-0.000 281 742 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 742(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
-0.000 281 742(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:

|-0.000 281 742| = 0.000 281 742


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: 0.000 281 742.

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) 0.000 281 742 × 2 = 0 + 0.000 563 484;
  • 2) 0.000 563 484 × 2 = 0 + 0.001 126 968;
  • 3) 0.001 126 968 × 2 = 0 + 0.002 253 936;
  • 4) 0.002 253 936 × 2 = 0 + 0.004 507 872;
  • 5) 0.004 507 872 × 2 = 0 + 0.009 015 744;
  • 6) 0.009 015 744 × 2 = 0 + 0.018 031 488;
  • 7) 0.018 031 488 × 2 = 0 + 0.036 062 976;
  • 8) 0.036 062 976 × 2 = 0 + 0.072 125 952;
  • 9) 0.072 125 952 × 2 = 0 + 0.144 251 904;
  • 10) 0.144 251 904 × 2 = 0 + 0.288 503 808;
  • 11) 0.288 503 808 × 2 = 0 + 0.577 007 616;
  • 12) 0.577 007 616 × 2 = 1 + 0.154 015 232;
  • 13) 0.154 015 232 × 2 = 0 + 0.308 030 464;
  • 14) 0.308 030 464 × 2 = 0 + 0.616 060 928;
  • 15) 0.616 060 928 × 2 = 1 + 0.232 121 856;
  • 16) 0.232 121 856 × 2 = 0 + 0.464 243 712;
  • 17) 0.464 243 712 × 2 = 0 + 0.928 487 424;
  • 18) 0.928 487 424 × 2 = 1 + 0.856 974 848;
  • 19) 0.856 974 848 × 2 = 1 + 0.713 949 696;
  • 20) 0.713 949 696 × 2 = 1 + 0.427 899 392;
  • 21) 0.427 899 392 × 2 = 0 + 0.855 798 784;
  • 22) 0.855 798 784 × 2 = 1 + 0.711 597 568;
  • 23) 0.711 597 568 × 2 = 1 + 0.423 195 136;
  • 24) 0.423 195 136 × 2 = 0 + 0.846 390 272;
  • 25) 0.846 390 272 × 2 = 1 + 0.692 780 544;
  • 26) 0.692 780 544 × 2 = 1 + 0.385 561 088;
  • 27) 0.385 561 088 × 2 = 0 + 0.771 122 176;
  • 28) 0.771 122 176 × 2 = 1 + 0.542 244 352;
  • 29) 0.542 244 352 × 2 = 1 + 0.084 488 704;
  • 30) 0.084 488 704 × 2 = 0 + 0.168 977 408;
  • 31) 0.168 977 408 × 2 = 0 + 0.337 954 816;
  • 32) 0.337 954 816 × 2 = 0 + 0.675 909 632;
  • 33) 0.675 909 632 × 2 = 1 + 0.351 819 264;
  • 34) 0.351 819 264 × 2 = 0 + 0.703 638 528;
  • 35) 0.703 638 528 × 2 = 1 + 0.407 277 056;
  • 36) 0.407 277 056 × 2 = 0 + 0.814 554 112;
  • 37) 0.814 554 112 × 2 = 1 + 0.629 108 224;
  • 38) 0.629 108 224 × 2 = 1 + 0.258 216 448;
  • 39) 0.258 216 448 × 2 = 0 + 0.516 432 896;
  • 40) 0.516 432 896 × 2 = 1 + 0.032 865 792;
  • 41) 0.032 865 792 × 2 = 0 + 0.065 731 584;
  • 42) 0.065 731 584 × 2 = 0 + 0.131 463 168;
  • 43) 0.131 463 168 × 2 = 0 + 0.262 926 336;
  • 44) 0.262 926 336 × 2 = 0 + 0.525 852 672;
  • 45) 0.525 852 672 × 2 = 1 + 0.051 705 344;
  • 46) 0.051 705 344 × 2 = 0 + 0.103 410 688;
  • 47) 0.103 410 688 × 2 = 0 + 0.206 821 376;
  • 48) 0.206 821 376 × 2 = 0 + 0.413 642 752;
  • 49) 0.413 642 752 × 2 = 0 + 0.827 285 504;
  • 50) 0.827 285 504 × 2 = 1 + 0.654 571 008;
  • 51) 0.654 571 008 × 2 = 1 + 0.309 142 016;
  • 52) 0.309 142 016 × 2 = 0 + 0.618 284 032;
  • 53) 0.618 284 032 × 2 = 1 + 0.236 568 064;
  • 54) 0.236 568 064 × 2 = 0 + 0.473 136 128;
  • 55) 0.473 136 128 × 2 = 0 + 0.946 272 256;
  • 56) 0.946 272 256 × 2 = 1 + 0.892 544 512;
  • 57) 0.892 544 512 × 2 = 1 + 0.785 089 024;
  • 58) 0.785 089 024 × 2 = 1 + 0.570 178 048;
  • 59) 0.570 178 048 × 2 = 1 + 0.140 356 096;
  • 60) 0.140 356 096 × 2 = 0 + 0.280 712 192;
  • 61) 0.280 712 192 × 2 = 0 + 0.561 424 384;
  • 62) 0.561 424 384 × 2 = 1 + 0.122 848 768;
  • 63) 0.122 848 768 × 2 = 0 + 0.245 697 536;
  • 64) 0.245 697 536 × 2 = 0 + 0.491 395 072;

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:


0.000 281 742(10) =

0.0000 0000 0001 0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100(2)

6. Positive number before normalization:

0.000 281 742(10) =

0.0000 0000 0001 0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100(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:


0.000 281 742(10) =

0.0000 0000 0001 0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100(2) =

0.0000 0000 0001 0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100(2) × 20 =

1.0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100(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 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100


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 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100 =

0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100


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 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100


Decimal number -0.000 281 742 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0110 1101 1000 1010 1101 0000 1000 0110 1001 1110 0100

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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