-0.016 738 891 601 562 496 547 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 547 8(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
-0.016 738 891 601 562 496 547 8(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:

|-0.016 738 891 601 562 496 547 8| = 0.016 738 891 601 562 496 547 8


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: 0.016 738 891 601 562 496 547 8.

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) 0.016 738 891 601 562 496 547 8 × 2 = 0 + 0.033 477 783 203 124 993 095 6;
  • 2) 0.033 477 783 203 124 993 095 6 × 2 = 0 + 0.066 955 566 406 249 986 191 2;
  • 3) 0.066 955 566 406 249 986 191 2 × 2 = 0 + 0.133 911 132 812 499 972 382 4;
  • 4) 0.133 911 132 812 499 972 382 4 × 2 = 0 + 0.267 822 265 624 999 944 764 8;
  • 5) 0.267 822 265 624 999 944 764 8 × 2 = 0 + 0.535 644 531 249 999 889 529 6;
  • 6) 0.535 644 531 249 999 889 529 6 × 2 = 1 + 0.071 289 062 499 999 779 059 2;
  • 7) 0.071 289 062 499 999 779 059 2 × 2 = 0 + 0.142 578 124 999 999 558 118 4;
  • 8) 0.142 578 124 999 999 558 118 4 × 2 = 0 + 0.285 156 249 999 999 116 236 8;
  • 9) 0.285 156 249 999 999 116 236 8 × 2 = 0 + 0.570 312 499 999 998 232 473 6;
  • 10) 0.570 312 499 999 998 232 473 6 × 2 = 1 + 0.140 624 999 999 996 464 947 2;
  • 11) 0.140 624 999 999 996 464 947 2 × 2 = 0 + 0.281 249 999 999 992 929 894 4;
  • 12) 0.281 249 999 999 992 929 894 4 × 2 = 0 + 0.562 499 999 999 985 859 788 8;
  • 13) 0.562 499 999 999 985 859 788 8 × 2 = 1 + 0.124 999 999 999 971 719 577 6;
  • 14) 0.124 999 999 999 971 719 577 6 × 2 = 0 + 0.249 999 999 999 943 439 155 2;
  • 15) 0.249 999 999 999 943 439 155 2 × 2 = 0 + 0.499 999 999 999 886 878 310 4;
  • 16) 0.499 999 999 999 886 878 310 4 × 2 = 0 + 0.999 999 999 999 773 756 620 8;
  • 17) 0.999 999 999 999 773 756 620 8 × 2 = 1 + 0.999 999 999 999 547 513 241 6;
  • 18) 0.999 999 999 999 547 513 241 6 × 2 = 1 + 0.999 999 999 999 095 026 483 2;
  • 19) 0.999 999 999 999 095 026 483 2 × 2 = 1 + 0.999 999 999 998 190 052 966 4;
  • 20) 0.999 999 999 998 190 052 966 4 × 2 = 1 + 0.999 999 999 996 380 105 932 8;
  • 21) 0.999 999 999 996 380 105 932 8 × 2 = 1 + 0.999 999 999 992 760 211 865 6;
  • 22) 0.999 999 999 992 760 211 865 6 × 2 = 1 + 0.999 999 999 985 520 423 731 2;
  • 23) 0.999 999 999 985 520 423 731 2 × 2 = 1 + 0.999 999 999 971 040 847 462 4;
  • 24) 0.999 999 999 971 040 847 462 4 × 2 = 1 + 0.999 999 999 942 081 694 924 8;
  • 25) 0.999 999 999 942 081 694 924 8 × 2 = 1 + 0.999 999 999 884 163 389 849 6;
  • 26) 0.999 999 999 884 163 389 849 6 × 2 = 1 + 0.999 999 999 768 326 779 699 2;
  • 27) 0.999 999 999 768 326 779 699 2 × 2 = 1 + 0.999 999 999 536 653 559 398 4;
  • 28) 0.999 999 999 536 653 559 398 4 × 2 = 1 + 0.999 999 999 073 307 118 796 8;
  • 29) 0.999 999 999 073 307 118 796 8 × 2 = 1 + 0.999 999 998 146 614 237 593 6;
