-0.000 282 596 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 596(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
-0.000 282 596(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:

|-0.000 282 596| = 0.000 282 596


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: 0.000 282 596.

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) 0.000 282 596 × 2 = 0 + 0.000 565 192;
  • 2) 0.000 565 192 × 2 = 0 + 0.001 130 384;
  • 3) 0.001 130 384 × 2 = 0 + 0.002 260 768;
  • 4) 0.002 260 768 × 2 = 0 + 0.004 521 536;
  • 5) 0.004 521 536 × 2 = 0 + 0.009 043 072;
  • 6) 0.009 043 072 × 2 = 0 + 0.018 086 144;
  • 7) 0.018 086 144 × 2 = 0 + 0.036 172 288;
  • 8) 0.036 172 288 × 2 = 0 + 0.072 344 576;
  • 9) 0.072 344 576 × 2 = 0 + 0.144 689 152;
  • 10) 0.144 689 152 × 2 = 0 + 0.289 378 304;
  • 11) 0.289 378 304 × 2 = 0 + 0.578 756 608;
  • 12) 0.578 756 608 × 2 = 1 + 0.157 513 216;
  • 13) 0.157 513 216 × 2 = 0 + 0.315 026 432;
  • 14) 0.315 026 432 × 2 = 0 + 0.630 052 864;
  • 15) 0.630 052 864 × 2 = 1 + 0.260 105 728;
  • 16) 0.260 105 728 × 2 = 0 + 0.520 211 456;
  • 17) 0.520 211 456 × 2 = 1 + 0.040 422 912;
  • 18) 0.040 422 912 × 2 = 0 + 0.080 845 824;
  • 19) 0.080 845 824 × 2 = 0 + 0.161 691 648;
  • 20) 0.161 691 648 × 2 = 0 + 0.323 383 296;
  • 21) 0.323 383 296 × 2 = 0 + 0.646 766 592;
  • 22) 0.646 766 592 × 2 = 1 + 0.293 533 184;
  • 23) 0.293 533 184 × 2 = 0 + 0.587 066 368;
  • 24) 0.587 066 368 × 2 = 1 + 0.174 132 736;
  • 25) 0.174 132 736 × 2 = 0 + 0.348 265 472;
  • 26) 0.348 265 472 × 2 = 0 + 0.696 530 944;
  • 27) 0.696 530 944 × 2 = 1 + 0.393 061 888;
  • 28) 0.393 061 888 × 2 = 0 + 0.786 123 776;
  • 29) 0.786 123 776 × 2 = 1 + 0.572 247 552;
  • 30) 0.572 247 552 × 2 = 1 + 0.144 495 104;
  • 31) 0.144 495 104 × 2 = 0 + 0.288 990 208;
  • 32) 0.288 990 208 × 2 = 0 + 0.577 980 416;
  • 33) 0.577 980 416 × 2 = 1 + 0.155 960 832;
  • 34) 0.155 960 832 × 2 = 0 + 0.311 921 664;
  • 35) 0.311 921 664 × 2 = 0 + 0.623 843 328;
  • 36) 0.623 843 328 × 2 = 1 + 0.247 686 656;
  • 37) 0.247 686 656 × 2 = 0 + 0.495 373 312;
  • 38) 0.495 373 312 × 2 = 0 + 0.990 746 624;
  • 39) 0.990 746 624 × 2 = 1 + 0.981 493 248;
  • 40) 0.981 493 248 × 2 = 1 + 0.962 986 496;
  • 41) 0.962 986 496 × 2 = 1 + 0.925 972 992;
  • 42) 0.925 972 992 × 2 = 1 + 0.851 945 984;
  • 43) 0.851 945 984 × 2 = 1 + 0.703 891 968;
  • 44) 0.703 891 968 × 2 = 1 + 0.407 783 936;
  • 45) 0.407 783 936 × 2 = 0 + 0.815 567 872;
  • 46) 0.815 567 872 × 2 = 1 + 0.631 135 744;
  • 47) 0.631 135 744 × 2 = 1 + 0.262 271 488;
  • 48) 0.262 271 488 × 2 = 0 + 0.524 542 976;
  • 49) 0.524 542 976 × 2 = 1 + 0.049 085 952;
  • 50) 0.049 085 952 × 2 = 0 + 0.098 171 904;
  • 51) 0.098 171 904 × 2 = 0 + 0.196 343 808;
  • 52) 0.196 343 808 × 2 = 0 + 0.392 687 616;
  • 53) 0.392 687 616 × 2 = 0 + 0.785 375 232;
  • 54) 0.785 375 232 × 2 = 1 + 0.570 750 464;
  • 55) 0.570 750 464 × 2 = 1 + 0.141 500 928;
  • 56) 0.141 500 928 × 2 = 0 + 0.283 001 856;
  • 57) 0.283 001 856 × 2 = 0 + 0.566 003 712;
  • 58) 0.566 003 712 × 2 = 1 + 0.132 007 424;
  • 59) 0.132 007 424 × 2 = 0 + 0.264 014 848;
  • 60) 0.264 014 848 × 2 = 0 + 0.528 029 696;
  • 61) 0.528 029 696 × 2 = 1 + 0.056 059 392;
  • 62) 0.056 059 392 × 2 = 0 + 0.112 118 784;
  • 63) 0.112 118 784 × 2 = 0 + 0.224 237 568;
  • 64) 0.224 237 568 × 2 = 0 + 0.448 475 136;

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:


0.000 282 596(10) =

0.0000 0000 0001 0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000(2)

6. Positive number before normalization:

0.000 282 596(10) =

0.0000 0000 0001 0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000(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:


0.000 282 596(10) =

0.0000 0000 0001 0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000(2) =

0.0000 0000 0001 0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000(2) × 20 =

1.0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000(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 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000


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 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000 =

0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000


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 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000


Decimal number -0.000 282 596 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 0010 1100 1001 0011 1111 0110 1000 0110 0100 1000

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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