-0.016 738 891 601 562 494 87 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 87(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
-0.016 738 891 601 562 494 87(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:

|-0.016 738 891 601 562 494 87| = 0.016 738 891 601 562 494 87


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: 0.016 738 891 601 562 494 87.

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) 0.016 738 891 601 562 494 87 × 2 = 0 + 0.033 477 783 203 124 989 74;
  • 2) 0.033 477 783 203 124 989 74 × 2 = 0 + 0.066 955 566 406 249 979 48;
  • 3) 0.066 955 566 406 249 979 48 × 2 = 0 + 0.133 911 132 812 499 958 96;
  • 4) 0.133 911 132 812 499 958 96 × 2 = 0 + 0.267 822 265 624 999 917 92;
  • 5) 0.267 822 265 624 999 917 92 × 2 = 0 + 0.535 644 531 249 999 835 84;
  • 6) 0.535 644 531 249 999 835 84 × 2 = 1 + 0.071 289 062 499 999 671 68;
  • 7) 0.071 289 062 499 999 671 68 × 2 = 0 + 0.142 578 124 999 999 343 36;
  • 8) 0.142 578 124 999 999 343 36 × 2 = 0 + 0.285 156 249 999 998 686 72;
  • 9) 0.285 156 249 999 998 686 72 × 2 = 0 + 0.570 312 499 999 997 373 44;
  • 10) 0.570 312 499 999 997 373 44 × 2 = 1 + 0.140 624 999 999 994 746 88;
  • 11) 0.140 624 999 999 994 746 88 × 2 = 0 + 0.281 249 999 999 989 493 76;
  • 12) 0.281 249 999 999 989 493 76 × 2 = 0 + 0.562 499 999 999 978 987 52;
  • 13) 0.562 499 999 999 978 987 52 × 2 = 1 + 0.124 999 999 999 957 975 04;
  • 14) 0.124 999 999 999 957 975 04 × 2 = 0 + 0.249 999 999 999 915 950 08;
  • 15) 0.249 999 999 999 915 950 08 × 2 = 0 + 0.499 999 999 999 831 900 16;
  • 16) 0.499 999 999 999 831 900 16 × 2 = 0 + 0.999 999 999 999 663 800 32;
  • 17) 0.999 999 999 999 663 800 32 × 2 = 1 + 0.999 999 999 999 327 600 64;
  • 18) 0.999 999 999 999 327 600 64 × 2 = 1 + 0.999 999 999 998 655 201 28;
  • 19) 0.999 999 999 998 655 201 28 × 2 = 1 + 0.999 999 999 997 310 402 56;
  • 20) 0.999 999 999 997 310 402 56 × 2 = 1 + 0.999 999 999 994 620 805 12;
  • 21) 0.999 999 999 994 620 805 12 × 2 = 1 + 0.999 999 999 989 241 610 24;
  • 22) 0.999 999 999 989 241 610 24 × 2 = 1 + 0.999 999 999 978 483 220 48;
  • 23) 0.999 999 999 978 483 220 48 × 2 = 1 + 0.999 999 999 956 966 440 96;
  • 24) 0.999 999 999 956 966 440 96 × 2 = 1 + 0.999 999 999 913 932 881 92;
  • 25) 0.999 999 999 913 932 881 92 × 2 = 1 + 0.999 999 999 827 865 763 84;
  • 26) 0.999 999 999 827 865 763 84 × 2 = 1 + 0.999 999 999 655 731 527 68;
  • 27) 0.999 999 999 655 731 527 68 × 2 = 1 + 0.999 999 999 311 463 055 36;
  • 28) 0.999 999 999 311 463 055 36 × 2 = 1 + 0.999 999 998 622 926 110 72;
  • 29) 0.999 999 998 622 926 110 72 × 2 = 1 + 0.999 999 997 245 852 221 44;
  • 30) 0.999 999 997 245 852 221 44 × 2 = 1 + 0.999 999 994 491 704 442 88;
  • 31) 0.999 999 994 491 704 442 88 × 2 = 1 + 0.999 999 988 983 408 885 76;
  • 32) 0.999 999 988 983 408 885 76 × 2 = 1 + 0.999 999 977 966 817 771 52;
  • 33) 0.999 999 977 966 817 771 52 × 2 = 1 + 0.999 999 955 933 635 543 04;
  • 34) 0.999 999 955 933 635 543 04 × 2 = 1 + 0.999 999 911 867 271 086 08;
  • 35) 0.999 999 911 867 271 086 08 × 2 = 1 + 0.999 999 823 734 542 172 16;
  • 36) 0.999 999 823 734 542 172 16 × 2 = 1 + 0.999 999 647 469 084 344 32;
  • 37) 0.999 999 647 469 084 344 32 × 2 = 1 + 0.999 999 294 938 168 688 64;
  • 38) 0.999 999 294 938 168 688 64 × 2 = 1 + 0.999 998 589 876 337 377 28;
  • 39) 0.999 998 589 876 337 377 28 × 2 = 1 + 0.999 997 179 752 674 754 56;
  • 40) 0.999 997 179 752 674 754 56 × 2 = 1 + 0.999 994 359 505 349 509 12;
  • 41) 0.999 994 359 505 349 509 12 × 2 = 1 + 0.999 988 719 010 699 018 24;
  • 42) 0.999 988 719 010 699 018 24 × 2 = 1 + 0.999 977 438 021 398 036 48;
  • 43) 0.999 977 438 021 398 036 48 × 2 = 1 + 0.999 954 876 042 796 072 96;
  • 44) 0.999 954 876 042 796 072 96 × 2 = 1 + 0.999 909 752 085 592 145 92;
  • 45) 0.999 909 752 085 592 145 92 × 2 = 1 + 0.999 819 504 171 184 291 84;
  • 46) 0.999 819 504 171 184 291 84 × 2 = 1 + 0.999 639 008 342 368 583 68;
  • 47) 0.999 639 008 342 368 583 68 × 2 = 1 + 0.999 278 016 684 737 167 36;
  • 48) 0.999 278 016 684 737 167 36 × 2 = 1 + 0.998 556 033 369 474 334 72;
  • 49) 0.998 556 033 369 474 334 72 × 2 = 1 + 0.997 112 066 738 948 669 44;
  • 50) 0.997 112 066 738 948 669 44 × 2 = 1 + 0.994 224 133 477 897 338 88;
  • 51) 0.994 224 133 477 897 338 88 × 2 = 1 + 0.988 448 266 955 794 677 76;
  • 52) 0.988 448 266 955 794 677 76 × 2 = 1 + 0.976 896 533 911 589 355 52;
  • 53) 0.976 896 533 911 589 355 52 × 2 = 1 + 0.953 793 067 823 178 711 04;
  • 54) 0.953 793 067 823 178 711 04 × 2 = 1 + 0.907 586 135 646 357 422 08;
  • 55) 0.907 586 135 646 357 422 08 × 2 = 1 + 0.815 172 271 292 714 844 16;
  • 56) 0.815 172 271 292 714 844 16 × 2 = 1 + 0.630 344 542 585 429 688 32;
  • 57) 0.630 344 542 585 429 688 32 × 2 = 1 + 0.260 689 085 170 859 376 64;
  • 58) 0.260 689 085 170 859 376 64 × 2 = 0 + 0.521 378 170 341 718 753 28;

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:


0.016 738 891 601 562 494 87(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 494 87(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:


0.016 738 891 601 562 494 87(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 -0.016 738 891 601 562 494 87 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