-0.000 282 005 934 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 934(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
-0.000 282 005 934(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:

|-0.000 282 005 934| = 0.000 282 005 934


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: 0.000 282 005 934.

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) 0.000 282 005 934 × 2 = 0 + 0.000 564 011 868;
  • 2) 0.000 564 011 868 × 2 = 0 + 0.001 128 023 736;
  • 3) 0.001 128 023 736 × 2 = 0 + 0.002 256 047 472;
  • 4) 0.002 256 047 472 × 2 = 0 + 0.004 512 094 944;
  • 5) 0.004 512 094 944 × 2 = 0 + 0.009 024 189 888;
  • 6) 0.009 024 189 888 × 2 = 0 + 0.018 048 379 776;
  • 7) 0.018 048 379 776 × 2 = 0 + 0.036 096 759 552;
  • 8) 0.036 096 759 552 × 2 = 0 + 0.072 193 519 104;
  • 9) 0.072 193 519 104 × 2 = 0 + 0.144 387 038 208;
  • 10) 0.144 387 038 208 × 2 = 0 + 0.288 774 076 416;
  • 11) 0.288 774 076 416 × 2 = 0 + 0.577 548 152 832;
  • 12) 0.577 548 152 832 × 2 = 1 + 0.155 096 305 664;
  • 13) 0.155 096 305 664 × 2 = 0 + 0.310 192 611 328;
  • 14) 0.310 192 611 328 × 2 = 0 + 0.620 385 222 656;
  • 15) 0.620 385 222 656 × 2 = 1 + 0.240 770 445 312;
  • 16) 0.240 770 445 312 × 2 = 0 + 0.481 540 890 624;
  • 17) 0.481 540 890 624 × 2 = 0 + 0.963 081 781 248;
  • 18) 0.963 081 781 248 × 2 = 1 + 0.926 163 562 496;
  • 19) 0.926 163 562 496 × 2 = 1 + 0.852 327 124 992;
  • 20) 0.852 327 124 992 × 2 = 1 + 0.704 654 249 984;
  • 21) 0.704 654 249 984 × 2 = 1 + 0.409 308 499 968;
  • 22) 0.409 308 499 968 × 2 = 0 + 0.818 616 999 936;
  • 23) 0.818 616 999 936 × 2 = 1 + 0.637 233 999 872;
  • 24) 0.637 233 999 872 × 2 = 1 + 0.274 467 999 744;
  • 25) 0.274 467 999 744 × 2 = 0 + 0.548 935 999 488;
  • 26) 0.548 935 999 488 × 2 = 1 + 0.097 871 998 976;
  • 27) 0.097 871 998 976 × 2 = 0 + 0.195 743 997 952;
  • 28) 0.195 743 997 952 × 2 = 0 + 0.391 487 995 904;
  • 29) 0.391 487 995 904 × 2 = 0 + 0.782 975 991 808;
  • 30) 0.782 975 991 808 × 2 = 1 + 0.565 951 983 616;
  • 31) 0.565 951 983 616 × 2 = 1 + 0.131 903 967 232;
  • 32) 0.131 903 967 232 × 2 = 0 + 0.263 807 934 464;
  • 33) 0.263 807 934 464 × 2 = 0 + 0.527 615 868 928;
  • 34) 0.527 615 868 928 × 2 = 1 + 0.055 231 737 856;
  • 35) 0.055 231 737 856 × 2 = 0 + 0.110 463 475 712;
  • 36) 0.110 463 475 712 × 2 = 0 + 0.220 926 951 424;
  • 37) 0.220 926 951 424 × 2 = 0 + 0.441 853 902 848;
  • 38) 0.441 853 902 848 × 2 = 0 + 0.883 707 805 696;
  • 39) 0.883 707 805 696 × 2 = 1 + 0.767 415 611 392;
  • 40) 0.767 415 611 392 × 2 = 1 + 0.534 831 222 784;
  • 41) 0.534 831 222 784 × 2 = 1 + 0.069 662 445 568;
  • 42) 0.069 662 445 568 × 2 = 0 + 0.139 324 891 136;
  • 43) 0.139 324 891 136 × 2 = 0 + 0.278 649 782 272;
  • 44) 0.278 649 782 272 × 2 = 0 + 0.557 299 564 544;
  • 45) 0.557 299 564 544 × 2 = 1 + 0.114 599 129 088;
  • 46) 0.114 599 129 088 × 2 = 0 + 0.229 198 258 176;
  • 47) 0.229 198 258 176 × 2 = 0 + 0.458 396 516 352;
  • 48) 0.458 396 516 352 × 2 = 0 + 0.916 793 032 704;
  • 49) 0.916 793 032 704 × 2 = 1 + 0.833 586 065 408;
  • 50) 0.833 586 065 408 × 2 = 1 + 0.667 172 130 816;
  • 51) 0.667 172 130 816 × 2 = 1 + 0.334 344 261 632;
  • 52) 0.334 344 261 632 × 2 = 0 + 0.668 688 523 264;
  • 53) 0.668 688 523 264 × 2 = 1 + 0.337 377 046 528;
  • 54) 0.337 377 046 528 × 2 = 0 + 0.674 754 093 056;
  • 55) 0.674 754 093 056 × 2 = 1 + 0.349 508 186 112;
  • 56) 0.349 508 186 112 × 2 = 0 + 0.699 016 372 224;
  • 57) 0.699 016 372 224 × 2 = 1 + 0.398 032 744 448;
  • 58) 0.398 032 744 448 × 2 = 0 + 0.796 065 488 896;
  • 59) 0.796 065 488 896 × 2 = 1 + 0.592 130 977 792;
  • 60) 0.592 130 977 792 × 2 = 1 + 0.184 261 955 584;
  • 61) 0.184 261 955 584 × 2 = 0 + 0.368 523 911 168;
  • 62) 0.368 523 911 168 × 2 = 0 + 0.737 047 822 336;
  • 63) 0.737 047 822 336 × 2 = 1 + 0.474 095 644 672;
  • 64) 0.474 095 644 672 × 2 = 0 + 0.948 191 289 344;

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:


0.000 282 005 934(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010(2)

6. Positive number before normalization:

0.000 282 005 934(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010(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:


0.000 282 005 934(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010(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 0100 0011 1000 1000 1110 1010 1011 0010


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 0100 0011 1000 1000 1110 1010 1011 0010 =

0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010


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 0100 0011 1000 1000 1110 1010 1011 0010


Decimal number -0.000 282 005 934 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0100 0011 1000 1000 1110 1010 1011 0010

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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