-0.016 738 891 601 562 496 531 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 32(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
-0.016 738 891 601 562 496 531 32(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:

|-0.016 738 891 601 562 496 531 32| = 0.016 738 891 601 562 496 531 32


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: 0.016 738 891 601 562 496 531 32.

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) 0.016 738 891 601 562 496 531 32 × 2 = 0 + 0.033 477 783 203 124 993 062 64;
  • 2) 0.033 477 783 203 124 993 062 64 × 2 = 0 + 0.066 955 566 406 249 986 125 28;
  • 3) 0.066 955 566 406 249 986 125 28 × 2 = 0 + 0.133 911 132 812 499 972 250 56;
  • 4) 0.133 911 132 812 499 972 250 56 × 2 = 0 + 0.267 822 265 624 999 944 501 12;
  • 5) 0.267 822 265 624 999 944 501 12 × 2 = 0 + 0.535 644 531 249 999 889 002 24;
  • 6) 0.535 644 531 249 999 889 002 24 × 2 = 1 + 0.071 289 062 499 999 778 004 48;
  • 7) 0.071 289 062 499 999 778 004 48 × 2 = 0 + 0.142 578 124 999 999 556 008 96;
  • 8) 0.142 578 124 999 999 556 008 96 × 2 = 0 + 0.285 156 249 999 999 112 017 92;
  • 9) 0.285 156 249 999 999 112 017 92 × 2 = 0 + 0.570 312 499 999 998 224 035 84;
  • 10) 0.570 312 499 999 998 224 035 84 × 2 = 1 + 0.140 624 999 999 996 448 071 68;
  • 11) 0.140 624 999 999 996 448 071 68 × 2 = 0 + 0.281 249 999 999 992 896 143 36;
  • 12) 0.281 249 999 999 992 896 143 36 × 2 = 0 + 0.562 499 999 999 985 792 286 72;
  • 13) 0.562 499 999 999 985 792 286 72 × 2 = 1 + 0.124 999 999 999 971 584 573 44;
  • 14) 0.124 999 999 999 971 584 573 44 × 2 = 0 + 0.249 999 999 999 943 169 146 88;
  • 15) 0.249 999 999 999 943 169 146 88 × 2 = 0 + 0.499 999 999 999 886 338 293 76;
  • 16) 0.499 999 999 999 886 338 293 76 × 2 = 0 + 0.999 999 999 999 772 676 587 52;
  • 17) 0.999 999 999 999 772 676 587 52 × 2 = 1 + 0.999 999 999 999 545 353 175 04;
  • 18) 0.999 999 999 999 545 353 175 04 × 2 = 1 + 0.999 999 999 999 090 706 350 08;
  • 19) 0.999 999 999 999 090 706 350 08 × 2 = 1 + 0.999 999 999 998 181 412 700 16;
  • 20) 0.999 999 999 998 181 412 700 16 × 2 = 1 + 0.999 999 999 996 362 825 400 32;
  • 21) 0.999 999 999 996 362 825 400 32 × 2 = 1 + 0.999 999 999 992 725 650 800 64;
  • 22) 0.999 999 999 992 725 650 800 64 × 2 = 1 + 0.999 999 999 985 451 301 601 28;
  • 23) 0.999 999 999 985 451 301 601 28 × 2 = 1 + 0.999 999 999 970 902 603 202 56;
  • 24) 0.999 999 999 970 902 603 202 56 × 2 = 1 + 0.999 999 999 941 805 206 405 12;
  • 25) 0.999 999 999 941 805 206 405 12 × 2 = 1 + 0.999 999 999 883 610 412 810 24;
  • 26) 0.999 999 999 883 610 412 810 24 × 2 = 1 + 0.999 999 999 767 220 825 620 48;
  • 27) 0.999 999 999 767 220 825 620 48 × 2 = 1 + 0.999 999 999 534 441 651 240 96;
  • 28) 0.999 999 999 534 441 651 240 96 × 2 = 1 + 0.999 999 999 068 883 302 481 92;
  • 29) 0.999 999 999 068 883 302 481 92 × 2 = 1 + 0.999 999 998 137 766 604 963 84;
