-0.000 282 442 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 442(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
-0.000 282 442(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:

|-0.000 282 442| = 0.000 282 442


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: 0.000 282 442.

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) 0.000 282 442 × 2 = 0 + 0.000 564 884;
  • 2) 0.000 564 884 × 2 = 0 + 0.001 129 768;
  • 3) 0.001 129 768 × 2 = 0 + 0.002 259 536;
  • 4) 0.002 259 536 × 2 = 0 + 0.004 519 072;
  • 5) 0.004 519 072 × 2 = 0 + 0.009 038 144;
  • 6) 0.009 038 144 × 2 = 0 + 0.018 076 288;
  • 7) 0.018 076 288 × 2 = 0 + 0.036 152 576;
  • 8) 0.036 152 576 × 2 = 0 + 0.072 305 152;
  • 9) 0.072 305 152 × 2 = 0 + 0.144 610 304;
  • 10) 0.144 610 304 × 2 = 0 + 0.289 220 608;
  • 11) 0.289 220 608 × 2 = 0 + 0.578 441 216;
  • 12) 0.578 441 216 × 2 = 1 + 0.156 882 432;
  • 13) 0.156 882 432 × 2 = 0 + 0.313 764 864;
  • 14) 0.313 764 864 × 2 = 0 + 0.627 529 728;
  • 15) 0.627 529 728 × 2 = 1 + 0.255 059 456;
  • 16) 0.255 059 456 × 2 = 0 + 0.510 118 912;
  • 17) 0.510 118 912 × 2 = 1 + 0.020 237 824;
  • 18) 0.020 237 824 × 2 = 0 + 0.040 475 648;
  • 19) 0.040 475 648 × 2 = 0 + 0.080 951 296;
  • 20) 0.080 951 296 × 2 = 0 + 0.161 902 592;
  • 21) 0.161 902 592 × 2 = 0 + 0.323 805 184;
  • 22) 0.323 805 184 × 2 = 0 + 0.647 610 368;
  • 23) 0.647 610 368 × 2 = 1 + 0.295 220 736;
  • 24) 0.295 220 736 × 2 = 0 + 0.590 441 472;
  • 25) 0.590 441 472 × 2 = 1 + 0.180 882 944;
  • 26) 0.180 882 944 × 2 = 0 + 0.361 765 888;
  • 27) 0.361 765 888 × 2 = 0 + 0.723 531 776;
  • 28) 0.723 531 776 × 2 = 1 + 0.447 063 552;
  • 29) 0.447 063 552 × 2 = 0 + 0.894 127 104;
  • 30) 0.894 127 104 × 2 = 1 + 0.788 254 208;
  • 31) 0.788 254 208 × 2 = 1 + 0.576 508 416;
  • 32) 0.576 508 416 × 2 = 1 + 0.153 016 832;
  • 33) 0.153 016 832 × 2 = 0 + 0.306 033 664;
  • 34) 0.306 033 664 × 2 = 0 + 0.612 067 328;
  • 35) 0.612 067 328 × 2 = 1 + 0.224 134 656;
  • 36) 0.224 134 656 × 2 = 0 + 0.448 269 312;
  • 37) 0.448 269 312 × 2 = 0 + 0.896 538 624;
  • 38) 0.896 538 624 × 2 = 1 + 0.793 077 248;
  • 39) 0.793 077 248 × 2 = 1 + 0.586 154 496;
  • 40) 0.586 154 496 × 2 = 1 + 0.172 308 992;
  • 41) 0.172 308 992 × 2 = 0 + 0.344 617 984;
  • 42) 0.344 617 984 × 2 = 0 + 0.689 235 968;
  • 43) 0.689 235 968 × 2 = 1 + 0.378 471 936;
  • 44) 0.378 471 936 × 2 = 0 + 0.756 943 872;
  • 45) 0.756 943 872 × 2 = 1 + 0.513 887 744;
  • 46) 0.513 887 744 × 2 = 1 + 0.027 775 488;
  • 47) 0.027 775 488 × 2 = 0 + 0.055 550 976;
  • 48) 0.055 550 976 × 2 = 0 + 0.111 101 952;
  • 49) 0.111 101 952 × 2 = 0 + 0.222 203 904;
  • 50) 0.222 203 904 × 2 = 0 + 0.444 407 808;
  • 51) 0.444 407 808 × 2 = 0 + 0.888 815 616;
  • 52) 0.888 815 616 × 2 = 1 + 0.777 631 232;
  • 53) 0.777 631 232 × 2 = 1 + 0.555 262 464;
  • 54) 0.555 262 464 × 2 = 1 + 0.110 524 928;
  • 55) 0.110 524 928 × 2 = 0 + 0.221 049 856;
  • 56) 0.221 049 856 × 2 = 0 + 0.442 099 712;
  • 57) 0.442 099 712 × 2 = 0 + 0.884 199 424;
  • 58) 0.884 199 424 × 2 = 1 + 0.768 398 848;
  • 59) 0.768 398 848 × 2 = 1 + 0.536 797 696;
  • 60) 0.536 797 696 × 2 = 1 + 0.073 595 392;
  • 61) 0.073 595 392 × 2 = 0 + 0.147 190 784;
  • 62) 0.147 190 784 × 2 = 0 + 0.294 381 568;
  • 63) 0.294 381 568 × 2 = 0 + 0.588 763 136;
  • 64) 0.588 763 136 × 2 = 1 + 0.177 526 272;

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:


0.000 282 442(10) =

0.0000 0000 0001 0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001(2)

6. Positive number before normalization:

0.000 282 442(10) =

0.0000 0000 0001 0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001(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:


0.000 282 442(10) =

0.0000 0000 0001 0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001(2) =

0.0000 0000 0001 0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001(2) × 20 =

1.0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001(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 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001


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 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001 =

0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001


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 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001


Decimal number -0.000 282 442 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0010 1001 0111 0010 0111 0010 1100 0001 1100 0111 0001

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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