-0.000 282 006 347 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 347(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
-0.000 282 006 347(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:

|-0.000 282 006 347| = 0.000 282 006 347


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: 0.000 282 006 347.

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) 0.000 282 006 347 × 2 = 0 + 0.000 564 012 694;
  • 2) 0.000 564 012 694 × 2 = 0 + 0.001 128 025 388;
  • 3) 0.001 128 025 388 × 2 = 0 + 0.002 256 050 776;
  • 4) 0.002 256 050 776 × 2 = 0 + 0.004 512 101 552;
  • 5) 0.004 512 101 552 × 2 = 0 + 0.009 024 203 104;
  • 6) 0.009 024 203 104 × 2 = 0 + 0.018 048 406 208;
  • 7) 0.018 048 406 208 × 2 = 0 + 0.036 096 812 416;
  • 8) 0.036 096 812 416 × 2 = 0 + 0.072 193 624 832;
  • 9) 0.072 193 624 832 × 2 = 0 + 0.144 387 249 664;
  • 10) 0.144 387 249 664 × 2 = 0 + 0.288 774 499 328;
  • 11) 0.288 774 499 328 × 2 = 0 + 0.577 548 998 656;
  • 12) 0.577 548 998 656 × 2 = 1 + 0.155 097 997 312;
  • 13) 0.155 097 997 312 × 2 = 0 + 0.310 195 994 624;
  • 14) 0.310 195 994 624 × 2 = 0 + 0.620 391 989 248;
  • 15) 0.620 391 989 248 × 2 = 1 + 0.240 783 978 496;
  • 16) 0.240 783 978 496 × 2 = 0 + 0.481 567 956 992;
  • 17) 0.481 567 956 992 × 2 = 0 + 0.963 135 913 984;
  • 18) 0.963 135 913 984 × 2 = 1 + 0.926 271 827 968;
  • 19) 0.926 271 827 968 × 2 = 1 + 0.852 543 655 936;
  • 20) 0.852 543 655 936 × 2 = 1 + 0.705 087 311 872;
  • 21) 0.705 087 311 872 × 2 = 1 + 0.410 174 623 744;
  • 22) 0.410 174 623 744 × 2 = 0 + 0.820 349 247 488;
  • 23) 0.820 349 247 488 × 2 = 1 + 0.640 698 494 976;
  • 24) 0.640 698 494 976 × 2 = 1 + 0.281 396 989 952;
  • 25) 0.281 396 989 952 × 2 = 0 + 0.562 793 979 904;
  • 26) 0.562 793 979 904 × 2 = 1 + 0.125 587 959 808;
  • 27) 0.125 587 959 808 × 2 = 0 + 0.251 175 919 616;
  • 28) 0.251 175 919 616 × 2 = 0 + 0.502 351 839 232;
  • 29) 0.502 351 839 232 × 2 = 1 + 0.004 703 678 464;
  • 30) 0.004 703 678 464 × 2 = 0 + 0.009 407 356 928;
  • 31) 0.009 407 356 928 × 2 = 0 + 0.018 814 713 856;
  • 32) 0.018 814 713 856 × 2 = 0 + 0.037 629 427 712;
  • 33) 0.037 629 427 712 × 2 = 0 + 0.075 258 855 424;
  • 34) 0.075 258 855 424 × 2 = 0 + 0.150 517 710 848;
  • 35) 0.150 517 710 848 × 2 = 0 + 0.301 035 421 696;
  • 36) 0.301 035 421 696 × 2 = 0 + 0.602 070 843 392;
  • 37) 0.602 070 843 392 × 2 = 1 + 0.204 141 686 784;
  • 38) 0.204 141 686 784 × 2 = 0 + 0.408 283 373 568;
  • 39) 0.408 283 373 568 × 2 = 0 + 0.816 566 747 136;
  • 40) 0.816 566 747 136 × 2 = 1 + 0.633 133 494 272;
  • 41) 0.633 133 494 272 × 2 = 1 + 0.266 266 988 544;
  • 42) 0.266 266 988 544 × 2 = 0 + 0.532 533 977 088;
  • 43) 0.532 533 977 088 × 2 = 1 + 0.065 067 954 176;
  • 44) 0.065 067 954 176 × 2 = 0 + 0.130 135 908 352;
  • 45) 0.130 135 908 352 × 2 = 0 + 0.260 271 816 704;
  • 46) 0.260 271 816 704 × 2 = 0 + 0.520 543 633 408;
  • 47) 0.520 543 633 408 × 2 = 1 + 0.041 087 266 816;
  • 48) 0.041 087 266 816 × 2 = 0 + 0.082 174 533 632;
  • 49) 0.082 174 533 632 × 2 = 0 + 0.164 349 067 264;
  • 50) 0.164 349 067 264 × 2 = 0 + 0.328 698 134 528;
  • 51) 0.328 698 134 528 × 2 = 0 + 0.657 396 269 056;
  • 52) 0.657 396 269 056 × 2 = 1 + 0.314 792 538 112;
  • 53) 0.314 792 538 112 × 2 = 0 + 0.629 585 076 224;
  • 54) 0.629 585 076 224 × 2 = 1 + 0.259 170 152 448;
  • 55) 0.259 170 152 448 × 2 = 0 + 0.518 340 304 896;
  • 56) 0.518 340 304 896 × 2 = 1 + 0.036 680 609 792;
  • 57) 0.036 680 609 792 × 2 = 0 + 0.073 361 219 584;
  • 58) 0.073 361 219 584 × 2 = 0 + 0.146 722 439 168;
  • 59) 0.146 722 439 168 × 2 = 0 + 0.293 444 878 336;
  • 60) 0.293 444 878 336 × 2 = 0 + 0.586 889 756 672;
  • 61) 0.586 889 756 672 × 2 = 1 + 0.173 779 513 344;
  • 62) 0.173 779 513 344 × 2 = 0 + 0.347 559 026 688;
  • 63) 0.347 559 026 688 × 2 = 0 + 0.695 118 053 376;
  • 64) 0.695 118 053 376 × 2 = 1 + 0.390 236 106 752;

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:


0.000 282 006 347(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001(2)

6. Positive number before normalization:

0.000 282 006 347(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001(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:


0.000 282 006 347(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001(2) × 20 =

1.0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001(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 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001


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 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001 =

0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001


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 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001


Decimal number -0.000 282 006 347 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0000 1001 1010 0010 0001 0101 0000 1001

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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