-0.016 738 891 601 562 496 530 305 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 305(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
-0.016 738 891 601 562 496 530 305(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:

|-0.016 738 891 601 562 496 530 305| = 0.016 738 891 601 562 496 530 305


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: 0.016 738 891 601 562 496 530 305.

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) 0.016 738 891 601 562 496 530 305 × 2 = 0 + 0.033 477 783 203 124 993 060 61;
  • 2) 0.033 477 783 203 124 993 060 61 × 2 = 0 + 0.066 955 566 406 249 986 121 22;
  • 3) 0.066 955 566 406 249 986 121 22 × 2 = 0 + 0.133 911 132 812 499 972 242 44;
  • 4) 0.133 911 132 812 499 972 242 44 × 2 = 0 + 0.267 822 265 624 999 944 484 88;
  • 5) 0.267 822 265 624 999 944 484 88 × 2 = 0 + 0.535 644 531 249 999 888 969 76;
  • 6) 0.535 644 531 249 999 888 969 76 × 2 = 1 + 0.071 289 062 499 999 777 939 52;
  • 7) 0.071 289 062 499 999 777 939 52 × 2 = 0 + 0.142 578 124 999 999 555 879 04;
  • 8) 0.142 578 124 999 999 555 879 04 × 2 = 0 + 0.285 156 249 999 999 111 758 08;
  • 9) 0.285 156 249 999 999 111 758 08 × 2 = 0 + 0.570 312 499 999 998 223 516 16;
  • 10) 0.570 312 499 999 998 223 516 16 × 2 = 1 + 0.140 624 999 999 996 447 032 32;
  • 11) 0.140 624 999 999 996 447 032 32 × 2 = 0 + 0.281 249 999 999 992 894 064 64;
  • 12) 0.281 249 999 999 992 894 064 64 × 2 = 0 + 0.562 499 999 999 985 788 129 28;
  • 13) 0.562 499 999 999 985 788 129 28 × 2 = 1 + 0.124 999 999 999 971 576 258 56;
  • 14) 0.124 999 999 999 971 576 258 56 × 2 = 0 + 0.249 999 999 999 943 152 517 12;
  • 15) 0.249 999 999 999 943 152 517 12 × 2 = 0 + 0.499 999 999 999 886 305 034 24;
  • 16) 0.499 999 999 999 886 305 034 24 × 2 = 0 + 0.999 999 999 999 772 610 068 48;
  • 17) 0.999 999 999 999 772 610 068 48 × 2 = 1 + 0.999 999 999 999 545 220 136 96;
  • 18) 0.999 999 999 999 545 220 136 96 × 2 = 1 + 0.999 999 999 999 090 440 273 92;
  • 19) 0.999 999 999 999 090 440 273 92 × 2 = 1 + 0.999 999 999 998 180 880 547 84;
  • 20) 0.999 999 999 998 180 880 547 84 × 2 = 1 + 0.999 999 999 996 361 761 095 68;
  • 21) 0.999 999 999 996 361 761 095 68 × 2 = 1 + 0.999 999 999 992 723 522 191 36;
  • 22) 0.999 999 999 992 723 522 191 36 × 2 = 1 + 0.999 999 999 985 447 044 382 72;
  • 23) 0.999 999 999 985 447 044 382 72 × 2 = 1 + 0.999 999 999 970 894 088 765 44;
  • 24) 0.999 999 999 970 894 088 765 44 × 2 = 1 + 0.999 999 999 941 788 177 530 88;
  • 25) 0.999 999 999 941 788 177 530 88 × 2 = 1 + 0.999 999 999 883 576 355 061 76;
  • 26) 0.999 999 999 883 576 355 061 76 × 2 = 1 + 0.999 999 999 767 152 710 123 52;
  • 27) 0.999 999 999 767 152 710 123 52 × 2 = 1 + 0.999 999 999 534 305 420 247 04;
  • 28) 0.999 999 999 534 305 420 247 04 × 2 = 1 + 0.999 999 999 068 610 840 494 08;
  • 29) 0.999 999 999 068 610 840 494 08 × 2 = 1 + 0.999 999 998 137 221 680 988 16;
