Convert -3.551 089 741 740 96 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 -3.551 089 741 740 96(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:

|-3.551 089 741 740 96| = 3.551 089 741 740 96

2. First, convert to the binary (base 2) the integer part: 3.
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;
  • 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.

3(10) =


11(2)


4. Convert to the binary (base 2) the fractional part: 0.551 089 741 740 96.

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) 0.551 089 741 740 96 × 2 = 1 + 0.102 179 483 481 92;
  • 2) 0.102 179 483 481 92 × 2 = 0 + 0.204 358 966 963 84;
  • 3) 0.204 358 966 963 84 × 2 = 0 + 0.408 717 933 927 68;
  • 4) 0.408 717 933 927 68 × 2 = 0 + 0.817 435 867 855 36;
  • 5) 0.817 435 867 855 36 × 2 = 1 + 0.634 871 735 710 72;
  • 6) 0.634 871 735 710 72 × 2 = 1 + 0.269 743 471 421 44;
  • 7) 0.269 743 471 421 44 × 2 = 0 + 0.539 486 942 842 88;
  • 8) 0.539 486 942 842 88 × 2 = 1 + 0.078 973 885 685 76;
  • 9) 0.078 973 885 685 76 × 2 = 0 + 0.157 947 771 371 52;
  • 10) 0.157 947 771 371 52 × 2 = 0 + 0.315 895 542 743 04;
  • 11) 0.315 895 542 743 04 × 2 = 0 + 0.631 791 085 486 08;
  • 12) 0.631 791 085 486 08 × 2 = 1 + 0.263 582 170 972 16;
  • 13) 0.263 582 170 972 16 × 2 = 0 + 0.527 164 341 944 32;
  • 14) 0.527 164 341 944 32 × 2 = 1 + 0.054 328 683 888 64;
  • 15) 0.054 328 683 888 64 × 2 = 0 + 0.108 657 367 777 28;
  • 16) 0.108 657 367 777 28 × 2 = 0 + 0.217 314 735 554 56;
  • 17) 0.217 314 735 554 56 × 2 = 0 + 0.434 629 471 109 12;
  • 18) 0.434 629 471 109 12 × 2 = 0 + 0.869 258 942 218 24;
  • 19) 0.869 258 942 218 24 × 2 = 1 + 0.738 517 884 436 48;
  • 20) 0.738 517 884 436 48 × 2 = 1 + 0.477 035 768 872 96;
  • 21) 0.477 035 768 872 96 × 2 = 0 + 0.954 071 537 745 92;
  • 22) 0.954 071 537 745 92 × 2 = 1 + 0.908 143 075 491 84;
  • 23) 0.908 143 075 491 84 × 2 = 1 + 0.816 286 150 983 68;
  • 24) 0.816 286 150 983 68 × 2 = 1 + 0.632 572 301 967 36;
  • 25) 0.632 572 301 967 36 × 2 = 1 + 0.265 144 603 934 72;
  • 26) 0.265 144 603 934 72 × 2 = 0 + 0.530 289 207 869 44;
  • 27) 0.530 289 207 869 44 × 2 = 1 + 0.060 578 415 738 88;
  • 28) 0.060 578 415 738 88 × 2 = 0 + 0.121 156 831 477 76;
  • 29) 0.121 156 831 477 76 × 2 = 0 + 0.242 313 662 955 52;
  • 30) 0.242 313 662 955 52 × 2 = 0 + 0.484 627 325 911 04;
  • 31) 0.484 627 325 911 04 × 2 = 0 + 0.969 254 651 822 08;
  • 32) 0.969 254 651 822 08 × 2 = 1 + 0.938 509 303 644 16;
  • 33) 0.938 509 303 644 16 × 2 = 1 + 0.877 018 607 288 32;
  • 34) 0.877 018 607 288 32 × 2 = 1 + 0.754 037 214 576 64;
  • 35) 0.754 037 214 576 64 × 2 = 1 + 0.508 074 429 153 28;
  • 36) 0.508 074 429 153 28 × 2 = 1 + 0.016 148 858 306 56;
  • 37) 0.016 148 858 306 56 × 2 = 0 + 0.032 297 716 613 12;
  • 38) 0.032 297 716 613 12 × 2 = 0 + 0.064 595 433 226 24;
  • 39) 0.064 595 433 226 24 × 2 = 0 + 0.129 190 866 452 48;
  • 40) 0.129 190 866 452 48 × 2 = 0 + 0.258 381 732 904 96;
  • 41) 0.258 381 732 904 96 × 2 = 0 + 0.516 763 465 809 92;
  • 42) 0.516 763 465 809 92 × 2 = 1 + 0.033 526 931 619 84;
  • 43) 0.033 526 931 619 84 × 2 = 0 + 0.067 053 863 239 68;
  • 44) 0.067 053 863 239 68 × 2 = 0 + 0.134 107 726 479 36;
  • 45) 0.134 107 726 479 36 × 2 = 0 + 0.268 215 452 958 72;
  • 46) 0.268 215 452 958 72 × 2 = 0 + 0.536 430 905 917 44;
  • 47) 0.536 430 905 917 44 × 2 = 1 + 0.072 861 811 834 88;
  • 48) 0.072 861 811 834 88 × 2 = 0 + 0.145 723 623 669 76;
  • 49) 0.145 723 623 669 76 × 2 = 0 + 0.291 447 247 339 52;
  • 50) 0.291 447 247 339 52 × 2 = 0 + 0.582 894 494 679 04;
  • 51) 0.582 894 494 679 04 × 2 = 1 + 0.165 788 989 358 08;
  • 52) 0.165 788 989 358 08 × 2 = 0 + 0.331 577 978 716 16;
  • 53) 0.331 577 978 716 16 × 2 = 0 + 0.663 155 957 432 32;

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:

0.551 089 741 740 96(10) =


0.1000 1101 0001 0100 0011 0111 1010 0001 1111 0000 0100 0010 0010 0(2)


6. Positive number before normalization:

3.551 089 741 740 96(10) =


11.1000 1101 0001 0100 0011 0111 1010 0001 1111 0000 0100 0010 0010 0(2)


7. Normalize the binary representation of the number.

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

3.551 089 741 740 96(10) =


11.1000 1101 0001 0100 0011 0111 1010 0001 1111 0000 0100 0010 0010 0(2) =


11.1000 1101 0001 0100 0011 0111 1010 0001 1111 0000 0100 0010 0010 0(2) × 20 =


1.1100 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001 00(2) × 21


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):
1.1100 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001 00


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) =


1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =


1024(10) =


100 0000 0000(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 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001 00 =


1100 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001


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 0000


Mantissa (52 bits) =
1100 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001


Conclusion:

Number -3.551 089 741 740 96 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0000 - 1100 0110 1000 1010 0001 1011 1101 0000 1111 1000 0010 0001 0001

(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
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

-3.551 089 741 740 97 = ? ... -3.551 089 741 740 95 = ?


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

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