-0.016 738 891 601 562 496 530 693 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 693(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
-0.016 738 891 601 562 496 530 693(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:

|-0.016 738 891 601 562 496 530 693| = 0.016 738 891 601 562 496 530 693


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: 0.016 738 891 601 562 496 530 693.

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) 0.016 738 891 601 562 496 530 693 × 2 = 0 + 0.033 477 783 203 124 993 061 386;
  • 2) 0.033 477 783 203 124 993 061 386 × 2 = 0 + 0.066 955 566 406 249 986 122 772;
  • 3) 0.066 955 566 406 249 986 122 772 × 2 = 0 + 0.133 911 132 812 499 972 245 544;
  • 4) 0.133 911 132 812 499 972 245 544 × 2 = 0 + 0.267 822 265 624 999 944 491 088;
  • 5) 0.267 822 265 624 999 944 491 088 × 2 = 0 + 0.535 644 531 249 999 888 982 176;
  • 6) 0.535 644 531 249 999 888 982 176 × 2 = 1 + 0.071 289 062 499 999 777 964 352;
  • 7) 0.071 289 062 499 999 777 964 352 × 2 = 0 + 0.142 578 124 999 999 555 928 704;
  • 8) 0.142 578 124 999 999 555 928 704 × 2 = 0 + 0.285 156 249 999 999 111 857 408;
  • 9) 0.285 156 249 999 999 111 857 408 × 2 = 0 + 0.570 312 499 999 998 223 714 816;
  • 10) 0.570 312 499 999 998 223 714 816 × 2 = 1 + 0.140 624 999 999 996 447 429 632;
  • 11) 0.140 624 999 999 996 447 429 632 × 2 = 0 + 0.281 249 999 999 992 894 859 264;
  • 12) 0.281 249 999 999 992 894 859 264 × 2 = 0 + 0.562 499 999 999 985 789 718 528;
  • 13) 0.562 499 999 999 985 789 718 528 × 2 = 1 + 0.124 999 999 999 971 579 437 056;
  • 14) 0.124 999 999 999 971 579 437 056 × 2 = 0 + 0.249 999 999 999 943 158 874 112;
  • 15) 0.249 999 999 999 943 158 874 112 × 2 = 0 + 0.499 999 999 999 886 317 748 224;
  • 16) 0.499 999 999 999 886 317 748 224 × 2 = 0 + 0.999 999 999 999 772 635 496 448;
  • 17) 0.999 999 999 999 772 635 496 448 × 2 = 1 + 0.999 999 999 999 545 270 992 896;
  • 18) 0.999 999 999 999 545 270 992 896 × 2 = 1 + 0.999 999 999 999 090 541 985 792;
  • 19) 0.999 999 999 999 090 541 985 792 × 2 = 1 + 0.999 999 999 998 181 083 971 584;
  • 20) 0.999 999 999 998 181 083 971 584 × 2 = 1 + 0.999 999 999 996 362 167 943 168;
  • 21) 0.999 999 999 996 362 167 943 168 × 2 = 1 + 0.999 999 999 992 724 335 886 336;
  • 22) 0.999 999 999 992 724 335 886 336 × 2 = 1 + 0.999 999 999 985 448 671 772 672;
  • 23) 0.999 999 999 985 448 671 772 672 × 2 = 1 + 0.999 999 999 970 897 343 545 344;
  • 24) 0.999 999 999 970 897 343 545 344 × 2 = 1 + 0.999 999 999 941 794 687 090 688;
  • 25) 0.999 999 999 941 794 687 090 688 × 2 = 1 + 0.999 999 999 883 589 374 181 376;
  • 26) 0.999 999 999 883 589 374 181 376 × 2 = 1 + 0.999 999 999 767 178 748 362 752;
  • 27) 0.999 999 999 767 178 748 362 752 × 2 = 1 + 0.999 999 999 534 357 496 725 504;
  • 28) 0.999 999 999 534 357 496 725 504 × 2 = 1 + 0.999 999 999 068 714 993 451 008;
  • 29) 0.999 999 999 068 714 993 451 008 × 2 = 1 + 0.999 999 998 137 429 986 902 016;
  • 30) 0.999 999 998 137 429 986 902 016 × 2 = 1 + 0.999 999 996 274 859 973 804 032;
