-0.016 738 891 601 562 496 544 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 544 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
-0.016 738 891 601 562 496 544 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:

|-0.016 738 891 601 562 496 544 6| = 0.016 738 891 601 562 496 544 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: 0.016 738 891 601 562 496 544 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) 0.016 738 891 601 562 496 544 6 × 2 = 0 + 0.033 477 783 203 124 993 089 2;
  • 2) 0.033 477 783 203 124 993 089 2 × 2 = 0 + 0.066 955 566 406 249 986 178 4;
  • 3) 0.066 955 566 406 249 986 178 4 × 2 = 0 + 0.133 911 132 812 499 972 356 8;
  • 4) 0.133 911 132 812 499 972 356 8 × 2 = 0 + 0.267 822 265 624 999 944 713 6;
  • 5) 0.267 822 265 624 999 944 713 6 × 2 = 0 + 0.535 644 531 249 999 889 427 2;
  • 6) 0.535 644 531 249 999 889 427 2 × 2 = 1 + 0.071 289 062 499 999 778 854 4;
  • 7) 0.071 289 062 499 999 778 854 4 × 2 = 0 + 0.142 578 124 999 999 557 708 8;
  • 8) 0.142 578 124 999 999 557 708 8 × 2 = 0 + 0.285 156 249 999 999 115 417 6;
  • 9) 0.285 156 249 999 999 115 417 6 × 2 = 0 + 0.570 312 499 999 998 230 835 2;
  • 10) 0.570 312 499 999 998 230 835 2 × 2 = 1 + 0.140 624 999 999 996 461 670 4;
  • 11) 0.140 624 999 999 996 461 670 4 × 2 = 0 + 0.281 249 999 999 992 923 340 8;
  • 12) 0.281 249 999 999 992 923 340 8 × 2 = 0 + 0.562 499 999 999 985 846 681 6;
  • 13) 0.562 499 999 999 985 846 681 6 × 2 = 1 + 0.124 999 999 999 971 693 363 2;
  • 14) 0.124 999 999 999 971 693 363 2 × 2 = 0 + 0.249 999 999 999 943 386 726 4;
  • 15) 0.249 999 999 999 943 386 726 4 × 2 = 0 + 0.499 999 999 999 886 773 452 8;
  • 16) 0.499 999 999 999 886 773 452 8 × 2 = 0 + 0.999 999 999 999 773 546 905 6;
  • 17) 0.999 999 999 999 773 546 905 6 × 2 = 1 + 0.999 999 999 999 547 093 811 2;
  • 18) 0.999 999 999 999 547 093 811 2 × 2 = 1 + 0.999 999 999 999 094 187 622 4;
  • 19) 0.999 999 999 999 094 187 622 4 × 2 = 1 + 0.999 999 999 998 188 375 244 8;
  • 20) 0.999 999 999 998 188 375 244 8 × 2 = 1 + 0.999 999 999 996 376 750 489 6;
  • 21) 0.999 999 999 996 376 750 489 6 × 2 = 1 + 0.999 999 999 992 753 500 979 2;
  • 22) 0.999 999 999 992 753 500 979 2 × 2 = 1 + 0.999 999 999 985 507 001 958 4;
  • 23) 0.999 999 999 985 507 001 958 4 × 2 = 1 + 0.999 999 999 971 014 003 916 8;
  • 24) 0.999 999 999 971 014 003 916 8 × 2 = 1 + 0.999 999 999 942 028 007 833 6;
  • 25) 0.999 999 999 942 028 007 833 6 × 2 = 1 + 0.999 999 999 884 056 015 667 2;
  • 26) 0.999 999 999 884 056 015 667 2 × 2 = 1 + 0.999 999 999 768 112 031 334 4;
  • 27) 0.999 999 999 768 112 031 334 4 × 2 = 1 + 0.999 999 999 536 224 062 668 8;
  • 28) 0.999 999 999 536 224 062 668 8 × 2 = 1 + 0.999 999 999 072 448 125 337 6;
  • 29) 0.999 999 999 072 448 125 337 6 × 2 = 1 + 0.999 999 998 144 896 250 675 2;
