-0.016 738 891 601 562 496 520 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 520 3(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
-0.016 738 891 601 562 496 520 3(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:

|-0.016 738 891 601 562 496 520 3| = 0.016 738 891 601 562 496 520 3


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: 0.016 738 891 601 562 496 520 3.

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) 0.016 738 891 601 562 496 520 3 × 2 = 0 + 0.033 477 783 203 124 993 040 6;
  • 2) 0.033 477 783 203 124 993 040 6 × 2 = 0 + 0.066 955 566 406 249 986 081 2;
  • 3) 0.066 955 566 406 249 986 081 2 × 2 = 0 + 0.133 911 132 812 499 972 162 4;
  • 4) 0.133 911 132 812 499 972 162 4 × 2 = 0 + 0.267 822 265 624 999 944 324 8;
  • 5) 0.267 822 265 624 999 944 324 8 × 2 = 0 + 0.535 644 531 249 999 888 649 6;
  • 6) 0.535 644 531 249 999 888 649 6 × 2 = 1 + 0.071 289 062 499 999 777 299 2;
  • 7) 0.071 289 062 499 999 777 299 2 × 2 = 0 + 0.142 578 124 999 999 554 598 4;
  • 8) 0.142 578 124 999 999 554 598 4 × 2 = 0 + 0.285 156 249 999 999 109 196 8;
  • 9) 0.285 156 249 999 999 109 196 8 × 2 = 0 + 0.570 312 499 999 998 218 393 6;
  • 10) 0.570 312 499 999 998 218 393 6 × 2 = 1 + 0.140 624 999 999 996 436 787 2;
  • 11) 0.140 624 999 999 996 436 787 2 × 2 = 0 + 0.281 249 999 999 992 873 574 4;
  • 12) 0.281 249 999 999 992 873 574 4 × 2 = 0 + 0.562 499 999 999 985 747 148 8;
  • 13) 0.562 499 999 999 985 747 148 8 × 2 = 1 + 0.124 999 999 999 971 494 297 6;
  • 14) 0.124 999 999 999 971 494 297 6 × 2 = 0 + 0.249 999 999 999 942 988 595 2;
  • 15) 0.249 999 999 999 942 988 595 2 × 2 = 0 + 0.499 999 999 999 885 977 190 4;
  • 16) 0.499 999 999 999 885 977 190 4 × 2 = 0 + 0.999 999 999 999 771 954 380 8;
  • 17) 0.999 999 999 999 771 954 380 8 × 2 = 1 + 0.999 999 999 999 543 908 761 6;
  • 18) 0.999 999 999 999 543 908 761 6 × 2 = 1 + 0.999 999 999 999 087 817 523 2;
  • 19) 0.999 999 999 999 087 817 523 2 × 2 = 1 + 0.999 999 999 998 175 635 046 4;
  • 20) 0.999 999 999 998 175 635 046 4 × 2 = 1 + 0.999 999 999 996 351 270 092 8;
  • 21) 0.999 999 999 996 351 270 092 8 × 2 = 1 + 0.999 999 999 992 702 540 185 6;
  • 22) 0.999 999 999 992 702 540 185 6 × 2 = 1 + 0.999 999 999 985 405 080 371 2;
  • 23) 0.999 999 999 985 405 080 371 2 × 2 = 1 + 0.999 999 999 970 810 160 742 4;
  • 24) 0.999 999 999 970 810 160 742 4 × 2 = 1 + 0.999 999 999 941 620 321 484 8;
  • 25) 0.999 999 999 941 620 321 484 8 × 2 = 1 + 0.999 999 999 883 240 642 969 6;
  • 26) 0.999 999 999 883 240 642 969 6 × 2 = 1 + 0.999 999 999 766 481 285 939 2;
  • 27) 0.999 999 999 766 481 285 939 2 × 2 = 1 + 0.999 999 999 532 962 571 878 4;
  • 28) 0.999 999 999 532 962 571 878 4 × 2 = 1 + 0.999 999 999 065 925 143 756 8;
  • 29) 0.999 999 999 065 925 143 756 8 × 2 = 1 + 0.999 999 998 131 850 287 513 6;
