-0.000 282 006 301 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 301(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
-0.000 282 006 301(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:

|-0.000 282 006 301| = 0.000 282 006 301


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: 0.000 282 006 301.

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) 0.000 282 006 301 × 2 = 0 + 0.000 564 012 602;
  • 2) 0.000 564 012 602 × 2 = 0 + 0.001 128 025 204;
  • 3) 0.001 128 025 204 × 2 = 0 + 0.002 256 050 408;
  • 4) 0.002 256 050 408 × 2 = 0 + 0.004 512 100 816;
  • 5) 0.004 512 100 816 × 2 = 0 + 0.009 024 201 632;
  • 6) 0.009 024 201 632 × 2 = 0 + 0.018 048 403 264;
  • 7) 0.018 048 403 264 × 2 = 0 + 0.036 096 806 528;
  • 8) 0.036 096 806 528 × 2 = 0 + 0.072 193 613 056;
  • 9) 0.072 193 613 056 × 2 = 0 + 0.144 387 226 112;
  • 10) 0.144 387 226 112 × 2 = 0 + 0.288 774 452 224;
  • 11) 0.288 774 452 224 × 2 = 0 + 0.577 548 904 448;
  • 12) 0.577 548 904 448 × 2 = 1 + 0.155 097 808 896;
  • 13) 0.155 097 808 896 × 2 = 0 + 0.310 195 617 792;
  • 14) 0.310 195 617 792 × 2 = 0 + 0.620 391 235 584;
  • 15) 0.620 391 235 584 × 2 = 1 + 0.240 782 471 168;
  • 16) 0.240 782 471 168 × 2 = 0 + 0.481 564 942 336;
  • 17) 0.481 564 942 336 × 2 = 0 + 0.963 129 884 672;
  • 18) 0.963 129 884 672 × 2 = 1 + 0.926 259 769 344;
  • 19) 0.926 259 769 344 × 2 = 1 + 0.852 519 538 688;
  • 20) 0.852 519 538 688 × 2 = 1 + 0.705 039 077 376;
  • 21) 0.705 039 077 376 × 2 = 1 + 0.410 078 154 752;
  • 22) 0.410 078 154 752 × 2 = 0 + 0.820 156 309 504;
  • 23) 0.820 156 309 504 × 2 = 1 + 0.640 312 619 008;
  • 24) 0.640 312 619 008 × 2 = 1 + 0.280 625 238 016;
  • 25) 0.280 625 238 016 × 2 = 0 + 0.561 250 476 032;
  • 26) 0.561 250 476 032 × 2 = 1 + 0.122 500 952 064;
  • 27) 0.122 500 952 064 × 2 = 0 + 0.245 001 904 128;
  • 28) 0.245 001 904 128 × 2 = 0 + 0.490 003 808 256;
  • 29) 0.490 003 808 256 × 2 = 0 + 0.980 007 616 512;
  • 30) 0.980 007 616 512 × 2 = 1 + 0.960 015 233 024;
  • 31) 0.960 015 233 024 × 2 = 1 + 0.920 030 466 048;
  • 32) 0.920 030 466 048 × 2 = 1 + 0.840 060 932 096;
  • 33) 0.840 060 932 096 × 2 = 1 + 0.680 121 864 192;
  • 34) 0.680 121 864 192 × 2 = 1 + 0.360 243 728 384;
  • 35) 0.360 243 728 384 × 2 = 0 + 0.720 487 456 768;
  • 36) 0.720 487 456 768 × 2 = 1 + 0.440 974 913 536;
  • 37) 0.440 974 913 536 × 2 = 0 + 0.881 949 827 072;
  • 38) 0.881 949 827 072 × 2 = 1 + 0.763 899 654 144;
  • 39) 0.763 899 654 144 × 2 = 1 + 0.527 799 308 288;
  • 40) 0.527 799 308 288 × 2 = 1 + 0.055 598 616 576;
  • 41) 0.055 598 616 576 × 2 = 0 + 0.111 197 233 152;
  • 42) 0.111 197 233 152 × 2 = 0 + 0.222 394 466 304;
  • 43) 0.222 394 466 304 × 2 = 0 + 0.444 788 932 608;
  • 44) 0.444 788 932 608 × 2 = 0 + 0.889 577 865 216;
  • 45) 0.889 577 865 216 × 2 = 1 + 0.779 155 730 432;
  • 46) 0.779 155 730 432 × 2 = 1 + 0.558 311 460 864;
  • 47) 0.558 311 460 864 × 2 = 1 + 0.116 622 921 728;
  • 48) 0.116 622 921 728 × 2 = 0 + 0.233 245 843 456;
  • 49) 0.233 245 843 456 × 2 = 0 + 0.466 491 686 912;
  • 50) 0.466 491 686 912 × 2 = 0 + 0.932 983 373 824;
  • 51) 0.932 983 373 824 × 2 = 1 + 0.865 966 747 648;
  • 52) 0.865 966 747 648 × 2 = 1 + 0.731 933 495 296;
  • 53) 0.731 933 495 296 × 2 = 1 + 0.463 866 990 592;
  • 54) 0.463 866 990 592 × 2 = 0 + 0.927 733 981 184;
  • 55) 0.927 733 981 184 × 2 = 1 + 0.855 467 962 368;
  • 56) 0.855 467 962 368 × 2 = 1 + 0.710 935 924 736;
  • 57) 0.710 935 924 736 × 2 = 1 + 0.421 871 849 472;
  • 58) 0.421 871 849 472 × 2 = 0 + 0.843 743 698 944;
  • 59) 0.843 743 698 944 × 2 = 1 + 0.687 487 397 888;
  • 60) 0.687 487 397 888 × 2 = 1 + 0.374 974 795 776;
  • 61) 0.374 974 795 776 × 2 = 0 + 0.749 949 591 552;
  • 62) 0.749 949 591 552 × 2 = 1 + 0.499 899 183 104;
  • 63) 0.499 899 183 104 × 2 = 0 + 0.999 798 366 208;
  • 64) 0.999 798 366 208 × 2 = 1 + 0.999 596 732 416;

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:


0.000 282 006 301(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101(2)

6. Positive number before normalization:

0.000 282 006 301(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101(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:


0.000 282 006 301(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101(2) × 20 =

1.0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101(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 0111 1101 0111 0000 1110 0011 1011 1011 0101


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 0111 1101 0111 0000 1110 0011 1011 1011 0101 =

0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101


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 0111 1101 0111 0000 1110 0011 1011 1011 0101


Decimal number -0.000 282 006 301 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1101 0111 0000 1110 0011 1011 1011 0101

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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