  • 30) 0.999 999 998 137 766 604 963 84 × 2 = 1 + 0.999 999 996 275 533 209 927 68;
  • 31) 0.999 999 996 275 533 209 927 68 × 2 = 1 + 0.999 999 992 551 066 419 855 36;
  • 32) 0.999 999 992 551 066 419 855 36 × 2 = 1 + 0.999 999 985 102 132 839 710 72;
  • 33) 0.999 999 985 102 132 839 710 72 × 2 = 1 + 0.999 999 970 204 265 679 421 44;
  • 34) 0.999 999 970 204 265 679 421 44 × 2 = 1 + 0.999 999 940 408 531 358 842 88;
  • 35) 0.999 999 940 408 531 358 842 88 × 2 = 1 + 0.999 999 880 817 062 717 685 76;
  • 36) 0.999 999 880 817 062 717 685 76 × 2 = 1 + 0.999 999 761 634 125 435 371 52;
  • 37) 0.999 999 761 634 125 435 371 52 × 2 = 1 + 0.999 999 523 268 250 870 743 04;
  • 38) 0.999 999 523 268 250 870 743 04 × 2 = 1 + 0.999 999 046 536 501 741 486 08;
  • 39) 0.999 999 046 536 501 741 486 08 × 2 = 1 + 0.999 998 093 073 003 482 972 16;
  • 40) 0.999 998 093 073 003 482 972 16 × 2 = 1 + 0.999 996 186 146 006 965 944 32;
  • 41) 0.999 996 186 146 006 965 944 32 × 2 = 1 + 0.999 992 372 292 013 931 888 64;
  • 42) 0.999 992 372 292 013 931 888 64 × 2 = 1 + 0.999 984 744 584 027 863 777 28;
  • 43) 0.999 984 744 584 027 863 777 28 × 2 = 1 + 0.999 969 489 168 055 727 554 56;
  • 44) 0.999 969 489 168 055 727 554 56 × 2 = 1 + 0.999 938 978 336 111 455 109 12;
  • 45) 0.999 938 978 336 111 455 109 12 × 2 = 1 + 0.999 877 956 672 222 910 218 24;
  • 46) 0.999 877 956 672 222 910 218 24 × 2 = 1 + 0.999 755 913 344 445 820 436 48;
  • 47) 0.999 755 913 344 445 820 436 48 × 2 = 1 + 0.999 511 826 688 891 640 872 96;
  • 48) 0.999 511 826 688 891 640 872 96 × 2 = 1 + 0.999 023 653 377 783 281 745 92;
  • 49) 0.999 023 653 377 783 281 745 92 × 2 = 1 + 0.998 047 306 755 566 563 491 84;
  • 50) 0.998 047 306 755 566 563 491 84 × 2 = 1 + 0.996 094 613 511 133 126 983 68;
  • 51) 0.996 094 613 511 133 126 983 68 × 2 = 1 + 0.992 189 227 022 266 253 967 36;
  • 52) 0.992 189 227 022 266 253 967 36 × 2 = 1 + 0.984 378 454 044 532 507 934 72;
  • 53) 0.984 378 454 044 532 507 934 72 × 2 = 1 + 0.968 756 908 089 065 015 869 44;
  • 54) 0.968 756 908 089 065 015 869 44 × 2 = 1 + 0.937 513 816 178 130 031 738 88;
  • 55) 0.937 513 816 178 130 031 738 88 × 2 = 1 + 0.875 027 632 356 260 063 477 76;
  • 56) 0.875 027 632 356 260 063 477 76 × 2 = 1 + 0.750 055 264 712 520 126 955 52;
  • 57) 0.750 055 264 712 520 126 955 52 × 2 = 1 + 0.500 110 529 425 040 253 911 04;
  • 58) 0.500 110 529 425 040 253 911 04 × 2 = 1 + 0.000 221 058 850 080 507 822 08;

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:


0.016 738 891 601 562 496 531 32(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 32(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:


0.016 738 891 601 562 496 531 32(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 -0.016 738 891 601 562 496 531 32 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 -0.016 738 891 601 562 496 531 94 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