  • 30) 0.999 999 998 146 614 237 593 6 × 2 = 1 + 0.999 999 996 293 228 475 187 2;
  • 31) 0.999 999 996 293 228 475 187 2 × 2 = 1 + 0.999 999 992 586 456 950 374 4;
  • 32) 0.999 999 992 586 456 950 374 4 × 2 = 1 + 0.999 999 985 172 913 900 748 8;
  • 33) 0.999 999 985 172 913 900 748 8 × 2 = 1 + 0.999 999 970 345 827 801 497 6;
  • 34) 0.999 999 970 345 827 801 497 6 × 2 = 1 + 0.999 999 940 691 655 602 995 2;
  • 35) 0.999 999 940 691 655 602 995 2 × 2 = 1 + 0.999 999 881 383 311 205 990 4;
  • 36) 0.999 999 881 383 311 205 990 4 × 2 = 1 + 0.999 999 762 766 622 411 980 8;
  • 37) 0.999 999 762 766 622 411 980 8 × 2 = 1 + 0.999 999 525 533 244 823 961 6;
  • 38) 0.999 999 525 533 244 823 961 6 × 2 = 1 + 0.999 999 051 066 489 647 923 2;
  • 39) 0.999 999 051 066 489 647 923 2 × 2 = 1 + 0.999 998 102 132 979 295 846 4;
  • 40) 0.999 998 102 132 979 295 846 4 × 2 = 1 + 0.999 996 204 265 958 591 692 8;
  • 41) 0.999 996 204 265 958 591 692 8 × 2 = 1 + 0.999 992 408 531 917 183 385 6;
  • 42) 0.999 992 408 531 917 183 385 6 × 2 = 1 + 0.999 984 817 063 834 366 771 2;
  • 43) 0.999 984 817 063 834 366 771 2 × 2 = 1 + 0.999 969 634 127 668 733 542 4;
  • 44) 0.999 969 634 127 668 733 542 4 × 2 = 1 + 0.999 939 268 255 337 467 084 8;
  • 45) 0.999 939 268 255 337 467 084 8 × 2 = 1 + 0.999 878 536 510 674 934 169 6;
  • 46) 0.999 878 536 510 674 934 169 6 × 2 = 1 + 0.999 757 073 021 349 868 339 2;
  • 47) 0.999 757 073 021 349 868 339 2 × 2 = 1 + 0.999 514 146 042 699 736 678 4;
  • 48) 0.999 514 146 042 699 736 678 4 × 2 = 1 + 0.999 028 292 085 399 473 356 8;
  • 49) 0.999 028 292 085 399 473 356 8 × 2 = 1 + 0.998 056 584 170 798 946 713 6;
  • 50) 0.998 056 584 170 798 946 713 6 × 2 = 1 + 0.996 113 168 341 597 893 427 2;
  • 51) 0.996 113 168 341 597 893 427 2 × 2 = 1 + 0.992 226 336 683 195 786 854 4;
  • 52) 0.992 226 336 683 195 786 854 4 × 2 = 1 + 0.984 452 673 366 391 573 708 8;
  • 53) 0.984 452 673 366 391 573 708 8 × 2 = 1 + 0.968 905 346 732 783 147 417 6;
  • 54) 0.968 905 346 732 783 147 417 6 × 2 = 1 + 0.937 810 693 465 566 294 835 2;
  • 55) 0.937 810 693 465 566 294 835 2 × 2 = 1 + 0.875 621 386 931 132 589 670 4;
  • 56) 0.875 621 386 931 132 589 670 4 × 2 = 1 + 0.751 242 773 862 265 179 340 8;
  • 57) 0.751 242 773 862 265 179 340 8 × 2 = 1 + 0.502 485 547 724 530 358 681 6;
  • 58) 0.502 485 547 724 530 358 681 6 × 2 = 1 + 0.004 971 095 449 060 717 363 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:


0.016 738 891 601 562 496 547 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 547 8(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:


0.016 738 891 601 562 496 547 8(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 -0.016 738 891 601 562 496 547 8 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 -0.016 738 891 601 562 496 555 8 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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