-0.000 282 005 906 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 906 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
-0.000 282 005 906 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:

|-0.000 282 005 906 9| = 0.000 282 005 906 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: 0.000 282 005 906 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) 0.000 282 005 906 9 × 2 = 0 + 0.000 564 011 813 8;
  • 2) 0.000 564 011 813 8 × 2 = 0 + 0.001 128 023 627 6;
  • 3) 0.001 128 023 627 6 × 2 = 0 + 0.002 256 047 255 2;
  • 4) 0.002 256 047 255 2 × 2 = 0 + 0.004 512 094 510 4;
  • 5) 0.004 512 094 510 4 × 2 = 0 + 0.009 024 189 020 8;
  • 6) 0.009 024 189 020 8 × 2 = 0 + 0.018 048 378 041 6;
  • 7) 0.018 048 378 041 6 × 2 = 0 + 0.036 096 756 083 2;
  • 8) 0.036 096 756 083 2 × 2 = 0 + 0.072 193 512 166 4;
  • 9) 0.072 193 512 166 4 × 2 = 0 + 0.144 387 024 332 8;
  • 10) 0.144 387 024 332 8 × 2 = 0 + 0.288 774 048 665 6;
  • 11) 0.288 774 048 665 6 × 2 = 0 + 0.577 548 097 331 2;
  • 12) 0.577 548 097 331 2 × 2 = 1 + 0.155 096 194 662 4;
  • 13) 0.155 096 194 662 4 × 2 = 0 + 0.310 192 389 324 8;
  • 14) 0.310 192 389 324 8 × 2 = 0 + 0.620 384 778 649 6;
  • 15) 0.620 384 778 649 6 × 2 = 1 + 0.240 769 557 299 2;
  • 16) 0.240 769 557 299 2 × 2 = 0 + 0.481 539 114 598 4;
  • 17) 0.481 539 114 598 4 × 2 = 0 + 0.963 078 229 196 8;
  • 18) 0.963 078 229 196 8 × 2 = 1 + 0.926 156 458 393 6;
  • 19) 0.926 156 458 393 6 × 2 = 1 + 0.852 312 916 787 2;
  • 20) 0.852 312 916 787 2 × 2 = 1 + 0.704 625 833 574 4;
  • 21) 0.704 625 833 574 4 × 2 = 1 + 0.409 251 667 148 8;
  • 22) 0.409 251 667 148 8 × 2 = 0 + 0.818 503 334 297 6;
  • 23) 0.818 503 334 297 6 × 2 = 1 + 0.637 006 668 595 2;
  • 24) 0.637 006 668 595 2 × 2 = 1 + 0.274 013 337 190 4;
  • 25) 0.274 013 337 190 4 × 2 = 0 + 0.548 026 674 380 8;
  • 26) 0.548 026 674 380 8 × 2 = 1 + 0.096 053 348 761 6;
  • 27) 0.096 053 348 761 6 × 2 = 0 + 0.192 106 697 523 2;
  • 28) 0.192 106 697 523 2 × 2 = 0 + 0.384 213 395 046 4;
  • 29) 0.384 213 395 046 4 × 2 = 0 + 0.768 426 790 092 8;
  • 30) 0.768 426 790 092 8 × 2 = 1 + 0.536 853 580 185 6;
  • 31) 0.536 853 580 185 6 × 2 = 1 + 0.073 707 160 371 2;
  • 32) 0.073 707 160 371 2 × 2 = 0 + 0.147 414 320 742 4;
  • 33) 0.147 414 320 742 4 × 2 = 0 + 0.294 828 641 484 8;
  • 34) 0.294 828 641 484 8 × 2 = 0 + 0.589 657 282 969 6;
  • 35) 0.589 657 282 969 6 × 2 = 1 + 0.179 314 565 939 2;
  • 36) 0.179 314 565 939 2 × 2 = 0 + 0.358 629 131 878 4;
  • 37) 0.358 629 131 878 4 × 2 = 0 + 0.717 258 263 756 8;
  • 38) 0.717 258 263 756 8 × 2 = 1 + 0.434 516 527 513 6;
  • 39) 0.434 516 527 513 6 × 2 = 0 + 0.869 033 055 027 2;
  • 40) 0.869 033 055 027 2 × 2 = 1 + 0.738 066 110 054 4;
  • 41) 0.738 066 110 054 4 × 2 = 1 + 0.476 132 220 108 8;
  • 42) 0.476 132 220 108 8 × 2 = 0 + 0.952 264 440 217 6;
  • 43) 0.952 264 440 217 6 × 2 = 1 + 0.904 528 880 435 2;
  • 44) 0.904 528 880 435 2 × 2 = 1 + 0.809 057 760 870 4;
  • 45) 0.809 057 760 870 4 × 2 = 1 + 0.618 115 521 740 8;
  • 46) 0.618 115 521 740 8 × 2 = 1 + 0.236 231 043 481 6;
  • 47) 0.236 231 043 481 6 × 2 = 0 + 0.472 462 086 963 2;
  • 48) 0.472 462 086 963 2 × 2 = 0 + 0.944 924 173 926 4;
  • 49) 0.944 924 173 926 4 × 2 = 1 + 0.889 848 347 852 8;
  • 50) 0.889 848 347 852 8 × 2 = 1 + 0.779 696 695 705 6;
  • 51) 0.779 696 695 705 6 × 2 = 1 + 0.559 393 391 411 2;
  • 52) 0.559 393 391 411 2 × 2 = 1 + 0.118 786 782 822 4;
  • 53) 0.118 786 782 822 4 × 2 = 0 + 0.237 573 565 644 8;
  • 54) 0.237 573 565 644 8 × 2 = 0 + 0.475 147 131 289 6;
  • 55) 0.475 147 131 289 6 × 2 = 0 + 0.950 294 262 579 2;
  • 56) 0.950 294 262 579 2 × 2 = 1 + 0.900 588 525 158 4;
  • 57) 0.900 588 525 158 4 × 2 = 1 + 0.801 177 050 316 8;
  • 58) 0.801 177 050 316 8 × 2 = 1 + 0.602 354 100 633 6;
  • 59) 0.602 354 100 633 6 × 2 = 1 + 0.204 708 201 267 2;
  • 60) 0.204 708 201 267 2 × 2 = 0 + 0.409 416 402 534 4;
  • 61) 0.409 416 402 534 4 × 2 = 0 + 0.818 832 805 068 8;
  • 62) 0.818 832 805 068 8 × 2 = 1 + 0.637 665 610 137 6;
  • 63) 0.637 665 610 137 6 × 2 = 1 + 0.275 331 220 275 2;
  • 64) 0.275 331 220 275 2 × 2 = 0 + 0.550 662 440 550 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:


0.000 282 005 906 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110(2)

6. Positive number before normalization:

0.000 282 005 906 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110(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:


0.000 282 005 906 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110(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 0010 0101 1011 1100 1111 0001 1110 0110


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 0010 0101 1011 1100 1111 0001 1110 0110 =

0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110


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 0010 0101 1011 1100 1111 0001 1110 0110


Decimal number -0.000 282 005 906 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0101 1011 1100 1111 0001 1110 0110

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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