  • 30) 0.999 999 998 131 850 287 513 6 × 2 = 1 + 0.999 999 996 263 700 575 027 2;
  • 31) 0.999 999 996 263 700 575 027 2 × 2 = 1 + 0.999 999 992 527 401 150 054 4;
  • 32) 0.999 999 992 527 401 150 054 4 × 2 = 1 + 0.999 999 985 054 802 300 108 8;
  • 33) 0.999 999 985 054 802 300 108 8 × 2 = 1 + 0.999 999 970 109 604 600 217 6;
  • 34) 0.999 999 970 109 604 600 217 6 × 2 = 1 + 0.999 999 940 219 209 200 435 2;
  • 35) 0.999 999 940 219 209 200 435 2 × 2 = 1 + 0.999 999 880 438 418 400 870 4;
  • 36) 0.999 999 880 438 418 400 870 4 × 2 = 1 + 0.999 999 760 876 836 801 740 8;
  • 37) 0.999 999 760 876 836 801 740 8 × 2 = 1 + 0.999 999 521 753 673 603 481 6;
  • 38) 0.999 999 521 753 673 603 481 6 × 2 = 1 + 0.999 999 043 507 347 206 963 2;
  • 39) 0.999 999 043 507 347 206 963 2 × 2 = 1 + 0.999 998 087 014 694 413 926 4;
  • 40) 0.999 998 087 014 694 413 926 4 × 2 = 1 + 0.999 996 174 029 388 827 852 8;
  • 41) 0.999 996 174 029 388 827 852 8 × 2 = 1 + 0.999 992 348 058 777 655 705 6;
  • 42) 0.999 992 348 058 777 655 705 6 × 2 = 1 + 0.999 984 696 117 555 311 411 2;
  • 43) 0.999 984 696 117 555 311 411 2 × 2 = 1 + 0.999 969 392 235 110 622 822 4;
  • 44) 0.999 969 392 235 110 622 822 4 × 2 = 1 + 0.999 938 784 470 221 245 644 8;
  • 45) 0.999 938 784 470 221 245 644 8 × 2 = 1 + 0.999 877 568 940 442 491 289 6;
  • 46) 0.999 877 568 940 442 491 289 6 × 2 = 1 + 0.999 755 137 880 884 982 579 2;
  • 47) 0.999 755 137 880 884 982 579 2 × 2 = 1 + 0.999 510 275 761 769 965 158 4;
  • 48) 0.999 510 275 761 769 965 158 4 × 2 = 1 + 0.999 020 551 523 539 930 316 8;
  • 49) 0.999 020 551 523 539 930 316 8 × 2 = 1 + 0.998 041 103 047 079 860 633 6;
  • 50) 0.998 041 103 047 079 860 633 6 × 2 = 1 + 0.996 082 206 094 159 721 267 2;
  • 51) 0.996 082 206 094 159 721 267 2 × 2 = 1 + 0.992 164 412 188 319 442 534 4;
  • 52) 0.992 164 412 188 319 442 534 4 × 2 = 1 + 0.984 328 824 376 638 885 068 8;
  • 53) 0.984 328 824 376 638 885 068 8 × 2 = 1 + 0.968 657 648 753 277 770 137 6;
  • 54) 0.968 657 648 753 277 770 137 6 × 2 = 1 + 0.937 315 297 506 555 540 275 2;
  • 55) 0.937 315 297 506 555 540 275 2 × 2 = 1 + 0.874 630 595 013 111 080 550 4;
  • 56) 0.874 630 595 013 111 080 550 4 × 2 = 1 + 0.749 261 190 026 222 161 100 8;
  • 57) 0.749 261 190 026 222 161 100 8 × 2 = 1 + 0.498 522 380 052 444 322 201 6;
  • 58) 0.498 522 380 052 444 322 201 6 × 2 = 0 + 0.997 044 760 104 888 644 403 2;

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:


0.016 738 891 601 562 496 520 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 520 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 520 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 520 3 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 1110

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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