-0.180 000 000 028 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 028 9(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
-0.180 000 000 028 9(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:

|-0.180 000 000 028 9| = 0.180 000 000 028 9


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: 0.180 000 000 028 9.

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) 0.180 000 000 028 9 × 2 = 0 + 0.360 000 000 057 8;
  • 2) 0.360 000 000 057 8 × 2 = 0 + 0.720 000 000 115 6;
  • 3) 0.720 000 000 115 6 × 2 = 1 + 0.440 000 000 231 2;
  • 4) 0.440 000 000 231 2 × 2 = 0 + 0.880 000 000 462 4;
  • 5) 0.880 000 000 462 4 × 2 = 1 + 0.760 000 000 924 8;
  • 6) 0.760 000 000 924 8 × 2 = 1 + 0.520 000 001 849 6;
  • 7) 0.520 000 001 849 6 × 2 = 1 + 0.040 000 003 699 2;
  • 8) 0.040 000 003 699 2 × 2 = 0 + 0.080 000 007 398 4;
  • 9) 0.080 000 007 398 4 × 2 = 0 + 0.160 000 014 796 8;
  • 10) 0.160 000 014 796 8 × 2 = 0 + 0.320 000 029 593 6;
  • 11) 0.320 000 029 593 6 × 2 = 0 + 0.640 000 059 187 2;
  • 12) 0.640 000 059 187 2 × 2 = 1 + 0.280 000 118 374 4;
  • 13) 0.280 000 118 374 4 × 2 = 0 + 0.560 000 236 748 8;
  • 14) 0.560 000 236 748 8 × 2 = 1 + 0.120 000 473 497 6;
  • 15) 0.120 000 473 497 6 × 2 = 0 + 0.240 000 946 995 2;
  • 16) 0.240 000 946 995 2 × 2 = 0 + 0.480 001 893 990 4;
  • 17) 0.480 001 893 990 4 × 2 = 0 + 0.960 003 787 980 8;
  • 18) 0.960 003 787 980 8 × 2 = 1 + 0.920 007 575 961 6;
  • 19) 0.920 007 575 961 6 × 2 = 1 + 0.840 015 151 923 2;
  • 20) 0.840 015 151 923 2 × 2 = 1 + 0.680 030 303 846 4;
  • 21) 0.680 030 303 846 4 × 2 = 1 + 0.360 060 607 692 8;
  • 22) 0.360 060 607 692 8 × 2 = 0 + 0.720 121 215 385 6;
  • 23) 0.720 121 215 385 6 × 2 = 1 + 0.440 242 430 771 2;
  • 24) 0.440 242 430 771 2 × 2 = 0 + 0.880 484 861 542 4;
  • 25) 0.880 484 861 542 4 × 2 = 1 + 0.760 969 723 084 8;
  • 26) 0.760 969 723 084 8 × 2 = 1 + 0.521 939 446 169 6;
  • 27) 0.521 939 446 169 6 × 2 = 1 + 0.043 878 892 339 2;
  • 28) 0.043 878 892 339 2 × 2 = 0 + 0.087 757 784 678 4;
  • 29) 0.087 757 784 678 4 × 2 = 0 + 0.175 515 569 356 8;
  • 30) 0.175 515 569 356 8 × 2 = 0 + 0.351 031 138 713 6;
  • 31) 0.351 031 138 713 6 × 2 = 0 + 0.702 062 277 427 2;
  • 32) 0.702 062 277 427 2 × 2 = 1 + 0.404 124 554 854 4;
  • 33) 0.404 124 554 854 4 × 2 = 0 + 0.808 249 109 708 8;
  • 34) 0.808 249 109 708 8 × 2 = 1 + 0.616 498 219 417 6;
  • 35) 0.616 498 219 417 6 × 2 = 1 + 0.232 996 438 835 2;
  • 36) 0.232 996 438 835 2 × 2 = 0 + 0.465 992 877 670 4;
  • 37) 0.465 992 877 670 4 × 2 = 0 + 0.931 985 755 340 8;
  • 38) 0.931 985 755 340 8 × 2 = 1 + 0.863 971 510 681 6;
  • 39) 0.863 971 510 681 6 × 2 = 1 + 0.727 943 021 363 2;
  • 40) 0.727 943 021 363 2 × 2 = 1 + 0.455 886 042 726 4;
  • 41) 0.455 886 042 726 4 × 2 = 0 + 0.911 772 085 452 8;
  • 42) 0.911 772 085 452 8 × 2 = 1 + 0.823 544 170 905 6;
  • 43) 0.823 544 170 905 6 × 2 = 1 + 0.647 088 341 811 2;
  • 44) 0.647 088 341 811 2 × 2 = 1 + 0.294 176 683 622 4;
  • 45) 0.294 176 683 622 4 × 2 = 0 + 0.588 353 367 244 8;
  • 46) 0.588 353 367 244 8 × 2 = 1 + 0.176 706 734 489 6;
  • 47) 0.176 706 734 489 6 × 2 = 0 + 0.353 413 468 979 2;
  • 48) 0.353 413 468 979 2 × 2 = 0 + 0.706 826 937 958 4;
  • 49) 0.706 826 937 958 4 × 2 = 1 + 0.413 653 875 916 8;
  • 50) 0.413 653 875 916 8 × 2 = 0 + 0.827 307 751 833 6;
  • 51) 0.827 307 751 833 6 × 2 = 1 + 0.654 615 503 667 2;
  • 52) 0.654 615 503 667 2 × 2 = 1 + 0.309 231 007 334 4;
  • 53) 0.309 231 007 334 4 × 2 = 0 + 0.618 462 014 668 8;
  • 54) 0.618 462 014 668 8 × 2 = 1 + 0.236 924 029 337 6;
  • 55) 0.236 924 029 337 6 × 2 = 0 + 0.473 848 058 675 2;

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:


0.180 000 000 028 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0111 0111 0100 1011 010(2)

6. Positive number before normalization:

0.180 000 000 028 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0111 0111 0100 1011 010(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:


0.180 000 000 028 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0111 0111 0100 1011 010(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0111 0111 0100 1011 010(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0011 1011 1010 0101 1010(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 1011 0011 1011 1010 0101 1010


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 1011 0011 1011 1010 0101 1010 =

0111 0000 1010 0011 1101 0111 0000 1011 0011 1011 1010 0101 1010


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 1011 0011 1011 1010 0101 1010


Decimal number -0.180 000 000 028 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 0011 1011 1010 0101 1010

» Convert decimal -0.180 000 000 032 1 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