-0.016 738 891 601 562 496 845 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 845(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
-0.016 738 891 601 562 496 845(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:

|-0.016 738 891 601 562 496 845| = 0.016 738 891 601 562 496 845


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: 0.016 738 891 601 562 496 845.

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) 0.016 738 891 601 562 496 845 × 2 = 0 + 0.033 477 783 203 124 993 69;
  • 2) 0.033 477 783 203 124 993 69 × 2 = 0 + 0.066 955 566 406 249 987 38;
  • 3) 0.066 955 566 406 249 987 38 × 2 = 0 + 0.133 911 132 812 499 974 76;
  • 4) 0.133 911 132 812 499 974 76 × 2 = 0 + 0.267 822 265 624 999 949 52;
  • 5) 0.267 822 265 624 999 949 52 × 2 = 0 + 0.535 644 531 249 999 899 04;
  • 6) 0.535 644 531 249 999 899 04 × 2 = 1 + 0.071 289 062 499 999 798 08;
  • 7) 0.071 289 062 499 999 798 08 × 2 = 0 + 0.142 578 124 999 999 596 16;
  • 8) 0.142 578 124 999 999 596 16 × 2 = 0 + 0.285 156 249 999 999 192 32;
  • 9) 0.285 156 249 999 999 192 32 × 2 = 0 + 0.570 312 499 999 998 384 64;
  • 10) 0.570 312 499 999 998 384 64 × 2 = 1 + 0.140 624 999 999 996 769 28;
  • 11) 0.140 624 999 999 996 769 28 × 2 = 0 + 0.281 249 999 999 993 538 56;
  • 12) 0.281 249 999 999 993 538 56 × 2 = 0 + 0.562 499 999 999 987 077 12;
  • 13) 0.562 499 999 999 987 077 12 × 2 = 1 + 0.124 999 999 999 974 154 24;
  • 14) 0.124 999 999 999 974 154 24 × 2 = 0 + 0.249 999 999 999 948 308 48;
  • 15) 0.249 999 999 999 948 308 48 × 2 = 0 + 0.499 999 999 999 896 616 96;
  • 16) 0.499 999 999 999 896 616 96 × 2 = 0 + 0.999 999 999 999 793 233 92;
  • 17) 0.999 999 999 999 793 233 92 × 2 = 1 + 0.999 999 999 999 586 467 84;
  • 18) 0.999 999 999 999 586 467 84 × 2 = 1 + 0.999 999 999 999 172 935 68;
  • 19) 0.999 999 999 999 172 935 68 × 2 = 1 + 0.999 999 999 998 345 871 36;
  • 20) 0.999 999 999 998 345 871 36 × 2 = 1 + 0.999 999 999 996 691 742 72;
  • 21) 0.999 999 999 996 691 742 72 × 2 = 1 + 0.999 999 999 993 383 485 44;
  • 22) 0.999 999 999 993 383 485 44 × 2 = 1 + 0.999 999 999 986 766 970 88;
  • 23) 0.999 999 999 986 766 970 88 × 2 = 1 + 0.999 999 999 973 533 941 76;
  • 24) 0.999 999 999 973 533 941 76 × 2 = 1 + 0.999 999 999 947 067 883 52;
  • 25) 0.999 999 999 947 067 883 52 × 2 = 1 + 0.999 999 999 894 135 767 04;
  • 26) 0.999 999 999 894 135 767 04 × 2 = 1 + 0.999 999 999 788 271 534 08;
  • 27) 0.999 999 999 788 271 534 08 × 2 = 1 + 0.999 999 999 576 543 068 16;
  • 28) 0.999 999 999 576 543 068 16 × 2 = 1 + 0.999 999 999 153 086 136 32;
  • 29) 0.999 999 999 153 086 136 32 × 2 = 1 + 0.999 999 998 306 172 272 64;
  • 30) 0.999 999 998 306 172 272 64 × 2 = 1 + 0.999 999 996 612 344 545 28;
  • 31) 0.999 999 996 612 344 545 28 × 2 = 1 + 0.999 999 993 224 689 090 56;
  • 32) 0.999 999 993 224 689 090 56 × 2 = 1 + 0.999 999 986 449 378 181 12;
  • 33) 0.999 999 986 449 378 181 12 × 2 = 1 + 0.999 999 972 898 756 362 24;
  • 34) 0.999 999 972 898 756 362 24 × 2 = 1 + 0.999 999 945 797 512 724 48;
  • 35) 0.999 999 945 797 512 724 48 × 2 = 1 + 0.999 999 891 595 025 448 96;
  • 36) 0.999 999 891 595 025 448 96 × 2 = 1 + 0.999 999 783 190 050 897 92;
  • 37) 0.999 999 783 190 050 897 92 × 2 = 1 + 0.999 999 566 380 101 795 84;
  • 38) 0.999 999 566 380 101 795 84 × 2 = 1 + 0.999 999 132 760 203 591 68;
  • 39) 0.999 999 132 760 203 591 68 × 2 = 1 + 0.999 998 265 520 407 183 36;
  • 40) 0.999 998 265 520 407 183 36 × 2 = 1 + 0.999 996 531 040 814 366 72;
  • 41) 0.999 996 531 040 814 366 72 × 2 = 1 + 0.999 993 062 081 628 733 44;
  • 42) 0.999 993 062 081 628 733 44 × 2 = 1 + 0.999 986 124 163 257 466 88;
  • 43) 0.999 986 124 163 257 466 88 × 2 = 1 + 0.999 972 248 326 514 933 76;
  • 44) 0.999 972 248 326 514 933 76 × 2 = 1 + 0.999 944 496 653 029 867 52;
  • 45) 0.999 944 496 653 029 867 52 × 2 = 1 + 0.999 888 993 306 059 735 04;
  • 46) 0.999 888 993 306 059 735 04 × 2 = 1 + 0.999 777 986 612 119 470 08;
  • 47) 0.999 777 986 612 119 470 08 × 2 = 1 + 0.999 555 973 224 238 940 16;
  • 48) 0.999 555 973 224 238 940 16 × 2 = 1 + 0.999 111 946 448 477 880 32;
  • 49) 0.999 111 946 448 477 880 32 × 2 = 1 + 0.998 223 892 896 955 760 64;
  • 50) 0.998 223 892 896 955 760 64 × 2 = 1 + 0.996 447 785 793 911 521 28;
  • 51) 0.996 447 785 793 911 521 28 × 2 = 1 + 0.992 895 571 587 823 042 56;
  • 52) 0.992 895 571 587 823 042 56 × 2 = 1 + 0.985 791 143 175 646 085 12;
  • 53) 0.985 791 143 175 646 085 12 × 2 = 1 + 0.971 582 286 351 292 170 24;
  • 54) 0.971 582 286 351 292 170 24 × 2 = 1 + 0.943 164 572 702 584 340 48;
  • 55) 0.943 164 572 702 584 340 48 × 2 = 1 + 0.886 329 145 405 168 680 96;
  • 56) 0.886 329 145 405 168 680 96 × 2 = 1 + 0.772 658 290 810 337 361 92;
  • 57) 0.772 658 290 810 337 361 92 × 2 = 1 + 0.545 316 581 620 674 723 84;
  • 58) 0.545 316 581 620 674 723 84 × 2 = 1 + 0.090 633 163 241 349 447 68;

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:


0.016 738 891 601 562 496 845(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 845(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:


0.016 738 891 601 562 496 845(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 -0.016 738 891 601 562 496 845 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