-0.016 738 891 601 562 496 538 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 538 6(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
-0.016 738 891 601 562 496 538 6(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:

|-0.016 738 891 601 562 496 538 6| = 0.016 738 891 601 562 496 538 6


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: 0.016 738 891 601 562 496 538 6.

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) 0.016 738 891 601 562 496 538 6 × 2 = 0 + 0.033 477 783 203 124 993 077 2;
  • 2) 0.033 477 783 203 124 993 077 2 × 2 = 0 + 0.066 955 566 406 249 986 154 4;
  • 3) 0.066 955 566 406 249 986 154 4 × 2 = 0 + 0.133 911 132 812 499 972 308 8;
  • 4) 0.133 911 132 812 499 972 308 8 × 2 = 0 + 0.267 822 265 624 999 944 617 6;
  • 5) 0.267 822 265 624 999 944 617 6 × 2 = 0 + 0.535 644 531 249 999 889 235 2;
  • 6) 0.535 644 531 249 999 889 235 2 × 2 = 1 + 0.071 289 062 499 999 778 470 4;
  • 7) 0.071 289 062 499 999 778 470 4 × 2 = 0 + 0.142 578 124 999 999 556 940 8;
  • 8) 0.142 578 124 999 999 556 940 8 × 2 = 0 + 0.285 156 249 999 999 113 881 6;
  • 9) 0.285 156 249 999 999 113 881 6 × 2 = 0 + 0.570 312 499 999 998 227 763 2;
  • 10) 0.570 312 499 999 998 227 763 2 × 2 = 1 + 0.140 624 999 999 996 455 526 4;
  • 11) 0.140 624 999 999 996 455 526 4 × 2 = 0 + 0.281 249 999 999 992 911 052 8;
  • 12) 0.281 249 999 999 992 911 052 8 × 2 = 0 + 0.562 499 999 999 985 822 105 6;
  • 13) 0.562 499 999 999 985 822 105 6 × 2 = 1 + 0.124 999 999 999 971 644 211 2;
  • 14) 0.124 999 999 999 971 644 211 2 × 2 = 0 + 0.249 999 999 999 943 288 422 4;
  • 15) 0.249 999 999 999 943 288 422 4 × 2 = 0 + 0.499 999 999 999 886 576 844 8;
  • 16) 0.499 999 999 999 886 576 844 8 × 2 = 0 + 0.999 999 999 999 773 153 689 6;
  • 17) 0.999 999 999 999 773 153 689 6 × 2 = 1 + 0.999 999 999 999 546 307 379 2;
  • 18) 0.999 999 999 999 546 307 379 2 × 2 = 1 + 0.999 999 999 999 092 614 758 4;
  • 19) 0.999 999 999 999 092 614 758 4 × 2 = 1 + 0.999 999 999 998 185 229 516 8;
  • 20) 0.999 999 999 998 185 229 516 8 × 2 = 1 + 0.999 999 999 996 370 459 033 6;
  • 21) 0.999 999 999 996 370 459 033 6 × 2 = 1 + 0.999 999 999 992 740 918 067 2;
  • 22) 0.999 999 999 992 740 918 067 2 × 2 = 1 + 0.999 999 999 985 481 836 134 4;
  • 23) 0.999 999 999 985 481 836 134 4 × 2 = 1 + 0.999 999 999 970 963 672 268 8;
  • 24) 0.999 999 999 970 963 672 268 8 × 2 = 1 + 0.999 999 999 941 927 344 537 6;
  • 25) 0.999 999 999 941 927 344 537 6 × 2 = 1 + 0.999 999 999 883 854 689 075 2;
  • 26) 0.999 999 999 883 854 689 075 2 × 2 = 1 + 0.999 999 999 767 709 378 150 4;
  • 27) 0.999 999 999 767 709 378 150 4 × 2 = 1 + 0.999 999 999 535 418 756 300 8;
  • 28) 0.999 999 999 535 418 756 300 8 × 2 = 1 + 0.999 999 999 070 837 512 601 6;
  • 29) 0.999 999 999 070 837 512 601 6 × 2 = 1 + 0.999 999 998 141 675 025 203 2;
