-0.016 738 891 601 562 491 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 491 9(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
-0.016 738 891 601 562 491 9(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:

|-0.016 738 891 601 562 491 9| = 0.016 738 891 601 562 491 9


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: 0.016 738 891 601 562 491 9.

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) 0.016 738 891 601 562 491 9 × 2 = 0 + 0.033 477 783 203 124 983 8;
  • 2) 0.033 477 783 203 124 983 8 × 2 = 0 + 0.066 955 566 406 249 967 6;
  • 3) 0.066 955 566 406 249 967 6 × 2 = 0 + 0.133 911 132 812 499 935 2;
  • 4) 0.133 911 132 812 499 935 2 × 2 = 0 + 0.267 822 265 624 999 870 4;
  • 5) 0.267 822 265 624 999 870 4 × 2 = 0 + 0.535 644 531 249 999 740 8;
  • 6) 0.535 644 531 249 999 740 8 × 2 = 1 + 0.071 289 062 499 999 481 6;
  • 7) 0.071 289 062 499 999 481 6 × 2 = 0 + 0.142 578 124 999 998 963 2;
  • 8) 0.142 578 124 999 998 963 2 × 2 = 0 + 0.285 156 249 999 997 926 4;
  • 9) 0.285 156 249 999 997 926 4 × 2 = 0 + 0.570 312 499 999 995 852 8;
  • 10) 0.570 312 499 999 995 852 8 × 2 = 1 + 0.140 624 999 999 991 705 6;
  • 11) 0.140 624 999 999 991 705 6 × 2 = 0 + 0.281 249 999 999 983 411 2;
  • 12) 0.281 249 999 999 983 411 2 × 2 = 0 + 0.562 499 999 999 966 822 4;
  • 13) 0.562 499 999 999 966 822 4 × 2 = 1 + 0.124 999 999 999 933 644 8;
  • 14) 0.124 999 999 999 933 644 8 × 2 = 0 + 0.249 999 999 999 867 289 6;
  • 15) 0.249 999 999 999 867 289 6 × 2 = 0 + 0.499 999 999 999 734 579 2;
  • 16) 0.499 999 999 999 734 579 2 × 2 = 0 + 0.999 999 999 999 469 158 4;
  • 17) 0.999 999 999 999 469 158 4 × 2 = 1 + 0.999 999 999 998 938 316 8;
  • 18) 0.999 999 999 998 938 316 8 × 2 = 1 + 0.999 999 999 997 876 633 6;
  • 19) 0.999 999 999 997 876 633 6 × 2 = 1 + 0.999 999 999 995 753 267 2;
  • 20) 0.999 999 999 995 753 267 2 × 2 = 1 + 0.999 999 999 991 506 534 4;
  • 21) 0.999 999 999 991 506 534 4 × 2 = 1 + 0.999 999 999 983 013 068 8;
  • 22) 0.999 999 999 983 013 068 8 × 2 = 1 + 0.999 999 999 966 026 137 6;
  • 23) 0.999 999 999 966 026 137 6 × 2 = 1 + 0.999 999 999 932 052 275 2;
  • 24) 0.999 999 999 932 052 275 2 × 2 = 1 + 0.999 999 999 864 104 550 4;
  • 25) 0.999 999 999 864 104 550 4 × 2 = 1 + 0.999 999 999 728 209 100 8;
  • 26) 0.999 999 999 728 209 100 8 × 2 = 1 + 0.999 999 999 456 418 201 6;
  • 27) 0.999 999 999 456 418 201 6 × 2 = 1 + 0.999 999 998 912 836 403 2;
  • 28) 0.999 999 998 912 836 403 2 × 2 = 1 + 0.999 999 997 825 672 806 4;
  • 29) 0.999 999 997 825 672 806 4 × 2 = 1 + 0.999 999 995 651 345 612 8;
  • 30) 0.999 999 995 651 345 612 8 × 2 = 1 + 0.999 999 991 302 691 225 6;
  • 31) 0.999 999 991 302 691 225 6 × 2 = 1 + 0.999 999 982 605 382 451 2;
  • 32) 0.999 999 982 605 382 451 2 × 2 = 1 + 0.999 999 965 210 764 902 4;
  • 33) 0.999 999 965 210 764 902 4 × 2 = 1 + 0.999 999 930 421 529 804 8;
  • 34) 0.999 999 930 421 529 804 8 × 2 = 1 + 0.999 999 860 843 059 609 6;
  • 35) 0.999 999 860 843 059 609 6 × 2 = 1 + 0.999 999 721 686 119 219 2;
  • 36) 0.999 999 721 686 119 219 2 × 2 = 1 + 0.999 999 443 372 238 438 4;
  • 37) 0.999 999 443 372 238 438 4 × 2 = 1 + 0.999 998 886 744 476 876 8;
  • 38) 0.999 998 886 744 476 876 8 × 2 = 1 + 0.999 997 773 488 953 753 6;
  • 39) 0.999 997 773 488 953 753 6 × 2 = 1 + 0.999 995 546 977 907 507 2;
  • 40) 0.999 995 546 977 907 507 2 × 2 = 1 + 0.999 991 093 955 815 014 4;
  • 41) 0.999 991 093 955 815 014 4 × 2 = 1 + 0.999 982 187 911 630 028 8;
  • 42) 0.999 982 187 911 630 028 8 × 2 = 1 + 0.999 964 375 823 260 057 6;
  • 43) 0.999 964 375 823 260 057 6 × 2 = 1 + 0.999 928 751 646 520 115 2;
  • 44) 0.999 928 751 646 520 115 2 × 2 = 1 + 0.999 857 503 293 040 230 4;
  • 45) 0.999 857 503 293 040 230 4 × 2 = 1 + 0.999 715 006 586 080 460 8;
  • 46) 0.999 715 006 586 080 460 8 × 2 = 1 + 0.999 430 013 172 160 921 6;
  • 47) 0.999 430 013 172 160 921 6 × 2 = 1 + 0.998 860 026 344 321 843 2;
  • 48) 0.998 860 026 344 321 843 2 × 2 = 1 + 0.997 720 052 688 643 686 4;
  • 49) 0.997 720 052 688 643 686 4 × 2 = 1 + 0.995 440 105 377 287 372 8;
  • 50) 0.995 440 105 377 287 372 8 × 2 = 1 + 0.990 880 210 754 574 745 6;
  • 51) 0.990 880 210 754 574 745 6 × 2 = 1 + 0.981 760 421 509 149 491 2;
  • 52) 0.981 760 421 509 149 491 2 × 2 = 1 + 0.963 520 843 018 298 982 4;
  • 53) 0.963 520 843 018 298 982 4 × 2 = 1 + 0.927 041 686 036 597 964 8;
  • 54) 0.927 041 686 036 597 964 8 × 2 = 1 + 0.854 083 372 073 195 929 6;
  • 55) 0.854 083 372 073 195 929 6 × 2 = 1 + 0.708 166 744 146 391 859 2;
  • 56) 0.708 166 744 146 391 859 2 × 2 = 1 + 0.416 333 488 292 783 718 4;
  • 57) 0.416 333 488 292 783 718 4 × 2 = 0 + 0.832 666 976 585 567 436 8;
  • 58) 0.832 666 976 585 567 436 8 × 2 = 1 + 0.665 333 953 171 134 873 6;

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:


0.016 738 891 601 562 491 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.016 738 891 601 562 491 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(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:


0.016 738 891 601 562 491 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101(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 1101


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 1101 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101


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 1101


Decimal number -0.016 738 891 601 562 491 9 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 1101

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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