0.000 000 050 248 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 050 248(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
0.000 000 050 248(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 050 248.

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) 0.000 000 050 248 × 2 = 0 + 0.000 000 100 496;
  • 2) 0.000 000 100 496 × 2 = 0 + 0.000 000 200 992;
  • 3) 0.000 000 200 992 × 2 = 0 + 0.000 000 401 984;
  • 4) 0.000 000 401 984 × 2 = 0 + 0.000 000 803 968;
  • 5) 0.000 000 803 968 × 2 = 0 + 0.000 001 607 936;
  • 6) 0.000 001 607 936 × 2 = 0 + 0.000 003 215 872;
  • 7) 0.000 003 215 872 × 2 = 0 + 0.000 006 431 744;
  • 8) 0.000 006 431 744 × 2 = 0 + 0.000 012 863 488;
  • 9) 0.000 012 863 488 × 2 = 0 + 0.000 025 726 976;
  • 10) 0.000 025 726 976 × 2 = 0 + 0.000 051 453 952;
  • 11) 0.000 051 453 952 × 2 = 0 + 0.000 102 907 904;
  • 12) 0.000 102 907 904 × 2 = 0 + 0.000 205 815 808;
  • 13) 0.000 205 815 808 × 2 = 0 + 0.000 411 631 616;
  • 14) 0.000 411 631 616 × 2 = 0 + 0.000 823 263 232;
  • 15) 0.000 823 263 232 × 2 = 0 + 0.001 646 526 464;
  • 16) 0.001 646 526 464 × 2 = 0 + 0.003 293 052 928;
  • 17) 0.003 293 052 928 × 2 = 0 + 0.006 586 105 856;
  • 18) 0.006 586 105 856 × 2 = 0 + 0.013 172 211 712;
  • 19) 0.013 172 211 712 × 2 = 0 + 0.026 344 423 424;
  • 20) 0.026 344 423 424 × 2 = 0 + 0.052 688 846 848;
  • 21) 0.052 688 846 848 × 2 = 0 + 0.105 377 693 696;
  • 22) 0.105 377 693 696 × 2 = 0 + 0.210 755 387 392;
  • 23) 0.210 755 387 392 × 2 = 0 + 0.421 510 774 784;
  • 24) 0.421 510 774 784 × 2 = 0 + 0.843 021 549 568;
  • 25) 0.843 021 549 568 × 2 = 1 + 0.686 043 099 136;
  • 26) 0.686 043 099 136 × 2 = 1 + 0.372 086 198 272;
  • 27) 0.372 086 198 272 × 2 = 0 + 0.744 172 396 544;
  • 28) 0.744 172 396 544 × 2 = 1 + 0.488 344 793 088;
  • 29) 0.488 344 793 088 × 2 = 0 + 0.976 689 586 176;
  • 30) 0.976 689 586 176 × 2 = 1 + 0.953 379 172 352;
  • 31) 0.953 379 172 352 × 2 = 1 + 0.906 758 344 704;
  • 32) 0.906 758 344 704 × 2 = 1 + 0.813 516 689 408;
  • 33) 0.813 516 689 408 × 2 = 1 + 0.627 033 378 816;
  • 34) 0.627 033 378 816 × 2 = 1 + 0.254 066 757 632;
  • 35) 0.254 066 757 632 × 2 = 0 + 0.508 133 515 264;
  • 36) 0.508 133 515 264 × 2 = 1 + 0.016 267 030 528;
  • 37) 0.016 267 030 528 × 2 = 0 + 0.032 534 061 056;
  • 38) 0.032 534 061 056 × 2 = 0 + 0.065 068 122 112;
  • 39) 0.065 068 122 112 × 2 = 0 + 0.130 136 244 224;
  • 40) 0.130 136 244 224 × 2 = 0 + 0.260 272 488 448;
  • 41) 0.260 272 488 448 × 2 = 0 + 0.520 544 976 896;
  • 42) 0.520 544 976 896 × 2 = 1 + 0.041 089 953 792;
  • 43) 0.041 089 953 792 × 2 = 0 + 0.082 179 907 584;
  • 44) 0.082 179 907 584 × 2 = 0 + 0.164 359 815 168;
  • 45) 0.164 359 815 168 × 2 = 0 + 0.328 719 630 336;
  • 46) 0.328 719 630 336 × 2 = 0 + 0.657 439 260 672;
  • 47) 0.657 439 260 672 × 2 = 1 + 0.314 878 521 344;
  • 48) 0.314 878 521 344 × 2 = 0 + 0.629 757 042 688;
  • 49) 0.629 757 042 688 × 2 = 1 + 0.259 514 085 376;
  • 50) 0.259 514 085 376 × 2 = 0 + 0.519 028 170 752;
  • 51) 0.519 028 170 752 × 2 = 1 + 0.038 056 341 504;
  • 52) 0.038 056 341 504 × 2 = 0 + 0.076 112 683 008;
  • 53) 0.076 112 683 008 × 2 = 0 + 0.152 225 366 016;
  • 54) 0.152 225 366 016 × 2 = 0 + 0.304 450 732 032;
  • 55) 0.304 450 732 032 × 2 = 0 + 0.608 901 464 064;
  • 56) 0.608 901 464 064 × 2 = 1 + 0.217 802 928 128;
  • 57) 0.217 802 928 128 × 2 = 0 + 0.435 605 856 256;
  • 58) 0.435 605 856 256 × 2 = 0 + 0.871 211 712 512;
  • 59) 0.871 211 712 512 × 2 = 1 + 0.742 423 425 024;
  • 60) 0.742 423 425 024 × 2 = 1 + 0.484 846 850 048;
  • 61) 0.484 846 850 048 × 2 = 0 + 0.969 693 700 096;
  • 62) 0.969 693 700 096 × 2 = 1 + 0.939 387 400 192;
  • 63) 0.939 387 400 192 × 2 = 1 + 0.878 774 800 384;
  • 64) 0.878 774 800 384 × 2 = 1 + 0.757 549 600 768;
  • 65) 0.757 549 600 768 × 2 = 1 + 0.515 099 201 536;
  • 66) 0.515 099 201 536 × 2 = 1 + 0.030 198 403 072;
  • 67) 0.030 198 403 072 × 2 = 0 + 0.060 396 806 144;
  • 68) 0.060 396 806 144 × 2 = 0 + 0.120 793 612 288;
  • 69) 0.120 793 612 288 × 2 = 0 + 0.241 587 224 576;
  • 70) 0.241 587 224 576 × 2 = 0 + 0.483 174 449 152;
  • 71) 0.483 174 449 152 × 2 = 0 + 0.966 348 898 304;
  • 72) 0.966 348 898 304 × 2 = 1 + 0.932 697 796 608;
  • 73) 0.932 697 796 608 × 2 = 1 + 0.865 395 593 216;
  • 74) 0.865 395 593 216 × 2 = 1 + 0.730 791 186 432;
  • 75) 0.730 791 186 432 × 2 = 1 + 0.461 582 372 864;
  • 76) 0.461 582 372 864 × 2 = 0 + 0.923 164 745 728;
  • 77) 0.923 164 745 728 × 2 = 1 + 0.846 329 491 456;

