-0.000 282 005 997 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 997(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
-0.000 282 005 997(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:

|-0.000 282 005 997| = 0.000 282 005 997


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: 0.000 282 005 997.

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) 0.000 282 005 997 × 2 = 0 + 0.000 564 011 994;
  • 2) 0.000 564 011 994 × 2 = 0 + 0.001 128 023 988;
  • 3) 0.001 128 023 988 × 2 = 0 + 0.002 256 047 976;
  • 4) 0.002 256 047 976 × 2 = 0 + 0.004 512 095 952;
  • 5) 0.004 512 095 952 × 2 = 0 + 0.009 024 191 904;
  • 6) 0.009 024 191 904 × 2 = 0 + 0.018 048 383 808;
  • 7) 0.018 048 383 808 × 2 = 0 + 0.036 096 767 616;
  • 8) 0.036 096 767 616 × 2 = 0 + 0.072 193 535 232;
  • 9) 0.072 193 535 232 × 2 = 0 + 0.144 387 070 464;
  • 10) 0.144 387 070 464 × 2 = 0 + 0.288 774 140 928;
  • 11) 0.288 774 140 928 × 2 = 0 + 0.577 548 281 856;
  • 12) 0.577 548 281 856 × 2 = 1 + 0.155 096 563 712;
  • 13) 0.155 096 563 712 × 2 = 0 + 0.310 193 127 424;
  • 14) 0.310 193 127 424 × 2 = 0 + 0.620 386 254 848;
  • 15) 0.620 386 254 848 × 2 = 1 + 0.240 772 509 696;
  • 16) 0.240 772 509 696 × 2 = 0 + 0.481 545 019 392;
  • 17) 0.481 545 019 392 × 2 = 0 + 0.963 090 038 784;
  • 18) 0.963 090 038 784 × 2 = 1 + 0.926 180 077 568;
  • 19) 0.926 180 077 568 × 2 = 1 + 0.852 360 155 136;
  • 20) 0.852 360 155 136 × 2 = 1 + 0.704 720 310 272;
  • 21) 0.704 720 310 272 × 2 = 1 + 0.409 440 620 544;
  • 22) 0.409 440 620 544 × 2 = 0 + 0.818 881 241 088;
  • 23) 0.818 881 241 088 × 2 = 1 + 0.637 762 482 176;
  • 24) 0.637 762 482 176 × 2 = 1 + 0.275 524 964 352;
  • 25) 0.275 524 964 352 × 2 = 0 + 0.551 049 928 704;
  • 26) 0.551 049 928 704 × 2 = 1 + 0.102 099 857 408;
  • 27) 0.102 099 857 408 × 2 = 0 + 0.204 199 714 816;
  • 28) 0.204 199 714 816 × 2 = 0 + 0.408 399 429 632;
  • 29) 0.408 399 429 632 × 2 = 0 + 0.816 798 859 264;
  • 30) 0.816 798 859 264 × 2 = 1 + 0.633 597 718 528;
  • 31) 0.633 597 718 528 × 2 = 1 + 0.267 195 437 056;
  • 32) 0.267 195 437 056 × 2 = 0 + 0.534 390 874 112;
  • 33) 0.534 390 874 112 × 2 = 1 + 0.068 781 748 224;
  • 34) 0.068 781 748 224 × 2 = 0 + 0.137 563 496 448;
  • 35) 0.137 563 496 448 × 2 = 0 + 0.275 126 992 896;
  • 36) 0.275 126 992 896 × 2 = 0 + 0.550 253 985 792;
  • 37) 0.550 253 985 792 × 2 = 1 + 0.100 507 971 584;
  • 38) 0.100 507 971 584 × 2 = 0 + 0.201 015 943 168;
  • 39) 0.201 015 943 168 × 2 = 0 + 0.402 031 886 336;
  • 40) 0.402 031 886 336 × 2 = 0 + 0.804 063 772 672;
  • 41) 0.804 063 772 672 × 2 = 1 + 0.608 127 545 344;
  • 42) 0.608 127 545 344 × 2 = 1 + 0.216 255 090 688;
  • 43) 0.216 255 090 688 × 2 = 0 + 0.432 510 181 376;
  • 44) 0.432 510 181 376 × 2 = 0 + 0.865 020 362 752;
  • 45) 0.865 020 362 752 × 2 = 1 + 0.730 040 725 504;
  • 46) 0.730 040 725 504 × 2 = 1 + 0.460 081 451 008;
  • 47) 0.460 081 451 008 × 2 = 0 + 0.920 162 902 016;
  • 48) 0.920 162 902 016 × 2 = 1 + 0.840 325 804 032;
  • 49) 0.840 325 804 032 × 2 = 1 + 0.680 651 608 064;
  • 50) 0.680 651 608 064 × 2 = 1 + 0.361 303 216 128;
  • 51) 0.361 303 216 128 × 2 = 0 + 0.722 606 432 256;
  • 52) 0.722 606 432 256 × 2 = 1 + 0.445 212 864 512;
  • 53) 0.445 212 864 512 × 2 = 0 + 0.890 425 729 024;
  • 54) 0.890 425 729 024 × 2 = 1 + 0.780 851 458 048;
  • 55) 0.780 851 458 048 × 2 = 1 + 0.561 702 916 096;
  • 56) 0.561 702 916 096 × 2 = 1 + 0.123 405 832 192;
  • 57) 0.123 405 832 192 × 2 = 0 + 0.246 811 664 384;
  • 58) 0.246 811 664 384 × 2 = 0 + 0.493 623 328 768;
  • 59) 0.493 623 328 768 × 2 = 0 + 0.987 246 657 536;
  • 60) 0.987 246 657 536 × 2 = 1 + 0.974 493 315 072;
  • 61) 0.974 493 315 072 × 2 = 1 + 0.948 986 630 144;
  • 62) 0.948 986 630 144 × 2 = 1 + 0.897 973 260 288;
  • 63) 0.897 973 260 288 × 2 = 1 + 0.795 946 520 576;
  • 64) 0.795 946 520 576 × 2 = 1 + 0.591 893 041 152;

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:


0.000 282 005 997(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111(2)

6. Positive number before normalization:

0.000 282 005 997(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111(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:


0.000 282 005 997(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111(2) × 20 =

1.0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111(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 1000 1000 1100 1101 1101 0111 0001 1111


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 1000 1000 1100 1101 1101 0111 0001 1111 =

0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111


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 1000 1000 1100 1101 1101 0111 0001 1111


Decimal number -0.000 282 005 997 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1000 1000 1100 1101 1101 0111 0001 1111

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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