-0.000 282 608 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 608(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
-0.000 282 608(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:

|-0.000 282 608| = 0.000 282 608


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: 0.000 282 608.

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) 0.000 282 608 × 2 = 0 + 0.000 565 216;
  • 2) 0.000 565 216 × 2 = 0 + 0.001 130 432;
  • 3) 0.001 130 432 × 2 = 0 + 0.002 260 864;
  • 4) 0.002 260 864 × 2 = 0 + 0.004 521 728;
  • 5) 0.004 521 728 × 2 = 0 + 0.009 043 456;
  • 6) 0.009 043 456 × 2 = 0 + 0.018 086 912;
  • 7) 0.018 086 912 × 2 = 0 + 0.036 173 824;
  • 8) 0.036 173 824 × 2 = 0 + 0.072 347 648;
  • 9) 0.072 347 648 × 2 = 0 + 0.144 695 296;
  • 10) 0.144 695 296 × 2 = 0 + 0.289 390 592;
  • 11) 0.289 390 592 × 2 = 0 + 0.578 781 184;
  • 12) 0.578 781 184 × 2 = 1 + 0.157 562 368;
  • 13) 0.157 562 368 × 2 = 0 + 0.315 124 736;
  • 14) 0.315 124 736 × 2 = 0 + 0.630 249 472;
  • 15) 0.630 249 472 × 2 = 1 + 0.260 498 944;
  • 16) 0.260 498 944 × 2 = 0 + 0.520 997 888;
  • 17) 0.520 997 888 × 2 = 1 + 0.041 995 776;
  • 18) 0.041 995 776 × 2 = 0 + 0.083 991 552;
  • 19) 0.083 991 552 × 2 = 0 + 0.167 983 104;
  • 20) 0.167 983 104 × 2 = 0 + 0.335 966 208;
  • 21) 0.335 966 208 × 2 = 0 + 0.671 932 416;
  • 22) 0.671 932 416 × 2 = 1 + 0.343 864 832;
  • 23) 0.343 864 832 × 2 = 0 + 0.687 729 664;
  • 24) 0.687 729 664 × 2 = 1 + 0.375 459 328;
  • 25) 0.375 459 328 × 2 = 0 + 0.750 918 656;
  • 26) 0.750 918 656 × 2 = 1 + 0.501 837 312;
  • 27) 0.501 837 312 × 2 = 1 + 0.003 674 624;
  • 28) 0.003 674 624 × 2 = 0 + 0.007 349 248;
  • 29) 0.007 349 248 × 2 = 0 + 0.014 698 496;
  • 30) 0.014 698 496 × 2 = 0 + 0.029 396 992;
  • 31) 0.029 396 992 × 2 = 0 + 0.058 793 984;
  • 32) 0.058 793 984 × 2 = 0 + 0.117 587 968;
  • 33) 0.117 587 968 × 2 = 0 + 0.235 175 936;
  • 34) 0.235 175 936 × 2 = 0 + 0.470 351 872;
  • 35) 0.470 351 872 × 2 = 0 + 0.940 703 744;
  • 36) 0.940 703 744 × 2 = 1 + 0.881 407 488;
  • 37) 0.881 407 488 × 2 = 1 + 0.762 814 976;
  • 38) 0.762 814 976 × 2 = 1 + 0.525 629 952;
  • 39) 0.525 629 952 × 2 = 1 + 0.051 259 904;
  • 40) 0.051 259 904 × 2 = 0 + 0.102 519 808;
  • 41) 0.102 519 808 × 2 = 0 + 0.205 039 616;
  • 42) 0.205 039 616 × 2 = 0 + 0.410 079 232;
  • 43) 0.410 079 232 × 2 = 0 + 0.820 158 464;
  • 44) 0.820 158 464 × 2 = 1 + 0.640 316 928;
  • 45) 0.640 316 928 × 2 = 1 + 0.280 633 856;
  • 46) 0.280 633 856 × 2 = 0 + 0.561 267 712;
  • 47) 0.561 267 712 × 2 = 1 + 0.122 535 424;
  • 48) 0.122 535 424 × 2 = 0 + 0.245 070 848;
  • 49) 0.245 070 848 × 2 = 0 + 0.490 141 696;
  • 50) 0.490 141 696 × 2 = 0 + 0.980 283 392;
  • 51) 0.980 283 392 × 2 = 1 + 0.960 566 784;
  • 52) 0.960 566 784 × 2 = 1 + 0.921 133 568;
  • 53) 0.921 133 568 × 2 = 1 + 0.842 267 136;
  • 54) 0.842 267 136 × 2 = 1 + 0.684 534 272;
  • 55) 0.684 534 272 × 2 = 1 + 0.369 068 544;
  • 56) 0.369 068 544 × 2 = 0 + 0.738 137 088;
  • 57) 0.738 137 088 × 2 = 1 + 0.476 274 176;
  • 58) 0.476 274 176 × 2 = 0 + 0.952 548 352;
  • 59) 0.952 548 352 × 2 = 1 + 0.905 096 704;
  • 60) 0.905 096 704 × 2 = 1 + 0.810 193 408;
  • 61) 0.810 193 408 × 2 = 1 + 0.620 386 816;
  • 62) 0.620 386 816 × 2 = 1 + 0.240 773 632;
  • 63) 0.240 773 632 × 2 = 0 + 0.481 547 264;
  • 64) 0.481 547 264 × 2 = 0 + 0.963 094 528;

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:


0.000 282 608(10) =

0.0000 0000 0001 0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100(2)

6. Positive number before normalization:

0.000 282 608(10) =

0.0000 0000 0001 0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100(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:


0.000 282 608(10) =

0.0000 0000 0001 0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100(2) =

0.0000 0000 0001 0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100(2) × 20 =

1.0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100(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 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100


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 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100 =

0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100


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 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100


Decimal number -0.000 282 608 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 0110 0000 0001 1110 0001 1010 0011 1110 1011 1100

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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