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).


4. 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:


0.000 000 050 248(10) =

0.0000 0000 0000 0000 0000 0000 1101 0111 1101 0000 0100 0010 1010 0001 0011 0111 1100 0001 1110 1(2)

5. Positive number before normalization:

0.000 000 050 248(10) =

0.0000 0000 0000 0000 0000 0000 1101 0111 1101 0000 0100 0010 1010 0001 0011 0111 1100 0001 1110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 050 248(10) =

0.0000 0000 0000 0000 0000 0000 1101 0111 1101 0000 0100 0010 1010 0001 0011 0111 1100 0001 1110 1(2) =

0.0000 0000 0000 0000 0000 0000 1101 0111 1101 0000 0100 0010 1010 0001 0011 0111 1100 0001 1110 1(2) × 20 =

1.1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101(2) × 2-25


7. 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 0 (a positive number)

Exponent (unadjusted): -25

Mantissa (not normalized):
1.1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-25 + 2(11-1) - 1 =

(-25 + 1 023)(10) =

998(10)


9. 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;
  • 998 ÷ 2 = 499 + 0;
  • 499 ÷ 2 = 249 + 1;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

998(10) =

011 1110 0110(2)


11. 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. 1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101 =

1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 0110

Mantissa (52 bits) =
1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101


Decimal number 0.000 000 050 248 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0110 - 1010 1111 1010 0000 1000 0101 0100 0010 0110 1111 1000 0011 1101

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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