-0.000 281 864 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 864(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
-0.000 281 864(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:

|-0.000 281 864| = 0.000 281 864


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: 0.000 281 864.

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) 0.000 281 864 × 2 = 0 + 0.000 563 728;
  • 2) 0.000 563 728 × 2 = 0 + 0.001 127 456;
  • 3) 0.001 127 456 × 2 = 0 + 0.002 254 912;
  • 4) 0.002 254 912 × 2 = 0 + 0.004 509 824;
  • 5) 0.004 509 824 × 2 = 0 + 0.009 019 648;
  • 6) 0.009 019 648 × 2 = 0 + 0.018 039 296;
  • 7) 0.018 039 296 × 2 = 0 + 0.036 078 592;
  • 8) 0.036 078 592 × 2 = 0 + 0.072 157 184;
  • 9) 0.072 157 184 × 2 = 0 + 0.144 314 368;
  • 10) 0.144 314 368 × 2 = 0 + 0.288 628 736;
  • 11) 0.288 628 736 × 2 = 0 + 0.577 257 472;
  • 12) 0.577 257 472 × 2 = 1 + 0.154 514 944;
  • 13) 0.154 514 944 × 2 = 0 + 0.309 029 888;
  • 14) 0.309 029 888 × 2 = 0 + 0.618 059 776;
  • 15) 0.618 059 776 × 2 = 1 + 0.236 119 552;
  • 16) 0.236 119 552 × 2 = 0 + 0.472 239 104;
  • 17) 0.472 239 104 × 2 = 0 + 0.944 478 208;
  • 18) 0.944 478 208 × 2 = 1 + 0.888 956 416;
  • 19) 0.888 956 416 × 2 = 1 + 0.777 912 832;
  • 20) 0.777 912 832 × 2 = 1 + 0.555 825 664;
  • 21) 0.555 825 664 × 2 = 1 + 0.111 651 328;
  • 22) 0.111 651 328 × 2 = 0 + 0.223 302 656;
  • 23) 0.223 302 656 × 2 = 0 + 0.446 605 312;
  • 24) 0.446 605 312 × 2 = 0 + 0.893 210 624;
  • 25) 0.893 210 624 × 2 = 1 + 0.786 421 248;
  • 26) 0.786 421 248 × 2 = 1 + 0.572 842 496;
  • 27) 0.572 842 496 × 2 = 1 + 0.145 684 992;
  • 28) 0.145 684 992 × 2 = 0 + 0.291 369 984;
  • 29) 0.291 369 984 × 2 = 0 + 0.582 739 968;
  • 30) 0.582 739 968 × 2 = 1 + 0.165 479 936;
  • 31) 0.165 479 936 × 2 = 0 + 0.330 959 872;
  • 32) 0.330 959 872 × 2 = 0 + 0.661 919 744;
  • 33) 0.661 919 744 × 2 = 1 + 0.323 839 488;
  • 34) 0.323 839 488 × 2 = 0 + 0.647 678 976;
  • 35) 0.647 678 976 × 2 = 1 + 0.295 357 952;
  • 36) 0.295 357 952 × 2 = 0 + 0.590 715 904;
  • 37) 0.590 715 904 × 2 = 1 + 0.181 431 808;
  • 38) 0.181 431 808 × 2 = 0 + 0.362 863 616;
  • 39) 0.362 863 616 × 2 = 0 + 0.725 727 232;
  • 40) 0.725 727 232 × 2 = 1 + 0.451 454 464;
  • 41) 0.451 454 464 × 2 = 0 + 0.902 908 928;
  • 42) 0.902 908 928 × 2 = 1 + 0.805 817 856;
  • 43) 0.805 817 856 × 2 = 1 + 0.611 635 712;
  • 44) 0.611 635 712 × 2 = 1 + 0.223 271 424;
  • 45) 0.223 271 424 × 2 = 0 + 0.446 542 848;
  • 46) 0.446 542 848 × 2 = 0 + 0.893 085 696;
  • 47) 0.893 085 696 × 2 = 1 + 0.786 171 392;
  • 48) 0.786 171 392 × 2 = 1 + 0.572 342 784;
  • 49) 0.572 342 784 × 2 = 1 + 0.144 685 568;
  • 50) 0.144 685 568 × 2 = 0 + 0.289 371 136;
  • 51) 0.289 371 136 × 2 = 0 + 0.578 742 272;
  • 52) 0.578 742 272 × 2 = 1 + 0.157 484 544;
  • 53) 0.157 484 544 × 2 = 0 + 0.314 969 088;
  • 54) 0.314 969 088 × 2 = 0 + 0.629 938 176;
  • 55) 0.629 938 176 × 2 = 1 + 0.259 876 352;
  • 56) 0.259 876 352 × 2 = 0 + 0.519 752 704;
  • 57) 0.519 752 704 × 2 = 1 + 0.039 505 408;
  • 58) 0.039 505 408 × 2 = 0 + 0.079 010 816;
  • 59) 0.079 010 816 × 2 = 0 + 0.158 021 632;
  • 60) 0.158 021 632 × 2 = 0 + 0.316 043 264;
  • 61) 0.316 043 264 × 2 = 0 + 0.632 086 528;
  • 62) 0.632 086 528 × 2 = 1 + 0.264 173 056;
  • 63) 0.264 173 056 × 2 = 0 + 0.528 346 112;
  • 64) 0.528 346 112 × 2 = 1 + 0.056 692 224;

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:


0.000 281 864(10) =

0.0000 0000 0001 0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 0101(2)

6. Positive number before normalization:

0.000 281 864(10) =

0.0000 0000 0001 0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 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:


0.000 281 864(10) =

0.0000 0000 0001 0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 0101(2) =

0.0000 0000 0001 0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 0101(2) × 20 =

1.0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 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 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 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 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 0101 =

0010 0111 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 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 1000 1110 0100 1010 1001 0111 0011 1001 0010 1000 0101


Decimal number -0.000 281 864 converted to 64 bit double precision IEEE 754 binary floating point representation:

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