-0.016 738 891 601 562 496 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 49(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 49(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 49| = 0.016 738 891 601 562 496 49


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 49.

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 49 × 2 = 0 + 0.033 477 783 203 124 992 98;
  • 2) 0.033 477 783 203 124 992 98 × 2 = 0 + 0.066 955 566 406 249 985 96;
  • 3) 0.066 955 566 406 249 985 96 × 2 = 0 + 0.133 911 132 812 499 971 92;
  • 4) 0.133 911 132 812 499 971 92 × 2 = 0 + 0.267 822 265 624 999 943 84;
  • 5) 0.267 822 265 624 999 943 84 × 2 = 0 + 0.535 644 531 249 999 887 68;
  • 6) 0.535 644 531 249 999 887 68 × 2 = 1 + 0.071 289 062 499 999 775 36;
  • 7) 0.071 289 062 499 999 775 36 × 2 = 0 + 0.142 578 124 999 999 550 72;
  • 8) 0.142 578 124 999 999 550 72 × 2 = 0 + 0.285 156 249 999 999 101 44;
  • 9) 0.285 156 249 999 999 101 44 × 2 = 0 + 0.570 312 499 999 998 202 88;
  • 10) 0.570 312 499 999 998 202 88 × 2 = 1 + 0.140 624 999 999 996 405 76;
  • 11) 0.140 624 999 999 996 405 76 × 2 = 0 + 0.281 249 999 999 992 811 52;
  • 12) 0.281 249 999 999 992 811 52 × 2 = 0 + 0.562 499 999 999 985 623 04;
  • 13) 0.562 499 999 999 985 623 04 × 2 = 1 + 0.124 999 999 999 971 246 08;
  • 14) 0.124 999 999 999 971 246 08 × 2 = 0 + 0.249 999 999 999 942 492 16;
  • 15) 0.249 999 999 999 942 492 16 × 2 = 0 + 0.499 999 999 999 884 984 32;
  • 16) 0.499 999 999 999 884 984 32 × 2 = 0 + 0.999 999 999 999 769 968 64;
  • 17) 0.999 999 999 999 769 968 64 × 2 = 1 + 0.999 999 999 999 539 937 28;
  • 18) 0.999 999 999 999 539 937 28 × 2 = 1 + 0.999 999 999 999 079 874 56;
  • 19) 0.999 999 999 999 079 874 56 × 2 = 1 + 0.999 999 999 998 159 749 12;
  • 20) 0.999 999 999 998 159 749 12 × 2 = 1 + 0.999 999 999 996 319 498 24;
  • 21) 0.999 999 999 996 319 498 24 × 2 = 1 + 0.999 999 999 992 638 996 48;
  • 22) 0.999 999 999 992 638 996 48 × 2 = 1 + 0.999 999 999 985 277 992 96;
  • 23) 0.999 999 999 985 277 992 96 × 2 = 1 + 0.999 999 999 970 555 985 92;
  • 24) 0.999 999 999 970 555 985 92 × 2 = 1 + 0.999 999 999 941 111 971 84;
  • 25) 0.999 999 999 941 111 971 84 × 2 = 1 + 0.999 999 999 882 223 943 68;
  • 26) 0.999 999 999 882 223 943 68 × 2 = 1 + 0.999 999 999 764 447 887 36;
  • 27) 0.999 999 999 764 447 887 36 × 2 = 1 + 0.999 999 999 528 895 774 72;
  • 28) 0.999 999 999 528 895 774 72 × 2 = 1 + 0.999 999 999 057 791 549 44;
  • 29) 0.999 999 999 057 791 549 44 × 2 = 1 + 0.999 999 998 115 583 098 88;
  • 30) 0.999 999 998 115 583 098 88 × 2 = 1 + 0.999 999 996 231 166 197 76;
  • 31) 0.999 999 996 231 166 197 76 × 2 = 1 + 0.999 999 992 462 332 395 52;
  • 32) 0.999 999 992 462 332 395 52 × 2 = 1 + 0.999 999 984 924 664 791 04;
  • 33) 0.999 999 984 924 664 791 04 × 2 = 1 + 0.999 999 969 849 329 582 08;
  • 34) 0.999 999 969 849 329 582 08 × 2 = 1 + 0.999 999 939 698 659 164 16;
  • 35) 0.999 999 939 698 659 164 16 × 2 = 1 + 0.999 999 879 397 318 328 32;
  • 36) 0.999 999 879 397 318 328 32 × 2 = 1 + 0.999 999 758 794 636 656 64;
  • 37) 0.999 999 758 794 636 656 64 × 2 = 1 + 0.999 999 517 589 273 313 28;
  • 38) 0.999 999 517 589 273 313 28 × 2 = 1 + 0.999 999 035 178 546 626 56;
  • 39) 0.999 999 035 178 546 626 56 × 2 = 1 + 0.999 998 070 357 093 253 12;
  • 40) 0.999 998 070 357 093 253 12 × 2 = 1 + 0.999 996 140 714 186 506 24;
  • 41) 0.999 996 140 714 186 506 24 × 2 = 1 + 0.999 992 281 428 373 012 48;
  • 42) 0.999 992 281 428 373 012 48 × 2 = 1 + 0.999 984 562 856 746 024 96;
  • 43) 0.999 984 562 856 746 024 96 × 2 = 1 + 0.999 969 125 713 492 049 92;
  • 44) 0.999 969 125 713 492 049 92 × 2 = 1 + 0.999 938 251 426 984 099 84;
  • 45) 0.999 938 251 426 984 099 84 × 2 = 1 + 0.999 876 502 853 968 199 68;
  • 46) 0.999 876 502 853 968 199 68 × 2 = 1 + 0.999 753 005 707 936 399 36;
  • 47) 0.999 753 005 707 936 399 36 × 2 = 1 + 0.999 506 011 415 872 798 72;
  • 48) 0.999 506 011 415 872 798 72 × 2 = 1 + 0.999 012 022 831 745 597 44;
  • 49) 0.999 012 022 831 745 597 44 × 2 = 1 + 0.998 024 045 663 491 194 88;
  • 50) 0.998 024 045 663 491 194 88 × 2 = 1 + 0.996 048 091 326 982 389 76;
  • 51) 0.996 048 091 326 982 389 76 × 2 = 1 + 0.992 096 182 653 964 779 52;
  • 52) 0.992 096 182 653 964 779 52 × 2 = 1 + 0.984 192 365 307 929 559 04;
  • 53) 0.984 192 365 307 929 559 04 × 2 = 1 + 0.968 384 730 615 859 118 08;
  • 54) 0.968 384 730 615 859 118 08 × 2 = 1 + 0.936 769 461 231 718 236 16;
  • 55) 0.936 769 461 231 718 236 16 × 2 = 1 + 0.873 538 922 463 436 472 32;
  • 56) 0.873 538 922 463 436 472 32 × 2 = 1 + 0.747 077 844 926 872 944 64;
  • 57) 0.747 077 844 926 872 944 64 × 2 = 1 + 0.494 155 689 853 745 889 28;
  • 58) 0.494 155 689 853 745 889 28 × 2 = 0 + 0.988 311 379 707 491 778 56;

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 49(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 49(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 49(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 49 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