-0.016 738 891 601 562 493 85 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 493 85(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
-0.016 738 891 601 562 493 85(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:

|-0.016 738 891 601 562 493 85| = 0.016 738 891 601 562 493 85


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: 0.016 738 891 601 562 493 85.

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) 0.016 738 891 601 562 493 85 × 2 = 0 + 0.033 477 783 203 124 987 7;
  • 2) 0.033 477 783 203 124 987 7 × 2 = 0 + 0.066 955 566 406 249 975 4;
  • 3) 0.066 955 566 406 249 975 4 × 2 = 0 + 0.133 911 132 812 499 950 8;
  • 4) 0.133 911 132 812 499 950 8 × 2 = 0 + 0.267 822 265 624 999 901 6;
  • 5) 0.267 822 265 624 999 901 6 × 2 = 0 + 0.535 644 531 249 999 803 2;
  • 6) 0.535 644 531 249 999 803 2 × 2 = 1 + 0.071 289 062 499 999 606 4;
  • 7) 0.071 289 062 499 999 606 4 × 2 = 0 + 0.142 578 124 999 999 212 8;
  • 8) 0.142 578 124 999 999 212 8 × 2 = 0 + 0.285 156 249 999 998 425 6;
  • 9) 0.285 156 249 999 998 425 6 × 2 = 0 + 0.570 312 499 999 996 851 2;
  • 10) 0.570 312 499 999 996 851 2 × 2 = 1 + 0.140 624 999 999 993 702 4;
  • 11) 0.140 624 999 999 993 702 4 × 2 = 0 + 0.281 249 999 999 987 404 8;
  • 12) 0.281 249 999 999 987 404 8 × 2 = 0 + 0.562 499 999 999 974 809 6;
  • 13) 0.562 499 999 999 974 809 6 × 2 = 1 + 0.124 999 999 999 949 619 2;
  • 14) 0.124 999 999 999 949 619 2 × 2 = 0 + 0.249 999 999 999 899 238 4;
  • 15) 0.249 999 999 999 899 238 4 × 2 = 0 + 0.499 999 999 999 798 476 8;
  • 16) 0.499 999 999 999 798 476 8 × 2 = 0 + 0.999 999 999 999 596 953 6;
  • 17) 0.999 999 999 999 596 953 6 × 2 = 1 + 0.999 999 999 999 193 907 2;
  • 18) 0.999 999 999 999 193 907 2 × 2 = 1 + 0.999 999 999 998 387 814 4;
  • 19) 0.999 999 999 998 387 814 4 × 2 = 1 + 0.999 999 999 996 775 628 8;
  • 20) 0.999 999 999 996 775 628 8 × 2 = 1 + 0.999 999 999 993 551 257 6;
  • 21) 0.999 999 999 993 551 257 6 × 2 = 1 + 0.999 999 999 987 102 515 2;
  • 22) 0.999 999 999 987 102 515 2 × 2 = 1 + 0.999 999 999 974 205 030 4;
  • 23) 0.999 999 999 974 205 030 4 × 2 = 1 + 0.999 999 999 948 410 060 8;
  • 24) 0.999 999 999 948 410 060 8 × 2 = 1 + 0.999 999 999 896 820 121 6;
  • 25) 0.999 999 999 896 820 121 6 × 2 = 1 + 0.999 999 999 793 640 243 2;
  • 26) 0.999 999 999 793 640 243 2 × 2 = 1 + 0.999 999 999 587 280 486 4;
  • 27) 0.999 999 999 587 280 486 4 × 2 = 1 + 0.999 999 999 174 560 972 8;
  • 28) 0.999 999 999 174 560 972 8 × 2 = 1 + 0.999 999 998 349 121 945 6;
  • 29) 0.999 999 998 349 121 945 6 × 2 = 1 + 0.999 999 996 698 243 891 2;
  • 30) 0.999 999 996 698 243 891 2 × 2 = 1 + 0.999 999 993 396 487 782 4;
  • 31) 0.999 999 993 396 487 782 4 × 2 = 1 + 0.999 999 986 792 975 564 8;
  • 32) 0.999 999 986 792 975 564 8 × 2 = 1 + 0.999 999 973 585 951 129 6;
  • 33) 0.999 999 973 585 951 129 6 × 2 = 1 + 0.999 999 947 171 902 259 2;
  • 34) 0.999 999 947 171 902 259 2 × 2 = 1 + 0.999 999 894 343 804 518 4;
  • 35) 0.999 999 894 343 804 518 4 × 2 = 1 + 0.999 999 788 687 609 036 8;
  • 36) 0.999 999 788 687 609 036 8 × 2 = 1 + 0.999 999 577 375 218 073 6;
  • 37) 0.999 999 577 375 218 073 6 × 2 = 1 + 0.999 999 154 750 436 147 2;
  • 38) 0.999 999 154 750 436 147 2 × 2 = 1 + 0.999 998 309 500 872 294 4;
  • 39) 0.999 998 309 500 872 294 4 × 2 = 1 + 0.999 996 619 001 744 588 8;
  • 40) 0.999 996 619 001 744 588 8 × 2 = 1 + 0.999 993 238 003 489 177 6;
  • 41) 0.999 993 238 003 489 177 6 × 2 = 1 + 0.999 986 476 006 978 355 2;
  • 42) 0.999 986 476 006 978 355 2 × 2 = 1 + 0.999 972 952 013 956 710 4;
  • 43) 0.999 972 952 013 956 710 4 × 2 = 1 + 0.999 945 904 027 913 420 8;
  • 44) 0.999 945 904 027 913 420 8 × 2 = 1 + 0.999 891 808 055 826 841 6;
  • 45) 0.999 891 808 055 826 841 6 × 2 = 1 + 0.999 783 616 111 653 683 2;
  • 46) 0.999 783 616 111 653 683 2 × 2 = 1 + 0.999 567 232 223 307 366 4;
  • 47) 0.999 567 232 223 307 366 4 × 2 = 1 + 0.999 134 464 446 614 732 8;
  • 48) 0.999 134 464 446 614 732 8 × 2 = 1 + 0.998 268 928 893 229 465 6;
  • 49) 0.998 268 928 893 229 465 6 × 2 = 1 + 0.996 537 857 786 458 931 2;
  • 50) 0.996 537 857 786 458 931 2 × 2 = 1 + 0.993 075 715 572 917 862 4;
  • 51) 0.993 075 715 572 917 862 4 × 2 = 1 + 0.986 151 431 145 835 724 8;
  • 52) 0.986 151 431 145 835 724 8 × 2 = 1 + 0.972 302 862 291 671 449 6;
  • 53) 0.972 302 862 291 671 449 6 × 2 = 1 + 0.944 605 724 583 342 899 2;
  • 54) 0.944 605 724 583 342 899 2 × 2 = 1 + 0.889 211 449 166 685 798 4;
  • 55) 0.889 211 449 166 685 798 4 × 2 = 1 + 0.778 422 898 333 371 596 8;
  • 56) 0.778 422 898 333 371 596 8 × 2 = 1 + 0.556 845 796 666 743 193 6;
  • 57) 0.556 845 796 666 743 193 6 × 2 = 1 + 0.113 691 593 333 486 387 2;
  • 58) 0.113 691 593 333 486 387 2 × 2 = 0 + 0.227 383 186 666 972 774 4;

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:


0.016 738 891 601 562 493 85(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 493 85(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:


0.016 738 891 601 562 493 85(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 -0.016 738 891 601 562 493 85 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