  • 30) 0.999 999 998 144 896 250 675 2 × 2 = 1 + 0.999 999 996 289 792 501 350 4;
  • 31) 0.999 999 996 289 792 501 350 4 × 2 = 1 + 0.999 999 992 579 585 002 700 8;
  • 32) 0.999 999 992 579 585 002 700 8 × 2 = 1 + 0.999 999 985 159 170 005 401 6;
  • 33) 0.999 999 985 159 170 005 401 6 × 2 = 1 + 0.999 999 970 318 340 010 803 2;
  • 34) 0.999 999 970 318 340 010 803 2 × 2 = 1 + 0.999 999 940 636 680 021 606 4;
  • 35) 0.999 999 940 636 680 021 606 4 × 2 = 1 + 0.999 999 881 273 360 043 212 8;
  • 36) 0.999 999 881 273 360 043 212 8 × 2 = 1 + 0.999 999 762 546 720 086 425 6;
  • 37) 0.999 999 762 546 720 086 425 6 × 2 = 1 + 0.999 999 525 093 440 172 851 2;
  • 38) 0.999 999 525 093 440 172 851 2 × 2 = 1 + 0.999 999 050 186 880 345 702 4;
  • 39) 0.999 999 050 186 880 345 702 4 × 2 = 1 + 0.999 998 100 373 760 691 404 8;
  • 40) 0.999 998 100 373 760 691 404 8 × 2 = 1 + 0.999 996 200 747 521 382 809 6;
  • 41) 0.999 996 200 747 521 382 809 6 × 2 = 1 + 0.999 992 401 495 042 765 619 2;
  • 42) 0.999 992 401 495 042 765 619 2 × 2 = 1 + 0.999 984 802 990 085 531 238 4;
  • 43) 0.999 984 802 990 085 531 238 4 × 2 = 1 + 0.999 969 605 980 171 062 476 8;
  • 44) 0.999 969 605 980 171 062 476 8 × 2 = 1 + 0.999 939 211 960 342 124 953 6;
  • 45) 0.999 939 211 960 342 124 953 6 × 2 = 1 + 0.999 878 423 920 684 249 907 2;
  • 46) 0.999 878 423 920 684 249 907 2 × 2 = 1 + 0.999 756 847 841 368 499 814 4;
  • 47) 0.999 756 847 841 368 499 814 4 × 2 = 1 + 0.999 513 695 682 736 999 628 8;
  • 48) 0.999 513 695 682 736 999 628 8 × 2 = 1 + 0.999 027 391 365 473 999 257 6;
  • 49) 0.999 027 391 365 473 999 257 6 × 2 = 1 + 0.998 054 782 730 947 998 515 2;
  • 50) 0.998 054 782 730 947 998 515 2 × 2 = 1 + 0.996 109 565 461 895 997 030 4;
  • 51) 0.996 109 565 461 895 997 030 4 × 2 = 1 + 0.992 219 130 923 791 994 060 8;
  • 52) 0.992 219 130 923 791 994 060 8 × 2 = 1 + 0.984 438 261 847 583 988 121 6;
  • 53) 0.984 438 261 847 583 988 121 6 × 2 = 1 + 0.968 876 523 695 167 976 243 2;
  • 54) 0.968 876 523 695 167 976 243 2 × 2 = 1 + 0.937 753 047 390 335 952 486 4;
  • 55) 0.937 753 047 390 335 952 486 4 × 2 = 1 + 0.875 506 094 780 671 904 972 8;
  • 56) 0.875 506 094 780 671 904 972 8 × 2 = 1 + 0.751 012 189 561 343 809 945 6;
  • 57) 0.751 012 189 561 343 809 945 6 × 2 = 1 + 0.502 024 379 122 687 619 891 2;
  • 58) 0.502 024 379 122 687 619 891 2 × 2 = 1 + 0.004 048 758 245 375 239 782 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:


0.016 738 891 601 562 496 544 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 544 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:


0.016 738 891 601 562 496 544 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 -0.016 738 891 601 562 496 544 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

Share

» Convert decimal -0.016 738 891 601 562 496 539 1 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