-0.000 282 005 919 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 919 3(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
-0.000 282 005 919 3(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:

|-0.000 282 005 919 3| = 0.000 282 005 919 3


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: 0.000 282 005 919 3.

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) 0.000 282 005 919 3 × 2 = 0 + 0.000 564 011 838 6;
  • 2) 0.000 564 011 838 6 × 2 = 0 + 0.001 128 023 677 2;
  • 3) 0.001 128 023 677 2 × 2 = 0 + 0.002 256 047 354 4;
  • 4) 0.002 256 047 354 4 × 2 = 0 + 0.004 512 094 708 8;
  • 5) 0.004 512 094 708 8 × 2 = 0 + 0.009 024 189 417 6;
  • 6) 0.009 024 189 417 6 × 2 = 0 + 0.018 048 378 835 2;
  • 7) 0.018 048 378 835 2 × 2 = 0 + 0.036 096 757 670 4;
  • 8) 0.036 096 757 670 4 × 2 = 0 + 0.072 193 515 340 8;
  • 9) 0.072 193 515 340 8 × 2 = 0 + 0.144 387 030 681 6;
  • 10) 0.144 387 030 681 6 × 2 = 0 + 0.288 774 061 363 2;
  • 11) 0.288 774 061 363 2 × 2 = 0 + 0.577 548 122 726 4;
  • 12) 0.577 548 122 726 4 × 2 = 1 + 0.155 096 245 452 8;
  • 13) 0.155 096 245 452 8 × 2 = 0 + 0.310 192 490 905 6;
  • 14) 0.310 192 490 905 6 × 2 = 0 + 0.620 384 981 811 2;
  • 15) 0.620 384 981 811 2 × 2 = 1 + 0.240 769 963 622 4;
  • 16) 0.240 769 963 622 4 × 2 = 0 + 0.481 539 927 244 8;
  • 17) 0.481 539 927 244 8 × 2 = 0 + 0.963 079 854 489 6;
  • 18) 0.963 079 854 489 6 × 2 = 1 + 0.926 159 708 979 2;
  • 19) 0.926 159 708 979 2 × 2 = 1 + 0.852 319 417 958 4;
  • 20) 0.852 319 417 958 4 × 2 = 1 + 0.704 638 835 916 8;
  • 21) 0.704 638 835 916 8 × 2 = 1 + 0.409 277 671 833 6;
  • 22) 0.409 277 671 833 6 × 2 = 0 + 0.818 555 343 667 2;
  • 23) 0.818 555 343 667 2 × 2 = 1 + 0.637 110 687 334 4;
  • 24) 0.637 110 687 334 4 × 2 = 1 + 0.274 221 374 668 8;
  • 25) 0.274 221 374 668 8 × 2 = 0 + 0.548 442 749 337 6;
  • 26) 0.548 442 749 337 6 × 2 = 1 + 0.096 885 498 675 2;
  • 27) 0.096 885 498 675 2 × 2 = 0 + 0.193 770 997 350 4;
  • 28) 0.193 770 997 350 4 × 2 = 0 + 0.387 541 994 700 8;
  • 29) 0.387 541 994 700 8 × 2 = 0 + 0.775 083 989 401 6;
  • 30) 0.775 083 989 401 6 × 2 = 1 + 0.550 167 978 803 2;
  • 31) 0.550 167 978 803 2 × 2 = 1 + 0.100 335 957 606 4;
  • 32) 0.100 335 957 606 4 × 2 = 0 + 0.200 671 915 212 8;
  • 33) 0.200 671 915 212 8 × 2 = 0 + 0.401 343 830 425 6;
  • 34) 0.401 343 830 425 6 × 2 = 0 + 0.802 687 660 851 2;
  • 35) 0.802 687 660 851 2 × 2 = 1 + 0.605 375 321 702 4;
  • 36) 0.605 375 321 702 4 × 2 = 1 + 0.210 750 643 404 8;
  • 37) 0.210 750 643 404 8 × 2 = 0 + 0.421 501 286 809 6;
  • 38) 0.421 501 286 809 6 × 2 = 0 + 0.843 002 573 619 2;
  • 39) 0.843 002 573 619 2 × 2 = 1 + 0.686 005 147 238 4;
  • 40) 0.686 005 147 238 4 × 2 = 1 + 0.372 010 294 476 8;
  • 41) 0.372 010 294 476 8 × 2 = 0 + 0.744 020 588 953 6;
  • 42) 0.744 020 588 953 6 × 2 = 1 + 0.488 041 177 907 2;
  • 43) 0.488 041 177 907 2 × 2 = 0 + 0.976 082 355 814 4;
  • 44) 0.976 082 355 814 4 × 2 = 1 + 0.952 164 711 628 8;
  • 45) 0.952 164 711 628 8 × 2 = 1 + 0.904 329 423 257 6;
  • 46) 0.904 329 423 257 6 × 2 = 1 + 0.808 658 846 515 2;
  • 47) 0.808 658 846 515 2 × 2 = 1 + 0.617 317 693 030 4;
  • 48) 0.617 317 693 030 4 × 2 = 1 + 0.234 635 386 060 8;
  • 49) 0.234 635 386 060 8 × 2 = 0 + 0.469 270 772 121 6;
  • 50) 0.469 270 772 121 6 × 2 = 0 + 0.938 541 544 243 2;
  • 51) 0.938 541 544 243 2 × 2 = 1 + 0.877 083 088 486 4;
  • 52) 0.877 083 088 486 4 × 2 = 1 + 0.754 166 176 972 8;
  • 53) 0.754 166 176 972 8 × 2 = 1 + 0.508 332 353 945 6;
  • 54) 0.508 332 353 945 6 × 2 = 1 + 0.016 664 707 891 2;
  • 55) 0.016 664 707 891 2 × 2 = 0 + 0.033 329 415 782 4;
  • 56) 0.033 329 415 782 4 × 2 = 0 + 0.066 658 831 564 8;
  • 57) 0.066 658 831 564 8 × 2 = 0 + 0.133 317 663 129 6;
  • 58) 0.133 317 663 129 6 × 2 = 0 + 0.266 635 326 259 2;
  • 59) 0.266 635 326 259 2 × 2 = 0 + 0.533 270 652 518 4;
  • 60) 0.533 270 652 518 4 × 2 = 1 + 0.066 541 305 036 8;
  • 61) 0.066 541 305 036 8 × 2 = 0 + 0.133 082 610 073 6;
  • 62) 0.133 082 610 073 6 × 2 = 0 + 0.266 165 220 147 2;
  • 63) 0.266 165 220 147 2 × 2 = 0 + 0.532 330 440 294 4;
  • 64) 0.532 330 440 294 4 × 2 = 1 + 0.064 660 880 588 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:


0.000 282 005 919 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001(2)

6. Positive number before normalization:

0.000 282 005 919 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 919 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001 =

0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001


Decimal number -0.000 282 005 919 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0011 0101 1111 0011 1100 0001 0001

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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