-0.620 527 518 458 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.620 527 518 458 97(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
-0.620 527 518 458 97(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:

|-0.620 527 518 458 97| = 0.620 527 518 458 97


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: 0.620 527 518 458 97.

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) 0.620 527 518 458 97 × 2 = 1 + 0.241 055 036 917 94;
  • 2) 0.241 055 036 917 94 × 2 = 0 + 0.482 110 073 835 88;
  • 3) 0.482 110 073 835 88 × 2 = 0 + 0.964 220 147 671 76;
  • 4) 0.964 220 147 671 76 × 2 = 1 + 0.928 440 295 343 52;
  • 5) 0.928 440 295 343 52 × 2 = 1 + 0.856 880 590 687 04;
  • 6) 0.856 880 590 687 04 × 2 = 1 + 0.713 761 181 374 08;
  • 7) 0.713 761 181 374 08 × 2 = 1 + 0.427 522 362 748 16;
  • 8) 0.427 522 362 748 16 × 2 = 0 + 0.855 044 725 496 32;
  • 9) 0.855 044 725 496 32 × 2 = 1 + 0.710 089 450 992 64;
  • 10) 0.710 089 450 992 64 × 2 = 1 + 0.420 178 901 985 28;
  • 11) 0.420 178 901 985 28 × 2 = 0 + 0.840 357 803 970 56;
  • 12) 0.840 357 803 970 56 × 2 = 1 + 0.680 715 607 941 12;
  • 13) 0.680 715 607 941 12 × 2 = 1 + 0.361 431 215 882 24;
  • 14) 0.361 431 215 882 24 × 2 = 0 + 0.722 862 431 764 48;
  • 15) 0.722 862 431 764 48 × 2 = 1 + 0.445 724 863 528 96;
  • 16) 0.445 724 863 528 96 × 2 = 0 + 0.891 449 727 057 92;
  • 17) 0.891 449 727 057 92 × 2 = 1 + 0.782 899 454 115 84;
  • 18) 0.782 899 454 115 84 × 2 = 1 + 0.565 798 908 231 68;
  • 19) 0.565 798 908 231 68 × 2 = 1 + 0.131 597 816 463 36;
  • 20) 0.131 597 816 463 36 × 2 = 0 + 0.263 195 632 926 72;
  • 21) 0.263 195 632 926 72 × 2 = 0 + 0.526 391 265 853 44;
  • 22) 0.526 391 265 853 44 × 2 = 1 + 0.052 782 531 706 88;
  • 23) 0.052 782 531 706 88 × 2 = 0 + 0.105 565 063 413 76;
  • 24) 0.105 565 063 413 76 × 2 = 0 + 0.211 130 126 827 52;
  • 25) 0.211 130 126 827 52 × 2 = 0 + 0.422 260 253 655 04;
  • 26) 0.422 260 253 655 04 × 2 = 0 + 0.844 520 507 310 08;
  • 27) 0.844 520 507 310 08 × 2 = 1 + 0.689 041 014 620 16;
  • 28) 0.689 041 014 620 16 × 2 = 1 + 0.378 082 029 240 32;
  • 29) 0.378 082 029 240 32 × 2 = 0 + 0.756 164 058 480 64;
  • 30) 0.756 164 058 480 64 × 2 = 1 + 0.512 328 116 961 28;
  • 31) 0.512 328 116 961 28 × 2 = 1 + 0.024 656 233 922 56;
  • 32) 0.024 656 233 922 56 × 2 = 0 + 0.049 312 467 845 12;
  • 33) 0.049 312 467 845 12 × 2 = 0 + 0.098 624 935 690 24;
  • 34) 0.098 624 935 690 24 × 2 = 0 + 0.197 249 871 380 48;
  • 35) 0.197 249 871 380 48 × 2 = 0 + 0.394 499 742 760 96;
  • 36) 0.394 499 742 760 96 × 2 = 0 + 0.788 999 485 521 92;
  • 37) 0.788 999 485 521 92 × 2 = 1 + 0.577 998 971 043 84;
  • 38) 0.577 998 971 043 84 × 2 = 1 + 0.155 997 942 087 68;
  • 39) 0.155 997 942 087 68 × 2 = 0 + 0.311 995 884 175 36;
  • 40) 0.311 995 884 175 36 × 2 = 0 + 0.623 991 768 350 72;
  • 41) 0.623 991 768 350 72 × 2 = 1 + 0.247 983 536 701 44;
  • 42) 0.247 983 536 701 44 × 2 = 0 + 0.495 967 073 402 88;
  • 43) 0.495 967 073 402 88 × 2 = 0 + 0.991 934 146 805 76;
  • 44) 0.991 934 146 805 76 × 2 = 1 + 0.983 868 293 611 52;
  • 45) 0.983 868 293 611 52 × 2 = 1 + 0.967 736 587 223 04;
  • 46) 0.967 736 587 223 04 × 2 = 1 + 0.935 473 174 446 08;
  • 47) 0.935 473 174 446 08 × 2 = 1 + 0.870 946 348 892 16;
  • 48) 0.870 946 348 892 16 × 2 = 1 + 0.741 892 697 784 32;
  • 49) 0.741 892 697 784 32 × 2 = 1 + 0.483 785 395 568 64;
  • 50) 0.483 785 395 568 64 × 2 = 0 + 0.967 570 791 137 28;
  • 51) 0.967 570 791 137 28 × 2 = 1 + 0.935 141 582 274 56;
  • 52) 0.935 141 582 274 56 × 2 = 1 + 0.870 283 164 549 12;
  • 53) 0.870 283 164 549 12 × 2 = 1 + 0.740 566 329 098 24;

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:


0.620 527 518 458 97(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 1001 1111 1011 1(2)

6. Positive number before normalization:

0.620 527 518 458 97(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 1001 1111 1011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.620 527 518 458 97(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 1001 1111 1011 1(2) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 1001 1111 1011 1(2) × 20 =

1.0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111(2) × 2-1


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): -1

Mantissa (not normalized):
1.0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-1 + 2(11-1) - 1 =

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111 =

0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111


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 1110

Mantissa (52 bits) =
0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111


Decimal number -0.620 527 518 458 97 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 0011 1101 1011 0101 1100 1000 0110 1100 0001 1001 0011 1111 0111

» Convert decimal -0.620 527 518 458 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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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