-0.016 738 891 601 562 496 533 85 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 533 85(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
-0.016 738 891 601 562 496 533 85(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:

|-0.016 738 891 601 562 496 533 85| = 0.016 738 891 601 562 496 533 85


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: 0.016 738 891 601 562 496 533 85.

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) 0.016 738 891 601 562 496 533 85 × 2 = 0 + 0.033 477 783 203 124 993 067 7;
  • 2) 0.033 477 783 203 124 993 067 7 × 2 = 0 + 0.066 955 566 406 249 986 135 4;
  • 3) 0.066 955 566 406 249 986 135 4 × 2 = 0 + 0.133 911 132 812 499 972 270 8;
  • 4) 0.133 911 132 812 499 972 270 8 × 2 = 0 + 0.267 822 265 624 999 944 541 6;
  • 5) 0.267 822 265 624 999 944 541 6 × 2 = 0 + 0.535 644 531 249 999 889 083 2;
  • 6) 0.535 644 531 249 999 889 083 2 × 2 = 1 + 0.071 289 062 499 999 778 166 4;
  • 7) 0.071 289 062 499 999 778 166 4 × 2 = 0 + 0.142 578 124 999 999 556 332 8;
  • 8) 0.142 578 124 999 999 556 332 8 × 2 = 0 + 0.285 156 249 999 999 112 665 6;
  • 9) 0.285 156 249 999 999 112 665 6 × 2 = 0 + 0.570 312 499 999 998 225 331 2;
  • 10) 0.570 312 499 999 998 225 331 2 × 2 = 1 + 0.140 624 999 999 996 450 662 4;
  • 11) 0.140 624 999 999 996 450 662 4 × 2 = 0 + 0.281 249 999 999 992 901 324 8;
  • 12) 0.281 249 999 999 992 901 324 8 × 2 = 0 + 0.562 499 999 999 985 802 649 6;
  • 13) 0.562 499 999 999 985 802 649 6 × 2 = 1 + 0.124 999 999 999 971 605 299 2;
  • 14) 0.124 999 999 999 971 605 299 2 × 2 = 0 + 0.249 999 999 999 943 210 598 4;
  • 15) 0.249 999 999 999 943 210 598 4 × 2 = 0 + 0.499 999 999 999 886 421 196 8;
  • 16) 0.499 999 999 999 886 421 196 8 × 2 = 0 + 0.999 999 999 999 772 842 393 6;
  • 17) 0.999 999 999 999 772 842 393 6 × 2 = 1 + 0.999 999 999 999 545 684 787 2;
  • 18) 0.999 999 999 999 545 684 787 2 × 2 = 1 + 0.999 999 999 999 091 369 574 4;
  • 19) 0.999 999 999 999 091 369 574 4 × 2 = 1 + 0.999 999 999 998 182 739 148 8;
  • 20) 0.999 999 999 998 182 739 148 8 × 2 = 1 + 0.999 999 999 996 365 478 297 6;
  • 21) 0.999 999 999 996 365 478 297 6 × 2 = 1 + 0.999 999 999 992 730 956 595 2;
  • 22) 0.999 999 999 992 730 956 595 2 × 2 = 1 + 0.999 999 999 985 461 913 190 4;
  • 23) 0.999 999 999 985 461 913 190 4 × 2 = 1 + 0.999 999 999 970 923 826 380 8;
  • 24) 0.999 999 999 970 923 826 380 8 × 2 = 1 + 0.999 999 999 941 847 652 761 6;
  • 25) 0.999 999 999 941 847 652 761 6 × 2 = 1 + 0.999 999 999 883 695 305 523 2;
  • 26) 0.999 999 999 883 695 305 523 2 × 2 = 1 + 0.999 999 999 767 390 611 046 4;
  • 27) 0.999 999 999 767 390 611 046 4 × 2 = 1 + 0.999 999 999 534 781 222 092 8;
  • 28) 0.999 999 999 534 781 222 092 8 × 2 = 1 + 0.999 999 999 069 562 444 185 6;
  • 29) 0.999 999 999 069 562 444 185 6 × 2 = 1 + 0.999 999 998 139 124 888 371 2;
