-0.000 282 005 667 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 667(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
-0.000 282 005 667(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:

|-0.000 282 005 667| = 0.000 282 005 667


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: 0.000 282 005 667.

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) 0.000 282 005 667 × 2 = 0 + 0.000 564 011 334;
  • 2) 0.000 564 011 334 × 2 = 0 + 0.001 128 022 668;
  • 3) 0.001 128 022 668 × 2 = 0 + 0.002 256 045 336;
  • 4) 0.002 256 045 336 × 2 = 0 + 0.004 512 090 672;
  • 5) 0.004 512 090 672 × 2 = 0 + 0.009 024 181 344;
  • 6) 0.009 024 181 344 × 2 = 0 + 0.018 048 362 688;
  • 7) 0.018 048 362 688 × 2 = 0 + 0.036 096 725 376;
  • 8) 0.036 096 725 376 × 2 = 0 + 0.072 193 450 752;
  • 9) 0.072 193 450 752 × 2 = 0 + 0.144 386 901 504;
  • 10) 0.144 386 901 504 × 2 = 0 + 0.288 773 803 008;
  • 11) 0.288 773 803 008 × 2 = 0 + 0.577 547 606 016;
  • 12) 0.577 547 606 016 × 2 = 1 + 0.155 095 212 032;
  • 13) 0.155 095 212 032 × 2 = 0 + 0.310 190 424 064;
  • 14) 0.310 190 424 064 × 2 = 0 + 0.620 380 848 128;
  • 15) 0.620 380 848 128 × 2 = 1 + 0.240 761 696 256;
  • 16) 0.240 761 696 256 × 2 = 0 + 0.481 523 392 512;
  • 17) 0.481 523 392 512 × 2 = 0 + 0.963 046 785 024;
  • 18) 0.963 046 785 024 × 2 = 1 + 0.926 093 570 048;
  • 19) 0.926 093 570 048 × 2 = 1 + 0.852 187 140 096;
  • 20) 0.852 187 140 096 × 2 = 1 + 0.704 374 280 192;
  • 21) 0.704 374 280 192 × 2 = 1 + 0.408 748 560 384;
  • 22) 0.408 748 560 384 × 2 = 0 + 0.817 497 120 768;
  • 23) 0.817 497 120 768 × 2 = 1 + 0.634 994 241 536;
  • 24) 0.634 994 241 536 × 2 = 1 + 0.269 988 483 072;
  • 25) 0.269 988 483 072 × 2 = 0 + 0.539 976 966 144;
  • 26) 0.539 976 966 144 × 2 = 1 + 0.079 953 932 288;
  • 27) 0.079 953 932 288 × 2 = 0 + 0.159 907 864 576;
  • 28) 0.159 907 864 576 × 2 = 0 + 0.319 815 729 152;
  • 29) 0.319 815 729 152 × 2 = 0 + 0.639 631 458 304;
  • 30) 0.639 631 458 304 × 2 = 1 + 0.279 262 916 608;
  • 31) 0.279 262 916 608 × 2 = 0 + 0.558 525 833 216;
  • 32) 0.558 525 833 216 × 2 = 1 + 0.117 051 666 432;
  • 33) 0.117 051 666 432 × 2 = 0 + 0.234 103 332 864;
  • 34) 0.234 103 332 864 × 2 = 0 + 0.468 206 665 728;
  • 35) 0.468 206 665 728 × 2 = 0 + 0.936 413 331 456;
  • 36) 0.936 413 331 456 × 2 = 1 + 0.872 826 662 912;
  • 37) 0.872 826 662 912 × 2 = 1 + 0.745 653 325 824;
  • 38) 0.745 653 325 824 × 2 = 1 + 0.491 306 651 648;
  • 39) 0.491 306 651 648 × 2 = 0 + 0.982 613 303 296;
  • 40) 0.982 613 303 296 × 2 = 1 + 0.965 226 606 592;
  • 41) 0.965 226 606 592 × 2 = 1 + 0.930 453 213 184;
  • 42) 0.930 453 213 184 × 2 = 1 + 0.860 906 426 368;
  • 43) 0.860 906 426 368 × 2 = 1 + 0.721 812 852 736;
  • 44) 0.721 812 852 736 × 2 = 1 + 0.443 625 705 472;
  • 45) 0.443 625 705 472 × 2 = 0 + 0.887 251 410 944;
  • 46) 0.887 251 410 944 × 2 = 1 + 0.774 502 821 888;
  • 47) 0.774 502 821 888 × 2 = 1 + 0.549 005 643 776;
  • 48) 0.549 005 643 776 × 2 = 1 + 0.098 011 287 552;
  • 49) 0.098 011 287 552 × 2 = 0 + 0.196 022 575 104;
  • 50) 0.196 022 575 104 × 2 = 0 + 0.392 045 150 208;
  • 51) 0.392 045 150 208 × 2 = 0 + 0.784 090 300 416;
  • 52) 0.784 090 300 416 × 2 = 1 + 0.568 180 600 832;
  • 53) 0.568 180 600 832 × 2 = 1 + 0.136 361 201 664;
  • 54) 0.136 361 201 664 × 2 = 0 + 0.272 722 403 328;
  • 55) 0.272 722 403 328 × 2 = 0 + 0.545 444 806 656;
  • 56) 0.545 444 806 656 × 2 = 1 + 0.090 889 613 312;
  • 57) 0.090 889 613 312 × 2 = 0 + 0.181 779 226 624;
  • 58) 0.181 779 226 624 × 2 = 0 + 0.363 558 453 248;
  • 59) 0.363 558 453 248 × 2 = 0 + 0.727 116 906 496;
  • 60) 0.727 116 906 496 × 2 = 1 + 0.454 233 812 992;
  • 61) 0.454 233 812 992 × 2 = 0 + 0.908 467 625 984;
  • 62) 0.908 467 625 984 × 2 = 1 + 0.816 935 251 968;
  • 63) 0.816 935 251 968 × 2 = 1 + 0.633 870 503 936;
  • 64) 0.633 870 503 936 × 2 = 1 + 0.267 741 007 872;

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:


0.000 282 005 667(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111(2)

6. Positive number before normalization:

0.000 282 005 667(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111(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:


0.000 282 005 667(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111(2) × 20 =

1.0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111(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 0001 1101 1111 0111 0001 1001 0001 0111


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 0001 1101 1111 0111 0001 1001 0001 0111 =

0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111


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 0001 1101 1111 0111 0001 1001 0001 0111


Decimal number -0.000 282 005 667 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0001 1101 1111 0111 0001 1001 0001 0111

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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