  • 31) 0.999 999 996 274 859 973 804 032 × 2 = 1 + 0.999 999 992 549 719 947 608 064;
  • 32) 0.999 999 992 549 719 947 608 064 × 2 = 1 + 0.999 999 985 099 439 895 216 128;
  • 33) 0.999 999 985 099 439 895 216 128 × 2 = 1 + 0.999 999 970 198 879 790 432 256;
  • 34) 0.999 999 970 198 879 790 432 256 × 2 = 1 + 0.999 999 940 397 759 580 864 512;
  • 35) 0.999 999 940 397 759 580 864 512 × 2 = 1 + 0.999 999 880 795 519 161 729 024;
  • 36) 0.999 999 880 795 519 161 729 024 × 2 = 1 + 0.999 999 761 591 038 323 458 048;
  • 37) 0.999 999 761 591 038 323 458 048 × 2 = 1 + 0.999 999 523 182 076 646 916 096;
  • 38) 0.999 999 523 182 076 646 916 096 × 2 = 1 + 0.999 999 046 364 153 293 832 192;
  • 39) 0.999 999 046 364 153 293 832 192 × 2 = 1 + 0.999 998 092 728 306 587 664 384;
  • 40) 0.999 998 092 728 306 587 664 384 × 2 = 1 + 0.999 996 185 456 613 175 328 768;
  • 41) 0.999 996 185 456 613 175 328 768 × 2 = 1 + 0.999 992 370 913 226 350 657 536;
  • 42) 0.999 992 370 913 226 350 657 536 × 2 = 1 + 0.999 984 741 826 452 701 315 072;
  • 43) 0.999 984 741 826 452 701 315 072 × 2 = 1 + 0.999 969 483 652 905 402 630 144;
  • 44) 0.999 969 483 652 905 402 630 144 × 2 = 1 + 0.999 938 967 305 810 805 260 288;
  • 45) 0.999 938 967 305 810 805 260 288 × 2 = 1 + 0.999 877 934 611 621 610 520 576;
  • 46) 0.999 877 934 611 621 610 520 576 × 2 = 1 + 0.999 755 869 223 243 221 041 152;
  • 47) 0.999 755 869 223 243 221 041 152 × 2 = 1 + 0.999 511 738 446 486 442 082 304;
  • 48) 0.999 511 738 446 486 442 082 304 × 2 = 1 + 0.999 023 476 892 972 884 164 608;
  • 49) 0.999 023 476 892 972 884 164 608 × 2 = 1 + 0.998 046 953 785 945 768 329 216;
  • 50) 0.998 046 953 785 945 768 329 216 × 2 = 1 + 0.996 093 907 571 891 536 658 432;
  • 51) 0.996 093 907 571 891 536 658 432 × 2 = 1 + 0.992 187 815 143 783 073 316 864;
  • 52) 0.992 187 815 143 783 073 316 864 × 2 = 1 + 0.984 375 630 287 566 146 633 728;
  • 53) 0.984 375 630 287 566 146 633 728 × 2 = 1 + 0.968 751 260 575 132 293 267 456;
  • 54) 0.968 751 260 575 132 293 267 456 × 2 = 1 + 0.937 502 521 150 264 586 534 912;
  • 55) 0.937 502 521 150 264 586 534 912 × 2 = 1 + 0.875 005 042 300 529 173 069 824;
  • 56) 0.875 005 042 300 529 173 069 824 × 2 = 1 + 0.750 010 084 601 058 346 139 648;
  • 57) 0.750 010 084 601 058 346 139 648 × 2 = 1 + 0.500 020 169 202 116 692 279 296;
  • 58) 0.500 020 169 202 116 692 279 296 × 2 = 1 + 0.000 040 338 404 233 384 558 592;

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:


0.016 738 891 601 562 496 530 693(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 693(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:


0.016 738 891 601 562 496 530 693(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 -0.016 738 891 601 562 496 530 693 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

Share

» Convert decimal -0.016 738 891 601 562 496 530 631 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: My manager only says "we" when referring to "my work". NotebookLM YouTube Video of 11.13 minutes

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