-0.000 282 005 771 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 771(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
-0.000 282 005 771(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:

|-0.000 282 005 771| = 0.000 282 005 771


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: 0.000 282 005 771.

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) 0.000 282 005 771 × 2 = 0 + 0.000 564 011 542;
  • 2) 0.000 564 011 542 × 2 = 0 + 0.001 128 023 084;
  • 3) 0.001 128 023 084 × 2 = 0 + 0.002 256 046 168;
  • 4) 0.002 256 046 168 × 2 = 0 + 0.004 512 092 336;
  • 5) 0.004 512 092 336 × 2 = 0 + 0.009 024 184 672;
  • 6) 0.009 024 184 672 × 2 = 0 + 0.018 048 369 344;
  • 7) 0.018 048 369 344 × 2 = 0 + 0.036 096 738 688;
  • 8) 0.036 096 738 688 × 2 = 0 + 0.072 193 477 376;
  • 9) 0.072 193 477 376 × 2 = 0 + 0.144 386 954 752;
  • 10) 0.144 386 954 752 × 2 = 0 + 0.288 773 909 504;
  • 11) 0.288 773 909 504 × 2 = 0 + 0.577 547 819 008;
  • 12) 0.577 547 819 008 × 2 = 1 + 0.155 095 638 016;
  • 13) 0.155 095 638 016 × 2 = 0 + 0.310 191 276 032;
  • 14) 0.310 191 276 032 × 2 = 0 + 0.620 382 552 064;
  • 15) 0.620 382 552 064 × 2 = 1 + 0.240 765 104 128;
  • 16) 0.240 765 104 128 × 2 = 0 + 0.481 530 208 256;
  • 17) 0.481 530 208 256 × 2 = 0 + 0.963 060 416 512;
  • 18) 0.963 060 416 512 × 2 = 1 + 0.926 120 833 024;
  • 19) 0.926 120 833 024 × 2 = 1 + 0.852 241 666 048;
  • 20) 0.852 241 666 048 × 2 = 1 + 0.704 483 332 096;
  • 21) 0.704 483 332 096 × 2 = 1 + 0.408 966 664 192;
  • 22) 0.408 966 664 192 × 2 = 0 + 0.817 933 328 384;
  • 23) 0.817 933 328 384 × 2 = 1 + 0.635 866 656 768;
  • 24) 0.635 866 656 768 × 2 = 1 + 0.271 733 313 536;
  • 25) 0.271 733 313 536 × 2 = 0 + 0.543 466 627 072;
  • 26) 0.543 466 627 072 × 2 = 1 + 0.086 933 254 144;
  • 27) 0.086 933 254 144 × 2 = 0 + 0.173 866 508 288;
  • 28) 0.173 866 508 288 × 2 = 0 + 0.347 733 016 576;
  • 29) 0.347 733 016 576 × 2 = 0 + 0.695 466 033 152;
  • 30) 0.695 466 033 152 × 2 = 1 + 0.390 932 066 304;
  • 31) 0.390 932 066 304 × 2 = 0 + 0.781 864 132 608;
  • 32) 0.781 864 132 608 × 2 = 1 + 0.563 728 265 216;
  • 33) 0.563 728 265 216 × 2 = 1 + 0.127 456 530 432;
  • 34) 0.127 456 530 432 × 2 = 0 + 0.254 913 060 864;
  • 35) 0.254 913 060 864 × 2 = 0 + 0.509 826 121 728;
  • 36) 0.509 826 121 728 × 2 = 1 + 0.019 652 243 456;
  • 37) 0.019 652 243 456 × 2 = 0 + 0.039 304 486 912;
  • 38) 0.039 304 486 912 × 2 = 0 + 0.078 608 973 824;
  • 39) 0.078 608 973 824 × 2 = 0 + 0.157 217 947 648;
  • 40) 0.157 217 947 648 × 2 = 0 + 0.314 435 895 296;
  • 41) 0.314 435 895 296 × 2 = 0 + 0.628 871 790 592;
  • 42) 0.628 871 790 592 × 2 = 1 + 0.257 743 581 184;
  • 43) 0.257 743 581 184 × 2 = 0 + 0.515 487 162 368;
  • 44) 0.515 487 162 368 × 2 = 1 + 0.030 974 324 736;
  • 45) 0.030 974 324 736 × 2 = 0 + 0.061 948 649 472;
  • 46) 0.061 948 649 472 × 2 = 0 + 0.123 897 298 944;
  • 47) 0.123 897 298 944 × 2 = 0 + 0.247 794 597 888;
  • 48) 0.247 794 597 888 × 2 = 0 + 0.495 589 195 776;
  • 49) 0.495 589 195 776 × 2 = 0 + 0.991 178 391 552;
  • 50) 0.991 178 391 552 × 2 = 1 + 0.982 356 783 104;
  • 51) 0.982 356 783 104 × 2 = 1 + 0.964 713 566 208;
  • 52) 0.964 713 566 208 × 2 = 1 + 0.929 427 132 416;
  • 53) 0.929 427 132 416 × 2 = 1 + 0.858 854 264 832;
  • 54) 0.858 854 264 832 × 2 = 1 + 0.717 708 529 664;
  • 55) 0.717 708 529 664 × 2 = 1 + 0.435 417 059 328;
  • 56) 0.435 417 059 328 × 2 = 0 + 0.870 834 118 656;
  • 57) 0.870 834 118 656 × 2 = 1 + 0.741 668 237 312;
  • 58) 0.741 668 237 312 × 2 = 1 + 0.483 336 474 624;
  • 59) 0.483 336 474 624 × 2 = 0 + 0.966 672 949 248;
  • 60) 0.966 672 949 248 × 2 = 1 + 0.933 345 898 496;
  • 61) 0.933 345 898 496 × 2 = 1 + 0.866 691 796 992;
  • 62) 0.866 691 796 992 × 2 = 1 + 0.733 383 593 984;
  • 63) 0.733 383 593 984 × 2 = 1 + 0.466 767 187 968;
  • 64) 0.466 767 187 968 × 2 = 0 + 0.933 534 375 936;

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:


0.000 282 005 771(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 1101 1110(2)

6. Positive number before normalization:

0.000 282 005 771(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 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:


0.000 282 005 771(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 1101 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 1101 1110(2) × 20 =

1.0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 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 0101 1001 0000 0101 0000 0111 1110 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 0101 1001 0000 0101 0000 0111 1110 1101 1110 =

0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 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 0101 1001 0000 0101 0000 0111 1110 1101 1110


Decimal number -0.000 282 005 771 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1001 0000 0101 0000 0111 1110 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