-0.180 000 000 038 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 038 4(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
-0.180 000 000 038 4(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:

|-0.180 000 000 038 4| = 0.180 000 000 038 4


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: 0.180 000 000 038 4.

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) 0.180 000 000 038 4 × 2 = 0 + 0.360 000 000 076 8;
  • 2) 0.360 000 000 076 8 × 2 = 0 + 0.720 000 000 153 6;
  • 3) 0.720 000 000 153 6 × 2 = 1 + 0.440 000 000 307 2;
  • 4) 0.440 000 000 307 2 × 2 = 0 + 0.880 000 000 614 4;
  • 5) 0.880 000 000 614 4 × 2 = 1 + 0.760 000 001 228 8;
  • 6) 0.760 000 001 228 8 × 2 = 1 + 0.520 000 002 457 6;
  • 7) 0.520 000 002 457 6 × 2 = 1 + 0.040 000 004 915 2;
  • 8) 0.040 000 004 915 2 × 2 = 0 + 0.080 000 009 830 4;
  • 9) 0.080 000 009 830 4 × 2 = 0 + 0.160 000 019 660 8;
  • 10) 0.160 000 019 660 8 × 2 = 0 + 0.320 000 039 321 6;
  • 11) 0.320 000 039 321 6 × 2 = 0 + 0.640 000 078 643 2;
  • 12) 0.640 000 078 643 2 × 2 = 1 + 0.280 000 157 286 4;
  • 13) 0.280 000 157 286 4 × 2 = 0 + 0.560 000 314 572 8;
  • 14) 0.560 000 314 572 8 × 2 = 1 + 0.120 000 629 145 6;
  • 15) 0.120 000 629 145 6 × 2 = 0 + 0.240 001 258 291 2;
  • 16) 0.240 001 258 291 2 × 2 = 0 + 0.480 002 516 582 4;
  • 17) 0.480 002 516 582 4 × 2 = 0 + 0.960 005 033 164 8;
  • 18) 0.960 005 033 164 8 × 2 = 1 + 0.920 010 066 329 6;
  • 19) 0.920 010 066 329 6 × 2 = 1 + 0.840 020 132 659 2;
  • 20) 0.840 020 132 659 2 × 2 = 1 + 0.680 040 265 318 4;
  • 21) 0.680 040 265 318 4 × 2 = 1 + 0.360 080 530 636 8;
  • 22) 0.360 080 530 636 8 × 2 = 0 + 0.720 161 061 273 6;
  • 23) 0.720 161 061 273 6 × 2 = 1 + 0.440 322 122 547 2;
  • 24) 0.440 322 122 547 2 × 2 = 0 + 0.880 644 245 094 4;
  • 25) 0.880 644 245 094 4 × 2 = 1 + 0.761 288 490 188 8;
  • 26) 0.761 288 490 188 8 × 2 = 1 + 0.522 576 980 377 6;
  • 27) 0.522 576 980 377 6 × 2 = 1 + 0.045 153 960 755 2;
  • 28) 0.045 153 960 755 2 × 2 = 0 + 0.090 307 921 510 4;
  • 29) 0.090 307 921 510 4 × 2 = 0 + 0.180 615 843 020 8;
  • 30) 0.180 615 843 020 8 × 2 = 0 + 0.361 231 686 041 6;
  • 31) 0.361 231 686 041 6 × 2 = 0 + 0.722 463 372 083 2;
  • 32) 0.722 463 372 083 2 × 2 = 1 + 0.444 926 744 166 4;
  • 33) 0.444 926 744 166 4 × 2 = 0 + 0.889 853 488 332 8;
  • 34) 0.889 853 488 332 8 × 2 = 1 + 0.779 706 976 665 6;
  • 35) 0.779 706 976 665 6 × 2 = 1 + 0.559 413 953 331 2;
  • 36) 0.559 413 953 331 2 × 2 = 1 + 0.118 827 906 662 4;
  • 37) 0.118 827 906 662 4 × 2 = 0 + 0.237 655 813 324 8;
  • 38) 0.237 655 813 324 8 × 2 = 0 + 0.475 311 626 649 6;
  • 39) 0.475 311 626 649 6 × 2 = 0 + 0.950 623 253 299 2;
  • 40) 0.950 623 253 299 2 × 2 = 1 + 0.901 246 506 598 4;
  • 41) 0.901 246 506 598 4 × 2 = 1 + 0.802 493 013 196 8;
  • 42) 0.802 493 013 196 8 × 2 = 1 + 0.604 986 026 393 6;
  • 43) 0.604 986 026 393 6 × 2 = 1 + 0.209 972 052 787 2;
  • 44) 0.209 972 052 787 2 × 2 = 0 + 0.419 944 105 574 4;
  • 45) 0.419 944 105 574 4 × 2 = 0 + 0.839 888 211 148 8;
  • 46) 0.839 888 211 148 8 × 2 = 1 + 0.679 776 422 297 6;
  • 47) 0.679 776 422 297 6 × 2 = 1 + 0.359 552 844 595 2;
  • 48) 0.359 552 844 595 2 × 2 = 0 + 0.719 105 689 190 4;
  • 49) 0.719 105 689 190 4 × 2 = 1 + 0.438 211 378 380 8;
  • 50) 0.438 211 378 380 8 × 2 = 0 + 0.876 422 756 761 6;
  • 51) 0.876 422 756 761 6 × 2 = 1 + 0.752 845 513 523 2;
  • 52) 0.752 845 513 523 2 × 2 = 1 + 0.505 691 027 046 4;
  • 53) 0.505 691 027 046 4 × 2 = 1 + 0.011 382 054 092 8;
  • 54) 0.011 382 054 092 8 × 2 = 0 + 0.022 764 108 185 6;
  • 55) 0.022 764 108 185 6 × 2 = 0 + 0.045 528 216 371 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:


0.180 000 000 038 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1110 0110 1011 100(2)

6. Positive number before normalization:

0.180 000 000 038 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1110 0110 1011 100(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:


0.180 000 000 038 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1110 0110 1011 100(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1110 0110 1011 100(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 1000 1111 0011 0101 1100(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 1000 1111 0011 0101 1100


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 1000 1111 0011 0101 1100 =

0111 0000 1010 0011 1101 0111 0000 1011 1000 1111 0011 0101 1100


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 1000 1111 0011 0101 1100


Decimal number -0.180 000 000 038 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 1000 1111 0011 0101 1100

» Convert decimal -0.180 000 000 045 4 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