999 999 999 996 384 719 667 2;
  • 21) 0.999 999 999 996 384 719 667 2 × 2 = 1 + 0.999 999 999 992 769 439 334 4;
  • 22) 0.999 999 999 992 769 439 334 4 × 2 = 1 + 0.999 999 999 985 538 878 668 8;
  • 23) 0.999 999 999 985 538 878 668 8 × 2 = 1 + 0.999 999 999 971 077 757 337 6;
  • 24) 0.999 999 999 971 077 757 337 6 × 2 = 1 + 0.999 999 999 942 155 514 675 2;
  • 25) 0.999 999 999 942 155 514 675 2 × 2 = 1 + 0.999 999 999 884 311 029 350 4;
  • 26) 0.999 999 999 884 311 029 350 4 × 2 = 1 + 0.999 999 999 768 622 058 700 8;
  • 27) 0.999 999 999 768 622 058 700 8 × 2 = 1 + 0.999 999 999 537 244 117 401 6;
  • 28) 0.999 999 999 537 244 117 401 6 × 2 = 1 + 0.999 999 999 074 488 234 803 2;
  • 29) 0.999 999 999 074 488 234 803 2 × 2 = 1 + 0.999 999 998 148 976 469 606 4;
  • 30) 0.999 999 998 148 976 469 606 4 × 2 = 1 + 0.999 999 996 297 952 939 212 8;
  • 31) 0.999 999 996 297 952 939 212 8 × 2 = 1 + 0.999 999 992 595 905 878 425 6;
  • 32) 0.999 999 992 595 905 878 425 6 × 2 = 1 + 0.999 999 985 191 811 756 851 2;
  • 33) 0.999 999 985 191 811 756 851 2 × 2 = 1 + 0.999 999 970 383 623 513 702 4;
  • 34) 0.999 999 970 383 623 513 702 4 × 2 = 1 + 0.999 999 940 767 247 027 404 8;
  • 35) 0.999 999 940 767 247 027 404 8 × 2 = 1 + 0.999 999 881 534 494 054 809 6;
  • 36) 0.999 999 881 534 494 054 809 6 × 2 = 1 + 0.999 999 763 068 988 109 619 2;
  • 37) 0.999 999 763 068 988 109 619 2 × 2 = 1 + 0.999 999 526 137 976 219 238 4;
  • 38) 0.999 999 526 137 976 219 238 4 × 2 = 1 + 0.999 999 052 275 952 438 476 8;
  • 39) 0.999 999 052 275 952 438 476 8 × 2 = 1 + 0.999 998 104 551 904 876 953 6;
  • 40) 0.999 998 104 551 904 876 953 6 × 2 = 1 + 0.999 996 209 103 809 753 907 2;
  • 41) 0.999 996 209 103 809 753 907 2 × 2 = 1 + 0.999 992 418 207 619 507 814 4;
  • 42) 0.999 992 418 207 619 507 814 4 × 2 = 1 + 0.999 984 836 415 239 015 628 8;
  • 43) 0.999 984 836 415 239 015 628 8 × 2 = 1 + 0.999 969 672 830 478 031 257 6;
  • 44) 0.999 969 672 830 478 031 257 6 × 2 = 1 + 0.999 939 345 660 956 062 515 2;
  • 45) 0.999 939 345 660 956 062 515 2 × 2 = 1 + 0.999 878 691 321 912 125 030 4;
  • 46) 0.999 878 691 321 912 125 030 4 × 2 = 1 + 0.999 757 382 643 824 250 060 8;
  • 47) 0.999 757 382 643 824 250 060 8 × 2 = 1 + 0.999 514 765 287 648 500 121 6;
  • 48) 0.999 514 765 287 648 500 121 6 × 2 = 1 + 0.999 029 530 575 297 000 243 2;
  • 49) 0.999 029 530 575 297 000 243 2 × 2 = 1 + 0.998 059 061 150 594 000 486 4;
  • 50) 0.998 059 061 150 594 000 486 4 × 2 = 1 + 0.996 118 122 301 188 000 972 8;
  • 51) 0.996 118 122 301 188 000 972 8 × 2 = 1 + 0.992 236 244 602 376 001 945 6;
  • 52) 0.992 236 244 602 376 001 945 6 × 2 = 1 + 0.984 472 489 204 752 003 891 2;
  • 53) 0.984 472 489 204 752 003 891 2 × 2 = 1 + 0.968 944 978 409 504 007 782 4;
  • 54) 0.968 944 978 409 504 007 782 4 × 2 = 1 + 0.937 889 956 819 008 015 564 8;
  • 55) 0.937 889 956 819 008 015 564 8 × 2 = 1 + 0.875 779 913 638 016 031 129 6;
  • 56) 0.875 779 913 638 016 031 129 6 × 2 = 1 + 0.751 559 827 276 032 062 259 2;
  • 57) 0.751 559 827 276 032 062 259 2 × 2 = 1 + 0.503 119 654 552 064 124 518 4;
  • 58) 0.503 119 654 552 064 124 518 4 × 2 = 1 + 0.006 239 309 104 128 249 036 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:


0.016 738 891 601 562 496 552 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 552 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:


0.016 738 891 601 562 496 552 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 -0.016 738 891 601 562 496 552 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