-0.000 282 538 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 538(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
-0.000 282 538(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:

|-0.000 282 538| = 0.000 282 538


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: 0.000 282 538.

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) 0.000 282 538 × 2 = 0 + 0.000 565 076;
  • 2) 0.000 565 076 × 2 = 0 + 0.001 130 152;
  • 3) 0.001 130 152 × 2 = 0 + 0.002 260 304;
  • 4) 0.002 260 304 × 2 = 0 + 0.004 520 608;
  • 5) 0.004 520 608 × 2 = 0 + 0.009 041 216;
  • 6) 0.009 041 216 × 2 = 0 + 0.018 082 432;
  • 7) 0.018 082 432 × 2 = 0 + 0.036 164 864;
  • 8) 0.036 164 864 × 2 = 0 + 0.072 329 728;
  • 9) 0.072 329 728 × 2 = 0 + 0.144 659 456;
  • 10) 0.144 659 456 × 2 = 0 + 0.289 318 912;
  • 11) 0.289 318 912 × 2 = 0 + 0.578 637 824;
  • 12) 0.578 637 824 × 2 = 1 + 0.157 275 648;
  • 13) 0.157 275 648 × 2 = 0 + 0.314 551 296;
  • 14) 0.314 551 296 × 2 = 0 + 0.629 102 592;
  • 15) 0.629 102 592 × 2 = 1 + 0.258 205 184;
  • 16) 0.258 205 184 × 2 = 0 + 0.516 410 368;
  • 17) 0.516 410 368 × 2 = 1 + 0.032 820 736;
  • 18) 0.032 820 736 × 2 = 0 + 0.065 641 472;
  • 19) 0.065 641 472 × 2 = 0 + 0.131 282 944;
  • 20) 0.131 282 944 × 2 = 0 + 0.262 565 888;
  • 21) 0.262 565 888 × 2 = 0 + 0.525 131 776;
  • 22) 0.525 131 776 × 2 = 1 + 0.050 263 552;
  • 23) 0.050 263 552 × 2 = 0 + 0.100 527 104;
  • 24) 0.100 527 104 × 2 = 0 + 0.201 054 208;
  • 25) 0.201 054 208 × 2 = 0 + 0.402 108 416;
  • 26) 0.402 108 416 × 2 = 0 + 0.804 216 832;
  • 27) 0.804 216 832 × 2 = 1 + 0.608 433 664;
  • 28) 0.608 433 664 × 2 = 1 + 0.216 867 328;
  • 29) 0.216 867 328 × 2 = 0 + 0.433 734 656;
  • 30) 0.433 734 656 × 2 = 0 + 0.867 469 312;
  • 31) 0.867 469 312 × 2 = 1 + 0.734 938 624;
  • 32) 0.734 938 624 × 2 = 1 + 0.469 877 248;
  • 33) 0.469 877 248 × 2 = 0 + 0.939 754 496;
  • 34) 0.939 754 496 × 2 = 1 + 0.879 508 992;
  • 35) 0.879 508 992 × 2 = 1 + 0.759 017 984;
  • 36) 0.759 017 984 × 2 = 1 + 0.518 035 968;
  • 37) 0.518 035 968 × 2 = 1 + 0.036 071 936;
  • 38) 0.036 071 936 × 2 = 0 + 0.072 143 872;
  • 39) 0.072 143 872 × 2 = 0 + 0.144 287 744;
  • 40) 0.144 287 744 × 2 = 0 + 0.288 575 488;
  • 41) 0.288 575 488 × 2 = 0 + 0.577 150 976;
  • 42) 0.577 150 976 × 2 = 1 + 0.154 301 952;
  • 43) 0.154 301 952 × 2 = 0 + 0.308 603 904;
  • 44) 0.308 603 904 × 2 = 0 + 0.617 207 808;
  • 45) 0.617 207 808 × 2 = 1 + 0.234 415 616;
  • 46) 0.234 415 616 × 2 = 0 + 0.468 831 232;
  • 47) 0.468 831 232 × 2 = 0 + 0.937 662 464;
  • 48) 0.937 662 464 × 2 = 1 + 0.875 324 928;
  • 49) 0.875 324 928 × 2 = 1 + 0.750 649 856;
  • 50) 0.750 649 856 × 2 = 1 + 0.501 299 712;
  • 51) 0.501 299 712 × 2 = 1 + 0.002 599 424;
  • 52) 0.002 599 424 × 2 = 0 + 0.005 198 848;
  • 53) 0.005 198 848 × 2 = 0 + 0.010 397 696;
  • 54) 0.010 397 696 × 2 = 0 + 0.020 795 392;
  • 55) 0.020 795 392 × 2 = 0 + 0.041 590 784;
  • 56) 0.041 590 784 × 2 = 0 + 0.083 181 568;
  • 57) 0.083 181 568 × 2 = 0 + 0.166 363 136;
  • 58) 0.166 363 136 × 2 = 0 + 0.332 726 272;
  • 59) 0.332 726 272 × 2 = 0 + 0.665 452 544;
  • 60) 0.665 452 544 × 2 = 1 + 0.330 905 088;
  • 61) 0.330 905 088 × 2 = 0 + 0.661 810 176;
  • 62) 0.661 810 176 × 2 = 1 + 0.323 620 352;
  • 63) 0.323 620 352 × 2 = 0 + 0.647 240 704;
  • 64) 0.647 240 704 × 2 = 1 + 0.294 481 408;

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:


0.000 282 538(10) =

0.0000 0000 0001 0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101(2)

6. Positive number before normalization:

0.000 282 538(10) =

0.0000 0000 0001 0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101(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:


0.000 282 538(10) =

0.0000 0000 0001 0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101(2) =

0.0000 0000 0001 0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101(2) × 20 =

1.0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101(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 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101


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 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101 =

0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101


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 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101


Decimal number -0.000 282 538 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0100 0011 0011 0111 1000 0100 1001 1110 0000 0001 0101

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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