-0.180 000 000 028 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 028 1(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 1(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 1| = 0.180 000 000 028 1


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 1.

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 1 × 2 = 0 + 0.360 000 000 056 2;
  • 2) 0.360 000 000 056 2 × 2 = 0 + 0.720 000 000 112 4;
  • 3) 0.720 000 000 112 4 × 2 = 1 + 0.440 000 000 224 8;
  • 4) 0.440 000 000 224 8 × 2 = 0 + 0.880 000 000 449 6;
  • 5) 0.880 000 000 449 6 × 2 = 1 + 0.760 000 000 899 2;
  • 6) 0.760 000 000 899 2 × 2 = 1 + 0.520 000 001 798 4;
  • 7) 0.520 000 001 798 4 × 2 = 1 + 0.040 000 003 596 8;
  • 8) 0.040 000 003 596 8 × 2 = 0 + 0.080 000 007 193 6;
  • 9) 0.080 000 007 193 6 × 2 = 0 + 0.160 000 014 387 2;
  • 10) 0.160 000 014 387 2 × 2 = 0 + 0.320 000 028 774 4;
  • 11) 0.320 000 028 774 4 × 2 = 0 + 0.640 000 057 548 8;
  • 12) 0.640 000 057 548 8 × 2 = 1 + 0.280 000 115 097 6;
  • 13) 0.280 000 115 097 6 × 2 = 0 + 0.560 000 230 195 2;
  • 14) 0.560 000 230 195 2 × 2 = 1 + 0.120 000 460 390 4;
  • 15) 0.120 000 460 390 4 × 2 = 0 + 0.240 000 920 780 8;
  • 16) 0.240 000 920 780 8 × 2 = 0 + 0.480 001 841 561 6;
  • 17) 0.480 001 841 561 6 × 2 = 0 + 0.960 003 683 123 2;
  • 18) 0.960 003 683 123 2 × 2 = 1 + 0.920 007 366 246 4;
  • 19) 0.920 007 366 246 4 × 2 = 1 + 0.840 014 732 492 8;
  • 20) 0.840 014 732 492 8 × 2 = 1 + 0.680 029 464 985 6;
  • 21) 0.680 029 464 985 6 × 2 = 1 + 0.360 058 929 971 2;
  • 22) 0.360 058 929 971 2 × 2 = 0 + 0.720 117 859 942 4;
  • 23) 0.720 117 859 942 4 × 2 = 1 + 0.440 235 719 884 8;
  • 24) 0.440 235 719 884 8 × 2 = 0 + 0.880 471 439 769 6;
  • 25) 0.880 471 439 769 6 × 2 = 1 + 0.760 942 879 539 2;
  • 26) 0.760 942 879 539 2 × 2 = 1 + 0.521 885 759 078 4;
  • 27) 0.521 885 759 078 4 × 2 = 1 + 0.043 771 518 156 8;
  • 28) 0.043 771 518 156 8 × 2 = 0 + 0.087 543 036 313 6;
  • 29) 0.087 543 036 313 6 × 2 = 0 + 0.175 086 072 627 2;
  • 30) 0.175 086 072 627 2 × 2 = 0 + 0.350 172 145 254 4;
  • 31) 0.350 172 145 254 4 × 2 = 0 + 0.700 344 290 508 8;
  • 32) 0.700 344 290 508 8 × 2 = 1 + 0.400 688 581 017 6;
  • 33) 0.400 688 581 017 6 × 2 = 0 + 0.801 377 162 035 2;
  • 34) 0.801 377 162 035 2 × 2 = 1 + 0.602 754 324 070 4;
  • 35) 0.602 754 324 070 4 × 2 = 1 + 0.205 508 648 140 8;
  • 36) 0.205 508 648 140 8 × 2 = 0 + 0.411 017 296 281 6;
  • 37) 0.411 017 296 281 6 × 2 = 0 + 0.822 034 592 563 2;
  • 38) 0.822 034 592 563 2 × 2 = 1 + 0.644 069 185 126 4;
  • 39) 0.644 069 185 126 4 × 2 = 1 + 0.288 138 370 252 8;
  • 40) 0.288 138 370 252 8 × 2 = 0 + 0.576 276 740 505 6;
  • 41) 0.576 276 740 505 6 × 2 = 1 + 0.152 553 481 011 2;
  • 42) 0.152 553 481 011 2 × 2 = 0 + 0.305 106 962 022 4;
  • 43) 0.305 106 962 022 4 × 2 = 0 + 0.610 213 924 044 8;
  • 44) 0.610 213 924 044 8 × 2 = 1 + 0.220 427 848 089 6;
  • 45) 0.220 427 848 089 6 × 2 = 0 + 0.440 855 696 179 2;
  • 46) 0.440 855 696 179 2 × 2 = 0 + 0.881 711 392 358 4;
  • 47) 0.881 711 392 358 4 × 2 = 1 + 0.763 422 784 716 8;
  • 48) 0.763 422 784 716 8 × 2 = 1 + 0.526 845 569 433 6;
  • 49) 0.526 845 569 433 6 × 2 = 1 + 0.053 691 138 867 2;
  • 50) 0.053 691 138 867 2 × 2 = 0 + 0.107 382 277 734 4;
  • 51) 0.107 382 277 734 4 × 2 = 0 + 0.214 764 555 468 8;
  • 52) 0.214 764 555 468 8 × 2 = 0 + 0.429 529 110 937 6;
  • 53) 0.429 529 110 937 6 × 2 = 0 + 0.859 058 221 875 2;
  • 54) 0.859 058 221 875 2 × 2 = 1 + 0.718 116 443 750 4;
  • 55) 0.718 116 443 750 4 × 2 = 1 + 0.436 232 887 500 8;

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 1(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0110 1001 0011 1000 011(2)

6. Positive number before normalization:

0.180 000 000 028 1(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0110 1001 0011 1000 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:


0.180 000 000 028 1(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0110 1001 0011 1000 011(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0110 1001 0011 1000 011(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0011 0100 1001 1100 0011(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 0100 1001 1100 0011


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 0100 1001 1100 0011 =

0111 0000 1010 0011 1101 0111 0000 1011 0011 0100 1001 1100 0011


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 0100 1001 1100 0011


Decimal number -0.180 000 000 028 1 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 0100 1001 1100 0011

» Convert decimal -0.180 000 000 036 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

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