Convert -28.063 134 375 446 047 39 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number -28.063 134 375 446 047 39(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-28.063 134 375 446 047 39| = 28.063 134 375 446 047 39

2. First, convert to the binary (base 2) the integer part: 28.
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;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

28(10) =


1 1100(2)


4. Convert to the binary (base 2) the fractional part: 0.063 134 375 446 047 39.

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) 0.063 134 375 446 047 39 × 2 = 0 + 0.126 268 750 892 094 78;
  • 2) 0.126 268 750 892 094 78 × 2 = 0 + 0.252 537 501 784 189 56;
  • 3) 0.252 537 501 784 189 56 × 2 = 0 + 0.505 075 003 568 379 12;
  • 4) 0.505 075 003 568 379 12 × 2 = 1 + 0.010 150 007 136 758 24;
  • 5) 0.010 150 007 136 758 24 × 2 = 0 + 0.020 300 014 273 516 48;
  • 6) 0.020 300 014 273 516 48 × 2 = 0 + 0.040 600 028 547 032 96;
  • 7) 0.040 600 028 547 032 96 × 2 = 0 + 0.081 200 057 094 065 92;
  • 8) 0.081 200 057 094 065 92 × 2 = 0 + 0.162 400 114 188 131 84;
  • 9) 0.162 400 114 188 131 84 × 2 = 0 + 0.324 800 228 376 263 68;
  • 10) 0.324 800 228 376 263 68 × 2 = 0 + 0.649 600 456 752 527 36;
  • 11) 0.649 600 456 752 527 36 × 2 = 1 + 0.299 200 913 505 054 72;
  • 12) 0.299 200 913 505 054 72 × 2 = 0 + 0.598 401 827 010 109 44;
  • 13) 0.598 401 827 010 109 44 × 2 = 1 + 0.196 803 654 020 218 88;
  • 14) 0.196 803 654 020 218 88 × 2 = 0 + 0.393 607 308 040 437 76;
  • 15) 0.393 607 308 040 437 76 × 2 = 0 + 0.787 214 616 080 875 52;
  • 16) 0.787 214 616 080 875 52 × 2 = 1 + 0.574 429 232 161 751 04;
  • 17) 0.574 429 232 161 751 04 × 2 = 1 + 0.148 858 464 323 502 08;
  • 18) 0.148 858 464 323 502 08 × 2 = 0 + 0.297 716 928 647 004 16;
  • 19) 0.297 716 928 647 004 16 × 2 = 0 + 0.595 433 857 294 008 32;
  • 20) 0.595 433 857 294 008 32 × 2 = 1 + 0.190 867 714 588 016 64;
  • 21) 0.190 867 714 588 016 64 × 2 = 0 + 0.381 735 429 176 033 28;
  • 22) 0.381 735 429 176 033 28 × 2 = 0 + 0.763 470 858 352 066 56;
  • 23) 0.763 470 858 352 066 56 × 2 = 1 + 0.526 941 716 704 133 12;
  • 24) 0.526 941 716 704 133 12 × 2 = 1 + 0.053 883 433 408 266 24;
  • 25) 0.053 883 433 408 266 24 × 2 = 0 + 0.107 766 866 816 532 48;
  • 26) 0.107 766 866 816 532 48 × 2 = 0 + 0.215 533 733 633 064 96;
  • 27) 0.215 533 733 633 064 96 × 2 = 0 + 0.431 067 467 266 129 92;
  • 28) 0.431 067 467 266 129 92 × 2 = 0 + 0.862 134 934 532 259 84;
  • 29) 0.862 134 934 532 259 84 × 2 = 1 + 0.724 269 869 064 519 68;
  • 30) 0.724 269 869 064 519 68 × 2 = 1 + 0.448 539 738 129 039 36;
  • 31) 0.448 539 738 129 039 36 × 2 = 0 + 0.897 079 476 258 078 72;
  • 32) 0.897 079 476 258 078 72 × 2 = 1 + 0.794 158 952 516 157 44;
  • 33) 0.794 158 952 516 157 44 × 2 = 1 + 0.588 317 905 032 314 88;
  • 34) 0.588 317 905 032 314 88 × 2 = 1 + 0.176 635 810 064 629 76;
  • 35) 0.176 635 810 064 629 76 × 2 = 0 + 0.353 271 620 129 259 52;
  • 36) 0.353 271 620 129 259 52 × 2 = 0 + 0.706 543 240 258 519 04;
  • 37) 0.706 543 240 258 519 04 × 2 = 1 + 0.413 086 480 517 038 08;
  • 38) 0.413 086 480 517 038 08 × 2 = 0 + 0.826 172 961 034 076 16;
  • 39) 0.826 172 961 034 076 16 × 2 = 1 + 0.652 345 922 068 152 32;
  • 40) 0.652 345 922 068 152 32 × 2 = 1 + 0.304 691 844 136 304 64;
  • 41) 0.304 691 844 136 304 64 × 2 = 0 + 0.609 383 688 272 609 28;
  • 42) 0.609 383 688 272 609 28 × 2 = 1 + 0.218 767 376 545 218 56;
  • 43) 0.218 767 376 545 218 56 × 2 = 0 + 0.437 534 753 090 437 12;
  • 44) 0.437 534 753 090 437 12 × 2 = 0 + 0.875 069 506 180 874 24;
  • 45) 0.875 069 506 180 874 24 × 2 = 1 + 0.750 139 012 361 748 48;
  • 46) 0.750 139 012 361 748 48 × 2 = 1 + 0.500 278 024 723 496 96;
  • 47) 0.500 278 024 723 496 96 × 2 = 1 + 0.000 556 049 446 993 92;
  • 48) 0.000 556 049 446 993 92 × 2 = 0 + 0.001 112 098 893 987 84;
  • 49) 0.001 112 098 893 987 84 × 2 = 0 + 0.002 224 197 787 975 68;
  • 50) 0.002 224 197 787 975 68 × 2 = 0 + 0.004 448 395 575 951 36;
  • 51) 0.004 448 395 575 951 36 × 2 = 0 + 0.008 896 791 151 902 72;
  • 52) 0.008 896 791 151 902 72 × 2 = 0 + 0.017 793 582 303 805 44;
  • 53) 0.017 793 582 303 805 44 × 2 = 0 + 0.035 587 164 607 610 88;

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...)


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:

0.063 134 375 446 047 39(10) =


0.0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 0(2)


6. Positive number before normalization:

28.063 134 375 446 047 39(10) =


1 1100.0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 0(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left so that only one non zero digit remains to the left of it:

28.063 134 375 446 047 39(10) =


1 1100.0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 0(2) =


1 1100.0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 0(2) × 20 =


1.1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 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):
1.1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0000 0


9. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1027(10) =


100 0000 0011(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110 0 0000 =


1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 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) =
100 0000 0011


Mantissa (52 bits) =
1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110


Conclusion:

Number -28.063 134 375 446 047 39 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0011 - 1100 0001 0000 0010 1001 1001 0011 0000 1101 1100 1011 0100 1110

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 1

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 0

      0

More operations of this kind:

-28.063 134 375 446 047 4 = ? ... -28.063 134 375 446 047 38 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

-28.063 134 375 446 047 39 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:47 UTC (GMT)
43 367.576 9 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:47 UTC (GMT)
-11.403 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:47 UTC (GMT)
640 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:47 UTC (GMT)
4.899 999 619 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
1.666 666 666 2 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
78.238 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
954 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
0.933 635 485 751 756 8 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
-45.4 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
1 993 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
98 773 896.874 583 651 234 57 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
78.235 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:46 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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