-0.000 282 005 880 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 880 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
-0.000 282 005 880 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:

|-0.000 282 005 880 9| = 0.000 282 005 880 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: 0.000 282 005 880 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) 0.000 282 005 880 9 × 2 = 0 + 0.000 564 011 761 8;
  • 2) 0.000 564 011 761 8 × 2 = 0 + 0.001 128 023 523 6;
  • 3) 0.001 128 023 523 6 × 2 = 0 + 0.002 256 047 047 2;
  • 4) 0.002 256 047 047 2 × 2 = 0 + 0.004 512 094 094 4;
  • 5) 0.004 512 094 094 4 × 2 = 0 + 0.009 024 188 188 8;
  • 6) 0.009 024 188 188 8 × 2 = 0 + 0.018 048 376 377 6;
  • 7) 0.018 048 376 377 6 × 2 = 0 + 0.036 096 752 755 2;
  • 8) 0.036 096 752 755 2 × 2 = 0 + 0.072 193 505 510 4;
  • 9) 0.072 193 505 510 4 × 2 = 0 + 0.144 387 011 020 8;
  • 10) 0.144 387 011 020 8 × 2 = 0 + 0.288 774 022 041 6;
  • 11) 0.288 774 022 041 6 × 2 = 0 + 0.577 548 044 083 2;
  • 12) 0.577 548 044 083 2 × 2 = 1 + 0.155 096 088 166 4;
  • 13) 0.155 096 088 166 4 × 2 = 0 + 0.310 192 176 332 8;
  • 14) 0.310 192 176 332 8 × 2 = 0 + 0.620 384 352 665 6;
  • 15) 0.620 384 352 665 6 × 2 = 1 + 0.240 768 705 331 2;
  • 16) 0.240 768 705 331 2 × 2 = 0 + 0.481 537 410 662 4;
  • 17) 0.481 537 410 662 4 × 2 = 0 + 0.963 074 821 324 8;
  • 18) 0.963 074 821 324 8 × 2 = 1 + 0.926 149 642 649 6;
  • 19) 0.926 149 642 649 6 × 2 = 1 + 0.852 299 285 299 2;
  • 20) 0.852 299 285 299 2 × 2 = 1 + 0.704 598 570 598 4;
  • 21) 0.704 598 570 598 4 × 2 = 1 + 0.409 197 141 196 8;
  • 22) 0.409 197 141 196 8 × 2 = 0 + 0.818 394 282 393 6;
  • 23) 0.818 394 282 393 6 × 2 = 1 + 0.636 788 564 787 2;
  • 24) 0.636 788 564 787 2 × 2 = 1 + 0.273 577 129 574 4;
  • 25) 0.273 577 129 574 4 × 2 = 0 + 0.547 154 259 148 8;
  • 26) 0.547 154 259 148 8 × 2 = 1 + 0.094 308 518 297 6;
  • 27) 0.094 308 518 297 6 × 2 = 0 + 0.188 617 036 595 2;
  • 28) 0.188 617 036 595 2 × 2 = 0 + 0.377 234 073 190 4;
  • 29) 0.377 234 073 190 4 × 2 = 0 + 0.754 468 146 380 8;
  • 30) 0.754 468 146 380 8 × 2 = 1 + 0.508 936 292 761 6;
  • 31) 0.508 936 292 761 6 × 2 = 1 + 0.017 872 585 523 2;
  • 32) 0.017 872 585 523 2 × 2 = 0 + 0.035 745 171 046 4;
  • 33) 0.035 745 171 046 4 × 2 = 0 + 0.071 490 342 092 8;
  • 34) 0.071 490 342 092 8 × 2 = 0 + 0.142 980 684 185 6;
  • 35) 0.142 980 684 185 6 × 2 = 0 + 0.285 961 368 371 2;
  • 36) 0.285 961 368 371 2 × 2 = 0 + 0.571 922 736 742 4;
  • 37) 0.571 922 736 742 4 × 2 = 1 + 0.143 845 473 484 8;
  • 38) 0.143 845 473 484 8 × 2 = 0 + 0.287 690 946 969 6;
  • 39) 0.287 690 946 969 6 × 2 = 0 + 0.575 381 893 939 2;
  • 40) 0.575 381 893 939 2 × 2 = 1 + 0.150 763 787 878 4;
  • 41) 0.150 763 787 878 4 × 2 = 0 + 0.301 527 575 756 8;
  • 42) 0.301 527 575 756 8 × 2 = 0 + 0.603 055 151 513 6;
  • 43) 0.603 055 151 513 6 × 2 = 1 + 0.206 110 303 027 2;
  • 44) 0.206 110 303 027 2 × 2 = 0 + 0.412 220 606 054 4;
  • 45) 0.412 220 606 054 4 × 2 = 0 + 0.824 441 212 108 8;
  • 46) 0.824 441 212 108 8 × 2 = 1 + 0.648 882 424 217 6;
  • 47) 0.648 882 424 217 6 × 2 = 1 + 0.297 764 848 435 2;
  • 48) 0.297 764 848 435 2 × 2 = 0 + 0.595 529 696 870 4;
  • 49) 0.595 529 696 870 4 × 2 = 1 + 0.191 059 393 740 8;
  • 50) 0.191 059 393 740 8 × 2 = 0 + 0.382 118 787 481 6;
  • 51) 0.382 118 787 481 6 × 2 = 0 + 0.764 237 574 963 2;
  • 52) 0.764 237 574 963 2 × 2 = 1 + 0.528 475 149 926 4;
  • 53) 0.528 475 149 926 4 × 2 = 1 + 0.056 950 299 852 8;
  • 54) 0.056 950 299 852 8 × 2 = 0 + 0.113 900 599 705 6;
  • 55) 0.113 900 599 705 6 × 2 = 0 + 0.227 801 199 411 2;
  • 56) 0.227 801 199 411 2 × 2 = 0 + 0.455 602 398 822 4;
  • 57) 0.455 602 398 822 4 × 2 = 0 + 0.911 204 797 644 8;
  • 58) 0.911 204 797 644 8 × 2 = 1 + 0.822 409 595 289 6;
  • 59) 0.822 409 595 289 6 × 2 = 1 + 0.644 819 190 579 2;
  • 60) 0.644 819 190 579 2 × 2 = 1 + 0.289 638 381 158 4;
  • 61) 0.289 638 381 158 4 × 2 = 0 + 0.579 276 762 316 8;
  • 62) 0.579 276 762 316 8 × 2 = 1 + 0.158 553 524 633 6;
  • 63) 0.158 553 524 633 6 × 2 = 0 + 0.317 107 049 267 2;
  • 64) 0.317 107 049 267 2 × 2 = 0 + 0.634 214 098 534 4;

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:


0.000 282 005 880 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100(2)

6. Positive number before normalization:

0.000 282 005 880 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100(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:


0.000 282 005 880 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100(2) × 20 =

1.0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100(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 0000 1001 0010 0110 1001 1000 0111 0100


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 0000 1001 0010 0110 1001 1000 0111 0100 =

0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100


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 0000 1001 0010 0110 1001 1000 0111 0100


Decimal number -0.000 282 005 880 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 1001 0010 0110 1001 1000 0111 0100

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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