-0.000 281 884 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 884(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
-0.000 281 884(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:

|-0.000 281 884| = 0.000 281 884


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: 0.000 281 884.

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) 0.000 281 884 × 2 = 0 + 0.000 563 768;
  • 2) 0.000 563 768 × 2 = 0 + 0.001 127 536;
  • 3) 0.001 127 536 × 2 = 0 + 0.002 255 072;
  • 4) 0.002 255 072 × 2 = 0 + 0.004 510 144;
  • 5) 0.004 510 144 × 2 = 0 + 0.009 020 288;
  • 6) 0.009 020 288 × 2 = 0 + 0.018 040 576;
  • 7) 0.018 040 576 × 2 = 0 + 0.036 081 152;
  • 8) 0.036 081 152 × 2 = 0 + 0.072 162 304;
  • 9) 0.072 162 304 × 2 = 0 + 0.144 324 608;
  • 10) 0.144 324 608 × 2 = 0 + 0.288 649 216;
  • 11) 0.288 649 216 × 2 = 0 + 0.577 298 432;
  • 12) 0.577 298 432 × 2 = 1 + 0.154 596 864;
  • 13) 0.154 596 864 × 2 = 0 + 0.309 193 728;
  • 14) 0.309 193 728 × 2 = 0 + 0.618 387 456;
  • 15) 0.618 387 456 × 2 = 1 + 0.236 774 912;
  • 16) 0.236 774 912 × 2 = 0 + 0.473 549 824;
  • 17) 0.473 549 824 × 2 = 0 + 0.947 099 648;
  • 18) 0.947 099 648 × 2 = 1 + 0.894 199 296;
  • 19) 0.894 199 296 × 2 = 1 + 0.788 398 592;
  • 20) 0.788 398 592 × 2 = 1 + 0.576 797 184;
  • 21) 0.576 797 184 × 2 = 1 + 0.153 594 368;
  • 22) 0.153 594 368 × 2 = 0 + 0.307 188 736;
  • 23) 0.307 188 736 × 2 = 0 + 0.614 377 472;
  • 24) 0.614 377 472 × 2 = 1 + 0.228 754 944;
  • 25) 0.228 754 944 × 2 = 0 + 0.457 509 888;
  • 26) 0.457 509 888 × 2 = 0 + 0.915 019 776;
  • 27) 0.915 019 776 × 2 = 1 + 0.830 039 552;
  • 28) 0.830 039 552 × 2 = 1 + 0.660 079 104;
  • 29) 0.660 079 104 × 2 = 1 + 0.320 158 208;
  • 30) 0.320 158 208 × 2 = 0 + 0.640 316 416;
  • 31) 0.640 316 416 × 2 = 1 + 0.280 632 832;
  • 32) 0.280 632 832 × 2 = 0 + 0.561 265 664;
  • 33) 0.561 265 664 × 2 = 1 + 0.122 531 328;
  • 34) 0.122 531 328 × 2 = 0 + 0.245 062 656;
  • 35) 0.245 062 656 × 2 = 0 + 0.490 125 312;
  • 36) 0.490 125 312 × 2 = 0 + 0.980 250 624;
  • 37) 0.980 250 624 × 2 = 1 + 0.960 501 248;
  • 38) 0.960 501 248 × 2 = 1 + 0.921 002 496;
  • 39) 0.921 002 496 × 2 = 1 + 0.842 004 992;
  • 40) 0.842 004 992 × 2 = 1 + 0.684 009 984;
  • 41) 0.684 009 984 × 2 = 1 + 0.368 019 968;
  • 42) 0.368 019 968 × 2 = 0 + 0.736 039 936;
  • 43) 0.736 039 936 × 2 = 1 + 0.472 079 872;
  • 44) 0.472 079 872 × 2 = 0 + 0.944 159 744;
  • 45) 0.944 159 744 × 2 = 1 + 0.888 319 488;
  • 46) 0.888 319 488 × 2 = 1 + 0.776 638 976;
  • 47) 0.776 638 976 × 2 = 1 + 0.553 277 952;
  • 48) 0.553 277 952 × 2 = 1 + 0.106 555 904;
  • 49) 0.106 555 904 × 2 = 0 + 0.213 111 808;
  • 50) 0.213 111 808 × 2 = 0 + 0.426 223 616;
  • 51) 0.426 223 616 × 2 = 0 + 0.852 447 232;
  • 52) 0.852 447 232 × 2 = 1 + 0.704 894 464;
  • 53) 0.704 894 464 × 2 = 1 + 0.409 788 928;
  • 54) 0.409 788 928 × 2 = 0 + 0.819 577 856;
  • 55) 0.819 577 856 × 2 = 1 + 0.639 155 712;
  • 56) 0.639 155 712 × 2 = 1 + 0.278 311 424;
  • 57) 0.278 311 424 × 2 = 0 + 0.556 622 848;
  • 58) 0.556 622 848 × 2 = 1 + 0.113 245 696;
  • 59) 0.113 245 696 × 2 = 0 + 0.226 491 392;
  • 60) 0.226 491 392 × 2 = 0 + 0.452 982 784;
  • 61) 0.452 982 784 × 2 = 0 + 0.905 965 568;
  • 62) 0.905 965 568 × 2 = 1 + 0.811 931 136;
  • 63) 0.811 931 136 × 2 = 1 + 0.623 862 272;
  • 64) 0.623 862 272 × 2 = 1 + 0.247 724 544;

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:


0.000 281 884(10) =

0.0000 0000 0001 0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111(2)

6. Positive number before normalization:

0.000 281 884(10) =

0.0000 0000 0001 0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111(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:


0.000 281 884(10) =

0.0000 0000 0001 0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111(2) =

0.0000 0000 0001 0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111(2) × 20 =

1.0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111(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 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111


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 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111 =

0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111


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 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111


Decimal number -0.000 281 884 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1001 0011 1010 1000 1111 1010 1111 0001 1011 0100 0111

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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