-0.381 966 011 250 103 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 103 99(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
-0.381 966 011 250 103 99(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:

|-0.381 966 011 250 103 99| = 0.381 966 011 250 103 99


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: 0.381 966 011 250 103 99.

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) 0.381 966 011 250 103 99 × 2 = 0 + 0.763 932 022 500 207 98;
  • 2) 0.763 932 022 500 207 98 × 2 = 1 + 0.527 864 045 000 415 96;
  • 3) 0.527 864 045 000 415 96 × 2 = 1 + 0.055 728 090 000 831 92;
  • 4) 0.055 728 090 000 831 92 × 2 = 0 + 0.111 456 180 001 663 84;
  • 5) 0.111 456 180 001 663 84 × 2 = 0 + 0.222 912 360 003 327 68;
  • 6) 0.222 912 360 003 327 68 × 2 = 0 + 0.445 824 720 006 655 36;
  • 7) 0.445 824 720 006 655 36 × 2 = 0 + 0.891 649 440 013 310 72;
  • 8) 0.891 649 440 013 310 72 × 2 = 1 + 0.783 298 880 026 621 44;
  • 9) 0.783 298 880 026 621 44 × 2 = 1 + 0.566 597 760 053 242 88;
  • 10) 0.566 597 760 053 242 88 × 2 = 1 + 0.133 195 520 106 485 76;
  • 11) 0.133 195 520 106 485 76 × 2 = 0 + 0.266 391 040 212 971 52;
  • 12) 0.266 391 040 212 971 52 × 2 = 0 + 0.532 782 080 425 943 04;
  • 13) 0.532 782 080 425 943 04 × 2 = 1 + 0.065 564 160 851 886 08;
  • 14) 0.065 564 160 851 886 08 × 2 = 0 + 0.131 128 321 703 772 16;
  • 15) 0.131 128 321 703 772 16 × 2 = 0 + 0.262 256 643 407 544 32;
  • 16) 0.262 256 643 407 544 32 × 2 = 0 + 0.524 513 286 815 088 64;
  • 17) 0.524 513 286 815 088 64 × 2 = 1 + 0.049 026 573 630 177 28;
  • 18) 0.049 026 573 630 177 28 × 2 = 0 + 0.098 053 147 260 354 56;
  • 19) 0.098 053 147 260 354 56 × 2 = 0 + 0.196 106 294 520 709 12;
  • 20) 0.196 106 294 520 709 12 × 2 = 0 + 0.392 212 589 041 418 24;
  • 21) 0.392 212 589 041 418 24 × 2 = 0 + 0.784 425 178 082 836 48;
  • 22) 0.784 425 178 082 836 48 × 2 = 1 + 0.568 850 356 165 672 96;
  • 23) 0.568 850 356 165 672 96 × 2 = 1 + 0.137 700 712 331 345 92;
  • 24) 0.137 700 712 331 345 92 × 2 = 0 + 0.275 401 424 662 691 84;
  • 25) 0.275 401 424 662 691 84 × 2 = 0 + 0.550 802 849 325 383 68;
  • 26) 0.550 802 849 325 383 68 × 2 = 1 + 0.101 605 698 650 767 36;
  • 27) 0.101 605 698 650 767 36 × 2 = 0 + 0.203 211 397 301 534 72;
  • 28) 0.203 211 397 301 534 72 × 2 = 0 + 0.406 422 794 603 069 44;
  • 29) 0.406 422 794 603 069 44 × 2 = 0 + 0.812 845 589 206 138 88;
  • 30) 0.812 845 589 206 138 88 × 2 = 1 + 0.625 691 178 412 277 76;
  • 31) 0.625 691 178 412 277 76 × 2 = 1 + 0.251 382 356 824 555 52;
  • 32) 0.251 382 356 824 555 52 × 2 = 0 + 0.502 764 713 649 111 04;
  • 33) 0.502 764 713 649 111 04 × 2 = 1 + 0.005 529 427 298 222 08;
  • 34) 0.005 529 427 298 222 08 × 2 = 0 + 0.011 058 854 596 444 16;
  • 35) 0.011 058 854 596 444 16 × 2 = 0 + 0.022 117 709 192 888 32;
  • 36) 0.022 117 709 192 888 32 × 2 = 0 + 0.044 235 418 385 776 64;
  • 37) 0.044 235 418 385 776 64 × 2 = 0 + 0.088 470 836 771 553 28;
  • 38) 0.088 470 836 771 553 28 × 2 = 0 + 0.176 941 673 543 106 56;
  • 39) 0.176 941 673 543 106 56 × 2 = 0 + 0.353 883 347 086 213 12;
  • 40) 0.353 883 347 086 213 12 × 2 = 0 + 0.707 766 694 172 426 24;
  • 41) 0.707 766 694 172 426 24 × 2 = 1 + 0.415 533 388 344 852 48;
  • 42) 0.415 533 388 344 852 48 × 2 = 0 + 0.831 066 776 689 704 96;
  • 43) 0.831 066 776 689 704 96 × 2 = 1 + 0.662 133 553 379 409 92;
  • 44) 0.662 133 553 379 409 92 × 2 = 1 + 0.324 267 106 758 819 84;
  • 45) 0.324 267 106 758 819 84 × 2 = 0 + 0.648 534 213 517 639 68;
  • 46) 0.648 534 213 517 639 68 × 2 = 1 + 0.297 068 427 035 279 36;
  • 47) 0.297 068 427 035 279 36 × 2 = 0 + 0.594 136 854 070 558 72;
  • 48) 0.594 136 854 070 558 72 × 2 = 1 + 0.188 273 708 141 117 44;
  • 49) 0.188 273 708 141 117 44 × 2 = 0 + 0.376 547 416 282 234 88;
  • 50) 0.376 547 416 282 234 88 × 2 = 0 + 0.753 094 832 564 469 76;
  • 51) 0.753 094 832 564 469 76 × 2 = 1 + 0.506 189 665 128 939 52;
  • 52) 0.506 189 665 128 939 52 × 2 = 1 + 0.012 379 330 257 879 04;
  • 53) 0.012 379 330 257 879 04 × 2 = 0 + 0.024 758 660 515 758 08;
  • 54) 0.024 758 660 515 758 08 × 2 = 0 + 0.049 517 321 031 516 16;

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:


0.381 966 011 250 103 99(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0011 00(2)

6. Positive number before normalization:

0.381 966 011 250 103 99(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0011 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 250 103 99(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0011 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0011 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 1100(2) × 2-2


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): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 1100 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 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 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 1100


Decimal number -0.381 966 011 250 103 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0100 1100

» Convert decimal -0.381 966 011 250 103 02 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