  • 30) 0.999 999 998 139 124 888 371 2 × 2 = 1 + 0.999 999 996 278 249 776 742 4;
  • 31) 0.999 999 996 278 249 776 742 4 × 2 = 1 + 0.999 999 992 556 499 553 484 8;
  • 32) 0.999 999 992 556 499 553 484 8 × 2 = 1 + 0.999 999 985 112 999 106 969 6;
  • 33) 0.999 999 985 112 999 106 969 6 × 2 = 1 + 0.999 999 970 225 998 213 939 2;
  • 34) 0.999 999 970 225 998 213 939 2 × 2 = 1 + 0.999 999 940 451 996 427 878 4;
  • 35) 0.999 999 940 451 996 427 878 4 × 2 = 1 + 0.999 999 880 903 992 855 756 8;
  • 36) 0.999 999 880 903 992 855 756 8 × 2 = 1 + 0.999 999 761 807 985 711 513 6;
  • 37) 0.999 999 761 807 985 711 513 6 × 2 = 1 + 0.999 999 523 615 971 423 027 2;
  • 38) 0.999 999 523 615 971 423 027 2 × 2 = 1 + 0.999 999 047 231 942 846 054 4;
  • 39) 0.999 999 047 231 942 846 054 4 × 2 = 1 + 0.999 998 094 463 885 692 108 8;
  • 40) 0.999 998 094 463 885 692 108 8 × 2 = 1 + 0.999 996 188 927 771 384 217 6;
  • 41) 0.999 996 188 927 771 384 217 6 × 2 = 1 + 0.999 992 377 855 542 768 435 2;
  • 42) 0.999 992 377 855 542 768 435 2 × 2 = 1 + 0.999 984 755 711 085 536 870 4;
  • 43) 0.999 984 755 711 085 536 870 4 × 2 = 1 + 0.999 969 511 422 171 073 740 8;
  • 44) 0.999 969 511 422 171 073 740 8 × 2 = 1 + 0.999 939 022 844 342 147 481 6;
  • 45) 0.999 939 022 844 342 147 481 6 × 2 = 1 + 0.999 878 045 688 684 294 963 2;
  • 46) 0.999 878 045 688 684 294 963 2 × 2 = 1 + 0.999 756 091 377 368 589 926 4;
  • 47) 0.999 756 091 377 368 589 926 4 × 2 = 1 + 0.999 512 182 754 737 179 852 8;
  • 48) 0.999 512 182 754 737 179 852 8 × 2 = 1 + 0.999 024 365 509 474 359 705 6;
  • 49) 0.999 024 365 509 474 359 705 6 × 2 = 1 + 0.998 048 731 018 948 719 411 2;
  • 50) 0.998 048 731 018 948 719 411 2 × 2 = 1 + 0.996 097 462 037 897 438 822 4;
  • 51) 0.996 097 462 037 897 438 822 4 × 2 = 1 + 0.992 194 924 075 794 877 644 8;
  • 52) 0.992 194 924 075 794 877 644 8 × 2 = 1 + 0.984 389 848 151 589 755 289 6;
  • 53) 0.984 389 848 151 589 755 289 6 × 2 = 1 + 0.968 779 696 303 179 510 579 2;
  • 54) 0.968 779 696 303 179 510 579 2 × 2 = 1 + 0.937 559 392 606 359 021 158 4;
  • 55) 0.937 559 392 606 359 021 158 4 × 2 = 1 + 0.875 118 785 212 718 042 316 8;
  • 56) 0.875 118 785 212 718 042 316 8 × 2 = 1 + 0.750 237 570 425 436 084 633 6;
  • 57) 0.750 237 570 425 436 084 633 6 × 2 = 1 + 0.500 475 140 850 872 169 267 2;
  • 58) 0.500 475 140 850 872 169 267 2 × 2 = 1 + 0.000 950 281 701 744 338 534 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:


0.016 738 891 601 562 496 533 85(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 533 85(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:


0.016 738 891 601 562 496 533 85(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 -0.016 738 891 601 562 496 533 85 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