-0.016 738 891 601 562 497 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 29(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
-0.016 738 891 601 562 497 29(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:

|-0.016 738 891 601 562 497 29| = 0.016 738 891 601 562 497 29


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: 0.016 738 891 601 562 497 29.

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) 0.016 738 891 601 562 497 29 × 2 = 0 + 0.033 477 783 203 124 994 58;
  • 2) 0.033 477 783 203 124 994 58 × 2 = 0 + 0.066 955 566 406 249 989 16;
  • 3) 0.066 955 566 406 249 989 16 × 2 = 0 + 0.133 911 132 812 499 978 32;
  • 4) 0.133 911 132 812 499 978 32 × 2 = 0 + 0.267 822 265 624 999 956 64;
  • 5) 0.267 822 265 624 999 956 64 × 2 = 0 + 0.535 644 531 249 999 913 28;
  • 6) 0.535 644 531 249 999 913 28 × 2 = 1 + 0.071 289 062 499 999 826 56;
  • 7) 0.071 289 062 499 999 826 56 × 2 = 0 + 0.142 578 124 999 999 653 12;
  • 8) 0.142 578 124 999 999 653 12 × 2 = 0 + 0.285 156 249 999 999 306 24;
  • 9) 0.285 156 249 999 999 306 24 × 2 = 0 + 0.570 312 499 999 998 612 48;
  • 10) 0.570 312 499 999 998 612 48 × 2 = 1 + 0.140 624 999 999 997 224 96;
  • 11) 0.140 624 999 999 997 224 96 × 2 = 0 + 0.281 249 999 999 994 449 92;
  • 12) 0.281 249 999 999 994 449 92 × 2 = 0 + 0.562 499 999 999 988 899 84;
  • 13) 0.562 499 999 999 988 899 84 × 2 = 1 + 0.124 999 999 999 977 799 68;
  • 14) 0.124 999 999 999 977 799 68 × 2 = 0 + 0.249 999 999 999 955 599 36;
  • 15) 0.249 999 999 999 955 599 36 × 2 = 0 + 0.499 999 999 999 911 198 72;
  • 16) 0.499 999 999 999 911 198 72 × 2 = 0 + 0.999 999 999 999 822 397 44;
  • 17) 0.999 999 999 999 822 397 44 × 2 = 1 + 0.999 999 999 999 644 794 88;
  • 18) 0.999 999 999 999 644 794 88 × 2 = 1 + 0.999 999 999 999 289 589 76;
  • 19) 0.999 999 999 999 289 589 76 × 2 = 1 + 0.999 999 999 998 579 179 52;
  • 20) 0.999 999 999 998 579 179 52 × 2 = 1 + 0.999 999 999 997 158 359 04;
  • 21) 0.999 999 999 997 158 359 04 × 2 = 1 + 0.999 999 999 994 316 718 08;
  • 22) 0.999 999 999 994 316 718 08 × 2 = 1 + 0.999 999 999 988 633 436 16;
  • 23) 0.999 999 999 988 633 436 16 × 2 = 1 + 0.999 999 999 977 266 872 32;
  • 24) 0.999 999 999 977 266 872 32 × 2 = 1 + 0.999 999 999 954 533 744 64;
  • 25) 0.999 999 999 954 533 744 64 × 2 = 1 + 0.999 999 999 909 067 489 28;
  • 26) 0.999 999 999 909 067 489 28 × 2 = 1 + 0.999 999 999 818 134 978 56;
  • 27) 0.999 999 999 818 134 978 56 × 2 = 1 + 0.999 999 999 636 269 957 12;
  • 28) 0.999 999 999 636 269 957 12 × 2 = 1 + 0.999 999 999 272 539 914 24;
  • 29) 0.999 999 999 272 539 914 24 × 2 = 1 + 0.999 999 998 545 079 828 48;
  • 30) 0.999 999 998 545 079 828 48 × 2 = 1 + 0.999 999 997 090 159 656 96;
  • 31) 0.999 999 997 090 159 656 96 × 2 = 1 + 0.999 999 994 180 319 313 92;
  • 32) 0.999 999 994 180 319 313 92 × 2 = 1 + 0.999 999 988 360 638 627 84;
  • 33) 0.999 999 988 360 638 627 84 × 2 = 1 + 0.999 999 976 721 277 255 68;
  • 34) 0.999 999 976 721 277 255 68 × 2 = 1 + 0.999 999 953 442 554 511 36;
  • 35) 0.999 999 953 442 554 511 36 × 2 = 1 + 0.999 999 906 885 109 022 72;
  • 36) 0.999 999 906 885 109 022 72 × 2 = 1 + 0.999 999 813 770 218 045 44;
  • 37) 0.999 999 813 770 218 045 44 × 2 = 1 + 0.999 999 627 540 436 090 88;
  • 38) 0.999 999 627 540 436 090 88 × 2 = 1 + 0.999 999 255 080 872 181 76;
  • 39) 0.999 999 255 080 872 181 76 × 2 = 1 + 0.999 998 510 161 744 363 52;
  • 40) 0.999 998 510 161 744 363 52 × 2 = 1 + 0.999 997 020 323 488 727 04;
  • 41) 0.999 997 020 323 488 727 04 × 2 = 1 + 0.999 994 040 646 977 454 08;
  • 42) 0.999 994 040 646 977 454 08 × 2 = 1 + 0.999 988 081 293 954 908 16;
  • 43) 0.999 988 081 293 954 908 16 × 2 = 1 + 0.999 976 162 587 909 816 32;
  • 44) 0.999 976 162 587 909 816 32 × 2 = 1 + 0.999 952 325 175 819 632 64;
  • 45) 0.999 952 325 175 819 632 64 × 2 = 1 + 0.999 904 650 351 639 265 28;
  • 46) 0.999 904 650 351 639 265 28 × 2 = 1 + 0.999 809 300 703 278 530 56;
  • 47) 0.999 809 300 703 278 530 56 × 2 = 1 + 0.999 618 601 406 557 061 12;
  • 48) 0.999 618 601 406 557 061 12 × 2 = 1 + 0.999 237 202 813 114 122 24;
  • 49) 0.999 237 202 813 114 122 24 × 2 = 1 + 0.998 474 405 626 228 244 48;
  • 50) 0.998 474 405 626 228 244 48 × 2 = 1 + 0.996 948 811 252 456 488 96;
  • 51) 0.996 948 811 252 456 488 96 × 2 = 1 + 0.993 897 622 504 912 977 92;
  • 52) 0.993 897 622 504 912 977 92 × 2 = 1 + 0.987 795 245 009 825 955 84;
  • 53) 0.987 795 245 009 825 955 84 × 2 = 1 + 0.975 590 490 019 651 911 68;
  • 54) 0.975 590 490 019 651 911 68 × 2 = 1 + 0.951 180 980 039 303 823 36;
  • 55) 0.951 180 980 039 303 823 36 × 2 = 1 + 0.902 361 960 078 607 646 72;
  • 56) 0.902 361 960 078 607 646 72 × 2 = 1 + 0.804 723 920 157 215 293 44;
  • 57) 0.804 723 920 157 215 293 44 × 2 = 1 + 0.609 447 840 314 430 586 88;
  • 58) 0.609 447 840 314 430 586 88 × 2 = 1 + 0.218 895 680 628 861 173 76;

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:


0.016 738 891 601 562 497 29(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 497 29(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:


0.016 738 891 601 562 497 29(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 -0.016 738 891 601 562 497 29 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