-0.000 282 006 244 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 244(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
-0.000 282 006 244(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:

|-0.000 282 006 244| = 0.000 282 006 244


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: 0.000 282 006 244.

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) 0.000 282 006 244 × 2 = 0 + 0.000 564 012 488;
  • 2) 0.000 564 012 488 × 2 = 0 + 0.001 128 024 976;
  • 3) 0.001 128 024 976 × 2 = 0 + 0.002 256 049 952;
  • 4) 0.002 256 049 952 × 2 = 0 + 0.004 512 099 904;
  • 5) 0.004 512 099 904 × 2 = 0 + 0.009 024 199 808;
  • 6) 0.009 024 199 808 × 2 = 0 + 0.018 048 399 616;
  • 7) 0.018 048 399 616 × 2 = 0 + 0.036 096 799 232;
  • 8) 0.036 096 799 232 × 2 = 0 + 0.072 193 598 464;
  • 9) 0.072 193 598 464 × 2 = 0 + 0.144 387 196 928;
  • 10) 0.144 387 196 928 × 2 = 0 + 0.288 774 393 856;
  • 11) 0.288 774 393 856 × 2 = 0 + 0.577 548 787 712;
  • 12) 0.577 548 787 712 × 2 = 1 + 0.155 097 575 424;
  • 13) 0.155 097 575 424 × 2 = 0 + 0.310 195 150 848;
  • 14) 0.310 195 150 848 × 2 = 0 + 0.620 390 301 696;
  • 15) 0.620 390 301 696 × 2 = 1 + 0.240 780 603 392;
  • 16) 0.240 780 603 392 × 2 = 0 + 0.481 561 206 784;
  • 17) 0.481 561 206 784 × 2 = 0 + 0.963 122 413 568;
  • 18) 0.963 122 413 568 × 2 = 1 + 0.926 244 827 136;
  • 19) 0.926 244 827 136 × 2 = 1 + 0.852 489 654 272;
  • 20) 0.852 489 654 272 × 2 = 1 + 0.704 979 308 544;
  • 21) 0.704 979 308 544 × 2 = 1 + 0.409 958 617 088;
  • 22) 0.409 958 617 088 × 2 = 0 + 0.819 917 234 176;
  • 23) 0.819 917 234 176 × 2 = 1 + 0.639 834 468 352;
  • 24) 0.639 834 468 352 × 2 = 1 + 0.279 668 936 704;
  • 25) 0.279 668 936 704 × 2 = 0 + 0.559 337 873 408;
  • 26) 0.559 337 873 408 × 2 = 1 + 0.118 675 746 816;
  • 27) 0.118 675 746 816 × 2 = 0 + 0.237 351 493 632;
  • 28) 0.237 351 493 632 × 2 = 0 + 0.474 702 987 264;
  • 29) 0.474 702 987 264 × 2 = 0 + 0.949 405 974 528;
  • 30) 0.949 405 974 528 × 2 = 1 + 0.898 811 949 056;
  • 31) 0.898 811 949 056 × 2 = 1 + 0.797 623 898 112;
  • 32) 0.797 623 898 112 × 2 = 1 + 0.595 247 796 224;
  • 33) 0.595 247 796 224 × 2 = 1 + 0.190 495 592 448;
  • 34) 0.190 495 592 448 × 2 = 0 + 0.380 991 184 896;
  • 35) 0.380 991 184 896 × 2 = 0 + 0.761 982 369 792;
  • 36) 0.761 982 369 792 × 2 = 1 + 0.523 964 739 584;
  • 37) 0.523 964 739 584 × 2 = 1 + 0.047 929 479 168;
  • 38) 0.047 929 479 168 × 2 = 0 + 0.095 858 958 336;
  • 39) 0.095 858 958 336 × 2 = 0 + 0.191 717 916 672;
  • 40) 0.191 717 916 672 × 2 = 0 + 0.383 435 833 344;
  • 41) 0.383 435 833 344 × 2 = 0 + 0.766 871 666 688;
  • 42) 0.766 871 666 688 × 2 = 1 + 0.533 743 333 376;
  • 43) 0.533 743 333 376 × 2 = 1 + 0.067 486 666 752;
  • 44) 0.067 486 666 752 × 2 = 0 + 0.134 973 333 504;
  • 45) 0.134 973 333 504 × 2 = 0 + 0.269 946 667 008;
  • 46) 0.269 946 667 008 × 2 = 0 + 0.539 893 334 016;
  • 47) 0.539 893 334 016 × 2 = 1 + 0.079 786 668 032;
  • 48) 0.079 786 668 032 × 2 = 0 + 0.159 573 336 064;
  • 49) 0.159 573 336 064 × 2 = 0 + 0.319 146 672 128;
  • 50) 0.319 146 672 128 × 2 = 0 + 0.638 293 344 256;
  • 51) 0.638 293 344 256 × 2 = 1 + 0.276 586 688 512;
  • 52) 0.276 586 688 512 × 2 = 0 + 0.553 173 377 024;
  • 53) 0.553 173 377 024 × 2 = 1 + 0.106 346 754 048;
  • 54) 0.106 346 754 048 × 2 = 0 + 0.212 693 508 096;
  • 55) 0.212 693 508 096 × 2 = 0 + 0.425 387 016 192;
  • 56) 0.425 387 016 192 × 2 = 0 + 0.850 774 032 384;
  • 57) 0.850 774 032 384 × 2 = 1 + 0.701 548 064 768;
  • 58) 0.701 548 064 768 × 2 = 1 + 0.403 096 129 536;
  • 59) 0.403 096 129 536 × 2 = 0 + 0.806 192 259 072;
  • 60) 0.806 192 259 072 × 2 = 1 + 0.612 384 518 144;
  • 61) 0.612 384 518 144 × 2 = 1 + 0.224 769 036 288;
  • 62) 0.224 769 036 288 × 2 = 0 + 0.449 538 072 576;
  • 63) 0.449 538 072 576 × 2 = 0 + 0.899 076 145 152;
  • 64) 0.899 076 145 152 × 2 = 1 + 0.798 152 290 304;

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:


0.000 282 006 244(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 1001(2)

6. Positive number before normalization:

0.000 282 006 244(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 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:


0.000 282 006 244(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 1001(2) × 20 =

1.0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 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 0111 1001 1000 0110 0010 0010 1000 1101 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 0111 1001 1000 0110 0010 0010 1000 1101 1001 =

0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 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 0111 1001 1000 0110 0010 0010 1000 1101 1001


Decimal number -0.000 282 006 244 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1001 1000 0110 0010 0010 1000 1101 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