Convert -0.180 000 000 000 000 04 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 -0.180 000 000 000 000 04(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:

|-0.180 000 000 000 000 04| = 0.180 000 000 000 000 04

2. First, convert to the binary (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 the binary (base 2) the fractional part: 0.180 000 000 000 000 04.

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) 0.180 000 000 000 000 04 × 2 = 0 + 0.360 000 000 000 000 08;
  • 2) 0.360 000 000 000 000 08 × 2 = 0 + 0.720 000 000 000 000 16;
  • 3) 0.720 000 000 000 000 16 × 2 = 1 + 0.440 000 000 000 000 32;
  • 4) 0.440 000 000 000 000 32 × 2 = 0 + 0.880 000 000 000 000 64;
  • 5) 0.880 000 000 000 000 64 × 2 = 1 + 0.760 000 000 000 001 28;
  • 6) 0.760 000 000 000 001 28 × 2 = 1 + 0.520 000 000 000 002 56;
  • 7) 0.520 000 000 000 002 56 × 2 = 1 + 0.040 000 000 000 005 12;
  • 8) 0.040 000 000 000 005 12 × 2 = 0 + 0.080 000 000 000 010 24;
  • 9) 0.080 000 000 000 010 24 × 2 = 0 + 0.160 000 000 000 020 48;
  • 10) 0.160 000 000 000 020 48 × 2 = 0 + 0.320 000 000 000 040 96;
  • 11) 0.320 000 000 000 040 96 × 2 = 0 + 0.640 000 000 000 081 92;
  • 12) 0.640 000 000 000 081 92 × 2 = 1 + 0.280 000 000 000 163 84;
  • 13) 0.280 000 000 000 163 84 × 2 = 0 + 0.560 000 000 000 327 68;
  • 14) 0.560 000 000 000 327 68 × 2 = 1 + 0.120 000 000 000 655 36;
  • 15) 0.120 000 000 000 655 36 × 2 = 0 + 0.240 000 000 001 310 72;
  • 16) 0.240 000 000 001 310 72 × 2 = 0 + 0.480 000 000 002 621 44;
  • 17) 0.480 000 000 002 621 44 × 2 = 0 + 0.960 000 000 005 242 88;
  • 18) 0.960 000 000 005 242 88 × 2 = 1 + 0.920 000 000 010 485 76;
  • 19) 0.920 000 000 010 485 76 × 2 = 1 + 0.840 000 000 020 971 52;
  • 20) 0.840 000 000 020 971 52 × 2 = 1 + 0.680 000 000 041 943 04;
  • 21) 0.680 000 000 041 943 04 × 2 = 1 + 0.360 000 000 083 886 08;
  • 22) 0.360 000 000 083 886 08 × 2 = 0 + 0.720 000 000 167 772 16;
  • 23) 0.720 000 000 167 772 16 × 2 = 1 + 0.440 000 000 335 544 32;
  • 24) 0.440 000 000 335 544 32 × 2 = 0 + 0.880 000 000 671 088 64;
  • 25) 0.880 000 000 671 088 64 × 2 = 1 + 0.760 000 001 342 177 28;
  • 26) 0.760 000 001 342 177 28 × 2 = 1 + 0.520 000 002 684 354 56;
  • 27) 0.520 000 002 684 354 56 × 2 = 1 + 0.040 000 005 368 709 12;
  • 28) 0.040 000 005 368 709 12 × 2 = 0 + 0.080 000 010 737 418 24;
  • 29) 0.080 000 010 737 418 24 × 2 = 0 + 0.160 000 021 474 836 48;
  • 30) 0.160 000 021 474 836 48 × 2 = 0 + 0.320 000 042 949 672 96;
  • 31) 0.320 000 042 949 672 96 × 2 = 0 + 0.640 000 085 899 345 92;
  • 32) 0.640 000 085 899 345 92 × 2 = 1 + 0.280 000 171 798 691 84;
  • 33) 0.280 000 171 798 691 84 × 2 = 0 + 0.560 000 343 597 383 68;
  • 34) 0.560 000 343 597 383 68 × 2 = 1 + 0.120 000 687 194 767 36;
  • 35) 0.120 000 687 194 767 36 × 2 = 0 + 0.240 001 374 389 534 72;
  • 36) 0.240 001 374 389 534 72 × 2 = 0 + 0.480 002 748 779 069 44;
  • 37) 0.480 002 748 779 069 44 × 2 = 0 + 0.960 005 497 558 138 88;
  • 38) 0.960 005 497 558 138 88 × 2 = 1 + 0.920 010 995 116 277 76;
  • 39) 0.920 010 995 116 277 76 × 2 = 1 + 0.840 021 990 232 555 52;
  • 40) 0.840 021 990 232 555 52 × 2 = 1 + 0.680 043 980 465 111 04;
  • 41) 0.680 043 980 465 111 04 × 2 = 1 + 0.360 087 960 930 222 08;
  • 42) 0.360 087 960 930 222 08 × 2 = 0 + 0.720 175 921 860 444 16;
  • 43) 0.720 175 921 860 444 16 × 2 = 1 + 0.440 351 843 720 888 32;
  • 44) 0.440 351 843 720 888 32 × 2 = 0 + 0.880 703 687 441 776 64;
  • 45) 0.880 703 687 441 776 64 × 2 = 1 + 0.761 407 374 883 553 28;
  • 46) 0.761 407 374 883 553 28 × 2 = 1 + 0.522 814 749 767 106 56;
  • 47) 0.522 814 749 767 106 56 × 2 = 1 + 0.045 629 499 534 213 12;
  • 48) 0.045 629 499 534 213 12 × 2 = 0 + 0.091 258 999 068 426 24;
  • 49) 0.091 258 999 068 426 24 × 2 = 0 + 0.182 517 998 136 852 48;
  • 50) 0.182 517 998 136 852 48 × 2 = 0 + 0.365 035 996 273 704 96;
  • 51) 0.365 035 996 273 704 96 × 2 = 0 + 0.730 071 992 547 409 92;
  • 52) 0.730 071 992 547 409 92 × 2 = 1 + 0.460 143 985 094 819 84;
  • 53) 0.460 143 985 094 819 84 × 2 = 0 + 0.920 287 970 189 639 68;
  • 54) 0.920 287 970 189 639 68 × 2 = 1 + 0.840 575 940 379 279 36;
  • 55) 0.840 575 940 379 279 36 × 2 = 1 + 0.681 151 880 758 558 72;

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:

0.180 000 000 000 000 04(10) =


0.0010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 011(2)


6. Positive number before normalization:

0.180 000 000 000 000 04(10) =


0.0010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 011(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right so that only one non zero digit remains to the left of it:

0.180 000 000 000 000 04(10) =


0.0010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 011(2) =


0.0010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 011(2) × 20 =


1.0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011(2) × 2-3


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): -3


Mantissa (not normalized):
1.0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-3 + 2(11-1) - 1 =


(-3 + 1 023)(10) =


1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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) =


1020(10) =


011 1111 1100(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. 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011 =


0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011


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 1100


Mantissa (52 bits) =
0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011


Conclusion:

Number -0.180 000 000 000 000 04 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1011

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 1

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 1

      0

More operations of this kind:

-0.180 000 000 000 000 05 = ? ... -0.180 000 000 000 000 03 = ?


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