-0.000 282 005 898 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 898 6(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
-0.000 282 005 898 6(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:

|-0.000 282 005 898 6| = 0.000 282 005 898 6


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: 0.000 282 005 898 6.

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) 0.000 282 005 898 6 × 2 = 0 + 0.000 564 011 797 2;
  • 2) 0.000 564 011 797 2 × 2 = 0 + 0.001 128 023 594 4;
  • 3) 0.001 128 023 594 4 × 2 = 0 + 0.002 256 047 188 8;
  • 4) 0.002 256 047 188 8 × 2 = 0 + 0.004 512 094 377 6;
  • 5) 0.004 512 094 377 6 × 2 = 0 + 0.009 024 188 755 2;
  • 6) 0.009 024 188 755 2 × 2 = 0 + 0.018 048 377 510 4;
  • 7) 0.018 048 377 510 4 × 2 = 0 + 0.036 096 755 020 8;
  • 8) 0.036 096 755 020 8 × 2 = 0 + 0.072 193 510 041 6;
  • 9) 0.072 193 510 041 6 × 2 = 0 + 0.144 387 020 083 2;
  • 10) 0.144 387 020 083 2 × 2 = 0 + 0.288 774 040 166 4;
  • 11) 0.288 774 040 166 4 × 2 = 0 + 0.577 548 080 332 8;
  • 12) 0.577 548 080 332 8 × 2 = 1 + 0.155 096 160 665 6;
  • 13) 0.155 096 160 665 6 × 2 = 0 + 0.310 192 321 331 2;
  • 14) 0.310 192 321 331 2 × 2 = 0 + 0.620 384 642 662 4;
  • 15) 0.620 384 642 662 4 × 2 = 1 + 0.240 769 285 324 8;
  • 16) 0.240 769 285 324 8 × 2 = 0 + 0.481 538 570 649 6;
  • 17) 0.481 538 570 649 6 × 2 = 0 + 0.963 077 141 299 2;
  • 18) 0.963 077 141 299 2 × 2 = 1 + 0.926 154 282 598 4;
  • 19) 0.926 154 282 598 4 × 2 = 1 + 0.852 308 565 196 8;
  • 20) 0.852 308 565 196 8 × 2 = 1 + 0.704 617 130 393 6;
  • 21) 0.704 617 130 393 6 × 2 = 1 + 0.409 234 260 787 2;
  • 22) 0.409 234 260 787 2 × 2 = 0 + 0.818 468 521 574 4;
  • 23) 0.818 468 521 574 4 × 2 = 1 + 0.636 937 043 148 8;
  • 24) 0.636 937 043 148 8 × 2 = 1 + 0.273 874 086 297 6;
  • 25) 0.273 874 086 297 6 × 2 = 0 + 0.547 748 172 595 2;
  • 26) 0.547 748 172 595 2 × 2 = 1 + 0.095 496 345 190 4;
  • 27) 0.095 496 345 190 4 × 2 = 0 + 0.190 992 690 380 8;
  • 28) 0.190 992 690 380 8 × 2 = 0 + 0.381 985 380 761 6;
  • 29) 0.381 985 380 761 6 × 2 = 0 + 0.763 970 761 523 2;
  • 30) 0.763 970 761 523 2 × 2 = 1 + 0.527 941 523 046 4;
  • 31) 0.527 941 523 046 4 × 2 = 1 + 0.055 883 046 092 8;
  • 32) 0.055 883 046 092 8 × 2 = 0 + 0.111 766 092 185 6;
  • 33) 0.111 766 092 185 6 × 2 = 0 + 0.223 532 184 371 2;
  • 34) 0.223 532 184 371 2 × 2 = 0 + 0.447 064 368 742 4;
  • 35) 0.447 064 368 742 4 × 2 = 0 + 0.894 128 737 484 8;
  • 36) 0.894 128 737 484 8 × 2 = 1 + 0.788 257 474 969 6;
  • 37) 0.788 257 474 969 6 × 2 = 1 + 0.576 514 949 939 2;
  • 38) 0.576 514 949 939 2 × 2 = 1 + 0.153 029 899 878 4;
  • 39) 0.153 029 899 878 4 × 2 = 0 + 0.306 059 799 756 8;
  • 40) 0.306 059 799 756 8 × 2 = 0 + 0.612 119 599 513 6;
  • 41) 0.612 119 599 513 6 × 2 = 1 + 0.224 239 199 027 2;
  • 42) 0.224 239 199 027 2 × 2 = 0 + 0.448 478 398 054 4;
  • 43) 0.448 478 398 054 4 × 2 = 0 + 0.896 956 796 108 8;
  • 44) 0.896 956 796 108 8 × 2 = 1 + 0.793 913 592 217 6;
  • 45) 0.793 913 592 217 6 × 2 = 1 + 0.587 827 184 435 2;
  • 46) 0.587 827 184 435 2 × 2 = 1 + 0.175 654 368 870 4;
  • 47) 0.175 654 368 870 4 × 2 = 0 + 0.351 308 737 740 8;
  • 48) 0.351 308 737 740 8 × 2 = 0 + 0.702 617 475 481 6;
  • 49) 0.702 617 475 481 6 × 2 = 1 + 0.405 234 950 963 2;
  • 50) 0.405 234 950 963 2 × 2 = 0 + 0.810 469 901 926 4;
  • 51) 0.810 469 901 926 4 × 2 = 1 + 0.620 939 803 852 8;
  • 52) 0.620 939 803 852 8 × 2 = 1 + 0.241 879 607 705 6;
  • 53) 0.241 879 607 705 6 × 2 = 0 + 0.483 759 215 411 2;
  • 54) 0.483 759 215 411 2 × 2 = 0 + 0.967 518 430 822 4;
  • 55) 0.967 518 430 822 4 × 2 = 1 + 0.935 036 861 644 8;
  • 56) 0.935 036 861 644 8 × 2 = 1 + 0.870 073 723 289 6;
  • 57) 0.870 073 723 289 6 × 2 = 1 + 0.740 147 446 579 2;
  • 58) 0.740 147 446 579 2 × 2 = 1 + 0.480 294 893 158 4;
  • 59) 0.480 294 893 158 4 × 2 = 0 + 0.960 589 786 316 8;
  • 60) 0.960 589 786 316 8 × 2 = 1 + 0.921 179 572 633 6;
  • 61) 0.921 179 572 633 6 × 2 = 1 + 0.842 359 145 267 2;
  • 62) 0.842 359 145 267 2 × 2 = 1 + 0.684 718 290 534 4;
  • 63) 0.684 718 290 534 4 × 2 = 1 + 0.369 436 581 068 8;
  • 64) 0.369 436 581 068 8 × 2 = 0 + 0.738 873 162 137 6;

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:


0.000 282 005 898 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110(2)

6. Positive number before normalization:

0.000 282 005 898 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110(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:


0.000 282 005 898 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110(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 0001 1100 1001 1100 1011 0011 1101 1110


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 0001 1100 1001 1100 1011 0011 1101 1110 =

0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110


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 0001 1100 1001 1100 1011 0011 1101 1110


Decimal number -0.000 282 005 898 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1100 1001 1100 1011 0011 1101 1110

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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