  • 30) 0.999 999 998 137 221 680 988 16 × 2 = 1 + 0.999 999 996 274 443 361 976 32;
  • 31) 0.999 999 996 274 443 361 976 32 × 2 = 1 + 0.999 999 992 548 886 723 952 64;
  • 32) 0.999 999 992 548 886 723 952 64 × 2 = 1 + 0.999 999 985 097 773 447 905 28;
  • 33) 0.999 999 985 097 773 447 905 28 × 2 = 1 + 0.999 999 970 195 546 895 810 56;
  • 34) 0.999 999 970 195 546 895 810 56 × 2 = 1 + 0.999 999 940 391 093 791 621 12;
  • 35) 0.999 999 940 391 093 791 621 12 × 2 = 1 + 0.999 999 880 782 187 583 242 24;
  • 36) 0.999 999 880 782 187 583 242 24 × 2 = 1 + 0.999 999 761 564 375 166 484 48;
  • 37) 0.999 999 761 564 375 166 484 48 × 2 = 1 + 0.999 999 523 128 750 332 968 96;
  • 38) 0.999 999 523 128 750 332 968 96 × 2 = 1 + 0.999 999 046 257 500 665 937 92;
  • 39) 0.999 999 046 257 500 665 937 92 × 2 = 1 + 0.999 998 092 515 001 331 875 84;
  • 40) 0.999 998 092 515 001 331 875 84 × 2 = 1 + 0.999 996 185 030 002 663 751 68;
  • 41) 0.999 996 185 030 002 663 751 68 × 2 = 1 + 0.999 992 370 060 005 327 503 36;
  • 42) 0.999 992 370 060 005 327 503 36 × 2 = 1 + 0.999 984 740 120 010 655 006 72;
  • 43) 0.999 984 740 120 010 655 006 72 × 2 = 1 + 0.999 969 480 240 021 310 013 44;
  • 44) 0.999 969 480 240 021 310 013 44 × 2 = 1 + 0.999 938 960 480 042 620 026 88;
  • 45) 0.999 938 960 480 042 620 026 88 × 2 = 1 + 0.999 877 920 960 085 240 053 76;
  • 46) 0.999 877 920 960 085 240 053 76 × 2 = 1 + 0.999 755 841 920 170 480 107 52;
  • 47) 0.999 755 841 920 170 480 107 52 × 2 = 1 + 0.999 511 683 840 340 960 215 04;
  • 48) 0.999 511 683 840 340 960 215 04 × 2 = 1 + 0.999 023 367 680 681 920 430 08;
  • 49) 0.999 023 367 680 681 920 430 08 × 2 = 1 + 0.998 046 735 361 363 840 860 16;
  • 50) 0.998 046 735 361 363 840 860 16 × 2 = 1 + 0.996 093 470 722 727 681 720 32;
  • 51) 0.996 093 470 722 727 681 720 32 × 2 = 1 + 0.992 186 941 445 455 363 440 64;
  • 52) 0.992 186 941 445 455 363 440 64 × 2 = 1 + 0.984 373 882 890 910 726 881 28;
  • 53) 0.984 373 882 890 910 726 881 28 × 2 = 1 + 0.968 747 765 781 821 453 762 56;
  • 54) 0.968 747 765 781 821 453 762 56 × 2 = 1 + 0.937 495 531 563 642 907 525 12;
  • 55) 0.937 495 531 563 642 907 525 12 × 2 = 1 + 0.874 991 063 127 285 815 050 24;
  • 56) 0.874 991 063 127 285 815 050 24 × 2 = 1 + 0.749 982 126 254 571 630 100 48;
  • 57) 0.749 982 126 254 571 630 100 48 × 2 = 1 + 0.499 964 252 509 143 260 200 96;
  • 58) 0.499 964 252 509 143 260 200 96 × 2 = 0 + 0.999 928 505 018 286 520 401 92;

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:


0.016 738 891 601 562 496 530 305(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 305(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 305(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 530 305 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 1110

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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