  • 30) 0.999 999 998 141 675 025 203 2 × 2 = 1 + 0.999 999 996 283 350 050 406 4;
  • 31) 0.999 999 996 283 350 050 406 4 × 2 = 1 + 0.999 999 992 566 700 100 812 8;
  • 32) 0.999 999 992 566 700 100 812 8 × 2 = 1 + 0.999 999 985 133 400 201 625 6;
  • 33) 0.999 999 985 133 400 201 625 6 × 2 = 1 + 0.999 999 970 266 800 403 251 2;
  • 34) 0.999 999 970 266 800 403 251 2 × 2 = 1 + 0.999 999 940 533 600 806 502 4;
  • 35) 0.999 999 940 533 600 806 502 4 × 2 = 1 + 0.999 999 881 067 201 613 004 8;
  • 36) 0.999 999 881 067 201 613 004 8 × 2 = 1 + 0.999 999 762 134 403 226 009 6;
  • 37) 0.999 999 762 134 403 226 009 6 × 2 = 1 + 0.999 999 524 268 806 452 019 2;
  • 38) 0.999 999 524 268 806 452 019 2 × 2 = 1 + 0.999 999 048 537 612 904 038 4;
  • 39) 0.999 999 048 537 612 904 038 4 × 2 = 1 + 0.999 998 097 075 225 808 076 8;
  • 40) 0.999 998 097 075 225 808 076 8 × 2 = 1 + 0.999 996 194 150 451 616 153 6;
  • 41) 0.999 996 194 150 451 616 153 6 × 2 = 1 + 0.999 992 388 300 903 232 307 2;
  • 42) 0.999 992 388 300 903 232 307 2 × 2 = 1 + 0.999 984 776 601 806 464 614 4;
  • 43) 0.999 984 776 601 806 464 614 4 × 2 = 1 + 0.999 969 553 203 612 929 228 8;
  • 44) 0.999 969 553 203 612 929 228 8 × 2 = 1 + 0.999 939 106 407 225 858 457 6;
  • 45) 0.999 939 106 407 225 858 457 6 × 2 = 1 + 0.999 878 212 814 451 716 915 2;
  • 46) 0.999 878 212 814 451 716 915 2 × 2 = 1 + 0.999 756 425 628 903 433 830 4;
  • 47) 0.999 756 425 628 903 433 830 4 × 2 = 1 + 0.999 512 851 257 806 867 660 8;
  • 48) 0.999 512 851 257 806 867 660 8 × 2 = 1 + 0.999 025 702 515 613 735 321 6;
  • 49) 0.999 025 702 515 613 735 321 6 × 2 = 1 + 0.998 051 405 031 227 470 643 2;
  • 50) 0.998 051 405 031 227 470 643 2 × 2 = 1 + 0.996 102 810 062 454 941 286 4;
  • 51) 0.996 102 810 062 454 941 286 4 × 2 = 1 + 0.992 205 620 124 909 882 572 8;
  • 52) 0.992 205 620 124 909 882 572 8 × 2 = 1 + 0.984 411 240 249 819 765 145 6;
  • 53) 0.984 411 240 249 819 765 145 6 × 2 = 1 + 0.968 822 480 499 639 530 291 2;
  • 54) 0.968 822 480 499 639 530 291 2 × 2 = 1 + 0.937 644 960 999 279 060 582 4;
  • 55) 0.937 644 960 999 279 060 582 4 × 2 = 1 + 0.875 289 921 998 558 121 164 8;
  • 56) 0.875 289 921 998 558 121 164 8 × 2 = 1 + 0.750 579 843 997 116 242 329 6;
  • 57) 0.750 579 843 997 116 242 329 6 × 2 = 1 + 0.501 159 687 994 232 484 659 2;
  • 58) 0.501 159 687 994 232 484 659 2 × 2 = 1 + 0.002 319 375 988 464 969 318 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:


0.016 738 891 601 562 496 538 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 538 6(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:


0.016 738 891 601 562 496 538 6(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 -0.016 738 891 601 562 496 538 6 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