-0.000 282 006 438 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 438(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
-0.000 282 006 438(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:

|-0.000 282 006 438| = 0.000 282 006 438


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: 0.000 282 006 438.

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) 0.000 282 006 438 × 2 = 0 + 0.000 564 012 876;
  • 2) 0.000 564 012 876 × 2 = 0 + 0.001 128 025 752;
  • 3) 0.001 128 025 752 × 2 = 0 + 0.002 256 051 504;
  • 4) 0.002 256 051 504 × 2 = 0 + 0.004 512 103 008;
  • 5) 0.004 512 103 008 × 2 = 0 + 0.009 024 206 016;
  • 6) 0.009 024 206 016 × 2 = 0 + 0.018 048 412 032;
  • 7) 0.018 048 412 032 × 2 = 0 + 0.036 096 824 064;
  • 8) 0.036 096 824 064 × 2 = 0 + 0.072 193 648 128;
  • 9) 0.072 193 648 128 × 2 = 0 + 0.144 387 296 256;
  • 10) 0.144 387 296 256 × 2 = 0 + 0.288 774 592 512;
  • 11) 0.288 774 592 512 × 2 = 0 + 0.577 549 185 024;
  • 12) 0.577 549 185 024 × 2 = 1 + 0.155 098 370 048;
  • 13) 0.155 098 370 048 × 2 = 0 + 0.310 196 740 096;
  • 14) 0.310 196 740 096 × 2 = 0 + 0.620 393 480 192;
  • 15) 0.620 393 480 192 × 2 = 1 + 0.240 786 960 384;
  • 16) 0.240 786 960 384 × 2 = 0 + 0.481 573 920 768;
  • 17) 0.481 573 920 768 × 2 = 0 + 0.963 147 841 536;
  • 18) 0.963 147 841 536 × 2 = 1 + 0.926 295 683 072;
  • 19) 0.926 295 683 072 × 2 = 1 + 0.852 591 366 144;
  • 20) 0.852 591 366 144 × 2 = 1 + 0.705 182 732 288;
  • 21) 0.705 182 732 288 × 2 = 1 + 0.410 365 464 576;
  • 22) 0.410 365 464 576 × 2 = 0 + 0.820 730 929 152;
  • 23) 0.820 730 929 152 × 2 = 1 + 0.641 461 858 304;
  • 24) 0.641 461 858 304 × 2 = 1 + 0.282 923 716 608;
  • 25) 0.282 923 716 608 × 2 = 0 + 0.565 847 433 216;
  • 26) 0.565 847 433 216 × 2 = 1 + 0.131 694 866 432;
  • 27) 0.131 694 866 432 × 2 = 0 + 0.263 389 732 864;
  • 28) 0.263 389 732 864 × 2 = 0 + 0.526 779 465 728;
  • 29) 0.526 779 465 728 × 2 = 1 + 0.053 558 931 456;
  • 30) 0.053 558 931 456 × 2 = 0 + 0.107 117 862 912;
  • 31) 0.107 117 862 912 × 2 = 0 + 0.214 235 725 824;
  • 32) 0.214 235 725 824 × 2 = 0 + 0.428 471 451 648;
  • 33) 0.428 471 451 648 × 2 = 0 + 0.856 942 903 296;
  • 34) 0.856 942 903 296 × 2 = 1 + 0.713 885 806 592;
  • 35) 0.713 885 806 592 × 2 = 1 + 0.427 771 613 184;
  • 36) 0.427 771 613 184 × 2 = 0 + 0.855 543 226 368;
  • 37) 0.855 543 226 368 × 2 = 1 + 0.711 086 452 736;
  • 38) 0.711 086 452 736 × 2 = 1 + 0.422 172 905 472;
  • 39) 0.422 172 905 472 × 2 = 0 + 0.844 345 810 944;
  • 40) 0.844 345 810 944 × 2 = 1 + 0.688 691 621 888;
  • 41) 0.688 691 621 888 × 2 = 1 + 0.377 383 243 776;
  • 42) 0.377 383 243 776 × 2 = 0 + 0.754 766 487 552;
  • 43) 0.754 766 487 552 × 2 = 1 + 0.509 532 975 104;
  • 44) 0.509 532 975 104 × 2 = 1 + 0.019 065 950 208;
  • 45) 0.019 065 950 208 × 2 = 0 + 0.038 131 900 416;
  • 46) 0.038 131 900 416 × 2 = 0 + 0.076 263 800 832;
  • 47) 0.076 263 800 832 × 2 = 0 + 0.152 527 601 664;
  • 48) 0.152 527 601 664 × 2 = 0 + 0.305 055 203 328;
  • 49) 0.305 055 203 328 × 2 = 0 + 0.610 110 406 656;
  • 50) 0.610 110 406 656 × 2 = 1 + 0.220 220 813 312;
  • 51) 0.220 220 813 312 × 2 = 0 + 0.440 441 626 624;
  • 52) 0.440 441 626 624 × 2 = 0 + 0.880 883 253 248;
  • 53) 0.880 883 253 248 × 2 = 1 + 0.761 766 506 496;
  • 54) 0.761 766 506 496 × 2 = 1 + 0.523 533 012 992;
  • 55) 0.523 533 012 992 × 2 = 1 + 0.047 066 025 984;
  • 56) 0.047 066 025 984 × 2 = 0 + 0.094 132 051 968;
  • 57) 0.094 132 051 968 × 2 = 0 + 0.188 264 103 936;
  • 58) 0.188 264 103 936 × 2 = 0 + 0.376 528 207 872;
  • 59) 0.376 528 207 872 × 2 = 0 + 0.753 056 415 744;
  • 60) 0.753 056 415 744 × 2 = 1 + 0.506 112 831 488;
  • 61) 0.506 112 831 488 × 2 = 1 + 0.012 225 662 976;
  • 62) 0.012 225 662 976 × 2 = 0 + 0.024 451 325 952;
  • 63) 0.024 451 325 952 × 2 = 0 + 0.048 902 651 904;
  • 64) 0.048 902 651 904 × 2 = 0 + 0.097 805 303 808;

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:


0.000 282 006 438(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000(2)

6. Positive number before normalization:

0.000 282 006 438(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000(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:


0.000 282 006 438(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000(2) × 20 =

1.0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000(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 1000 0110 1101 1011 0000 0100 1110 0001 1000


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 1000 0110 1101 1011 0000 0100 1110 0001 1000 =

0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000


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 1000 0110 1101 1011 0000 0100 1110 0001 1000


Decimal number -0.000 282 006 438 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0110 1101 1011 0000 0100 1110 0001 1000

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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