-0.000 282 179 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 179(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
-0.000 282 179(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:

|-0.000 282 179| = 0.000 282 179


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: 0.000 282 179.

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) 0.000 282 179 × 2 = 0 + 0.000 564 358;
  • 2) 0.000 564 358 × 2 = 0 + 0.001 128 716;
  • 3) 0.001 128 716 × 2 = 0 + 0.002 257 432;
  • 4) 0.002 257 432 × 2 = 0 + 0.004 514 864;
  • 5) 0.004 514 864 × 2 = 0 + 0.009 029 728;
  • 6) 0.009 029 728 × 2 = 0 + 0.018 059 456;
  • 7) 0.018 059 456 × 2 = 0 + 0.036 118 912;
  • 8) 0.036 118 912 × 2 = 0 + 0.072 237 824;
  • 9) 0.072 237 824 × 2 = 0 + 0.144 475 648;
  • 10) 0.144 475 648 × 2 = 0 + 0.288 951 296;
  • 11) 0.288 951 296 × 2 = 0 + 0.577 902 592;
  • 12) 0.577 902 592 × 2 = 1 + 0.155 805 184;
  • 13) 0.155 805 184 × 2 = 0 + 0.311 610 368;
  • 14) 0.311 610 368 × 2 = 0 + 0.623 220 736;
  • 15) 0.623 220 736 × 2 = 1 + 0.246 441 472;
  • 16) 0.246 441 472 × 2 = 0 + 0.492 882 944;
  • 17) 0.492 882 944 × 2 = 0 + 0.985 765 888;
  • 18) 0.985 765 888 × 2 = 1 + 0.971 531 776;
  • 19) 0.971 531 776 × 2 = 1 + 0.943 063 552;
  • 20) 0.943 063 552 × 2 = 1 + 0.886 127 104;
  • 21) 0.886 127 104 × 2 = 1 + 0.772 254 208;
  • 22) 0.772 254 208 × 2 = 1 + 0.544 508 416;
  • 23) 0.544 508 416 × 2 = 1 + 0.089 016 832;
  • 24) 0.089 016 832 × 2 = 0 + 0.178 033 664;
  • 25) 0.178 033 664 × 2 = 0 + 0.356 067 328;
  • 26) 0.356 067 328 × 2 = 0 + 0.712 134 656;
  • 27) 0.712 134 656 × 2 = 1 + 0.424 269 312;
  • 28) 0.424 269 312 × 2 = 0 + 0.848 538 624;
  • 29) 0.848 538 624 × 2 = 1 + 0.697 077 248;
  • 30) 0.697 077 248 × 2 = 1 + 0.394 154 496;
  • 31) 0.394 154 496 × 2 = 0 + 0.788 308 992;
  • 32) 0.788 308 992 × 2 = 1 + 0.576 617 984;
  • 33) 0.576 617 984 × 2 = 1 + 0.153 235 968;
  • 34) 0.153 235 968 × 2 = 0 + 0.306 471 936;
  • 35) 0.306 471 936 × 2 = 0 + 0.612 943 872;
  • 36) 0.612 943 872 × 2 = 1 + 0.225 887 744;
  • 37) 0.225 887 744 × 2 = 0 + 0.451 775 488;
  • 38) 0.451 775 488 × 2 = 0 + 0.903 550 976;
  • 39) 0.903 550 976 × 2 = 1 + 0.807 101 952;
  • 40) 0.807 101 952 × 2 = 1 + 0.614 203 904;
  • 41) 0.614 203 904 × 2 = 1 + 0.228 407 808;
  • 42) 0.228 407 808 × 2 = 0 + 0.456 815 616;
  • 43) 0.456 815 616 × 2 = 0 + 0.913 631 232;
  • 44) 0.913 631 232 × 2 = 1 + 0.827 262 464;
  • 45) 0.827 262 464 × 2 = 1 + 0.654 524 928;
  • 46) 0.654 524 928 × 2 = 1 + 0.309 049 856;
  • 47) 0.309 049 856 × 2 = 0 + 0.618 099 712;
  • 48) 0.618 099 712 × 2 = 1 + 0.236 199 424;
  • 49) 0.236 199 424 × 2 = 0 + 0.472 398 848;
  • 50) 0.472 398 848 × 2 = 0 + 0.944 797 696;
  • 51) 0.944 797 696 × 2 = 1 + 0.889 595 392;
  • 52) 0.889 595 392 × 2 = 1 + 0.779 190 784;
  • 53) 0.779 190 784 × 2 = 1 + 0.558 381 568;
  • 54) 0.558 381 568 × 2 = 1 + 0.116 763 136;
  • 55) 0.116 763 136 × 2 = 0 + 0.233 526 272;
  • 56) 0.233 526 272 × 2 = 0 + 0.467 052 544;
  • 57) 0.467 052 544 × 2 = 0 + 0.934 105 088;
  • 58) 0.934 105 088 × 2 = 1 + 0.868 210 176;
  • 59) 0.868 210 176 × 2 = 1 + 0.736 420 352;
  • 60) 0.736 420 352 × 2 = 1 + 0.472 840 704;
  • 61) 0.472 840 704 × 2 = 0 + 0.945 681 408;
  • 62) 0.945 681 408 × 2 = 1 + 0.891 362 816;
  • 63) 0.891 362 816 × 2 = 1 + 0.782 725 632;
  • 64) 0.782 725 632 × 2 = 1 + 0.565 451 264;

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:


0.000 282 179(10) =

0.0000 0000 0001 0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 0111(2)

6. Positive number before normalization:

0.000 282 179(10) =

0.0000 0000 0001 0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 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:


0.000 282 179(10) =

0.0000 0000 0001 0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 0111(2) =

0.0000 0000 0001 0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 0111(2) × 20 =

1.0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 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 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 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 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 0111 =

0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 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 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 0111


Decimal number -0.000 282 179 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1110 0010 1101 1001 0011 1001 1101 0011 1100 0111 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