-0.016 738 891 601 562 496 83 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 83(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 83(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 83| = 0.016 738 891 601 562 496 83


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 83.

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 83 × 2 = 0 + 0.033 477 783 203 124 993 66;
  • 2) 0.033 477 783 203 124 993 66 × 2 = 0 + 0.066 955 566 406 249 987 32;
  • 3) 0.066 955 566 406 249 987 32 × 2 = 0 + 0.133 911 132 812 499 974 64;
  • 4) 0.133 911 132 812 499 974 64 × 2 = 0 + 0.267 822 265 624 999 949 28;
  • 5) 0.267 822 265 624 999 949 28 × 2 = 0 + 0.535 644 531 249 999 898 56;
  • 6) 0.535 644 531 249 999 898 56 × 2 = 1 + 0.071 289 062 499 999 797 12;
  • 7) 0.071 289 062 499 999 797 12 × 2 = 0 + 0.142 578 124 999 999 594 24;
  • 8) 0.142 578 124 999 999 594 24 × 2 = 0 + 0.285 156 249 999 999 188 48;
  • 9) 0.285 156 249 999 999 188 48 × 2 = 0 + 0.570 312 499 999 998 376 96;
  • 10) 0.570 312 499 999 998 376 96 × 2 = 1 + 0.140 624 999 999 996 753 92;
  • 11) 0.140 624 999 999 996 753 92 × 2 = 0 + 0.281 249 999 999 993 507 84;
  • 12) 0.281 249 999 999 993 507 84 × 2 = 0 + 0.562 499 999 999 987 015 68;
  • 13) 0.562 499 999 999 987 015 68 × 2 = 1 + 0.124 999 999 999 974 031 36;
  • 14) 0.124 999 999 999 974 031 36 × 2 = 0 + 0.249 999 999 999 948 062 72;
  • 15) 0.249 999 999 999 948 062 72 × 2 = 0 + 0.499 999 999 999 896 125 44;
  • 16) 0.499 999 999 999 896 125 44 × 2 = 0 + 0.999 999 999 999 792 250 88;
  • 17) 0.999 999 999 999 792 250 88 × 2 = 1 + 0.999 999 999 999 584 501 76;
  • 18) 0.999 999 999 999 584 501 76 × 2 = 1 + 0.999 999 999 999 169 003 52;
  • 19) 0.999 999 999 999 169 003 52 × 2 = 1 + 0.999 999 999 998 338 007 04;
  • 20) 0.999 999 999 998 338 007 04 × 2 = 1 + 0.999 999 999 996 676 014 08;
  • 21) 0.999 999 999 996 676 014 08 × 2 = 1 + 0.999 999 999 993 352 028 16;
  • 22) 0.999 999 999 993 352 028 16 × 2 = 1 + 0.999 999 999 986 704 056 32;
  • 23) 0.999 999 999 986 704 056 32 × 2 = 1 + 0.999 999 999 973 408 112 64;
  • 24) 0.999 999 999 973 408 112 64 × 2 = 1 + 0.999 999 999 946 816 225 28;
  • 25) 0.999 999 999 946 816 225 28 × 2 = 1 + 0.999 999 999 893 632 450 56;
  • 26) 0.999 999 999 893 632 450 56 × 2 = 1 + 0.999 999 999 787 264 901 12;
  • 27) 0.999 999 999 787 264 901 12 × 2 = 1 + 0.999 999 999 574 529 802 24;
  • 28) 0.999 999 999 574 529 802 24 × 2 = 1 + 0.999 999 999 149 059 604 48;
  • 29) 0.999 999 999 149 059 604 48 × 2 = 1 + 0.999 999 998 298 119 208 96;
  • 30) 0.999 999 998 298 119 208 96 × 2 = 1 + 0.999 999 996 596 238 417 92;
  • 31) 0.999 999 996 596 238 417 92 × 2 = 1 + 0.999 999 993 192 476 835 84;
  • 32) 0.999 999 993 192 476 835 84 × 2 = 1 + 0.999 999 986 384 953 671 68;
  • 33) 0.999 999 986 384 953 671 68 × 2 = 1 + 0.999 999 972 769 907 343 36;
  • 34) 0.999 999 972 769 907 343 36 × 2 = 1 + 0.999 999 945 539 814 686 72;
  • 35) 0.999 999 945 539 814 686 72 × 2 = 1 + 0.999 999 891 079 629 373 44;
  • 36) 0.999 999 891 079 629 373 44 × 2 = 1 + 0.999 999 782 159 258 746 88;
  • 37) 0.999 999 782 159 258 746 88 × 2 = 1 + 0.999 999 564 318 517 493 76;
  • 38) 0.999 999 564 318 517 493 76 × 2 = 1 + 0.999 999 128 637 034 987 52;
  • 39) 0.999 999 128 637 034 987 52 × 2 = 1 + 0.999 998 257 274 069 975 04;
  • 40) 0.999 998 257 274 069 975 04 × 2 = 1 + 0.999 996 514 548 139 950 08;
  • 41) 0.999 996 514 548 139 950 08 × 2 = 1 + 0.999 993 029 096 279 900 16;
  • 42) 0.999 993 029 096 279 900 16 × 2 = 1 + 0.999 986 058 192 559 800 32;
  • 43) 0.999 986 058 192 559 800 32 × 2 = 1 + 0.999 972 116 385 119 600 64;
  • 44) 0.999 972 116 385 119 600 64 × 2 = 1 + 0.999 944 232 770 239 201 28;
  • 45) 0.999 944 232 770 239 201 28 × 2 = 1 + 0.999 888 465 540 478 402 56;
  • 46) 0.999 888 465 540 478 402 56 × 2 = 1 + 0.999 776 931 080 956 805 12;
  • 47) 0.999 776 931 080 956 805 12 × 2 = 1 + 0.999 553 862 161 913 610 24;
  • 48) 0.999 553 862 161 913 610 24 × 2 = 1 + 0.999 107 724 323 827 220 48;
  • 49) 0.999 107 724 323 827 220 48 × 2 = 1 + 0.998 215 448 647 654 440 96;
  • 50) 0.998 215 448 647 654 440 96 × 2 = 1 + 0.996 430 897 295 308 881 92;
  • 51) 0.996 430 897 295 308 881 92 × 2 = 1 + 0.992 861 794 590 617 763 84;
  • 52) 0.992 861 794 590 617 763 84 × 2 = 1 + 0.985 723 589 181 235 527 68;
  • 53) 0.985 723 589 181 235 527 68 × 2 = 1 + 0.971 447 178 362 471 055 36;
  • 54) 0.971 447 178 362 471 055 36 × 2 = 1 + 0.942 894 356 724 942 110 72;
  • 55) 0.942 894 356 724 942 110 72 × 2 = 1 + 0.885 788 713 449 884 221 44;
  • 56) 0.885 788 713 449 884 221 44 × 2 = 1 + 0.771 577 426 899 768 442 88;
  • 57) 0.771 577 426 899 768 442 88 × 2 = 1 + 0.543 154 853 799 536 885 76;
  • 58) 0.543 154 853 799 536 885 76 × 2 = 1 + 0.086 309 707 599 073 771 52;

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 83(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 83(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 83(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 83 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