3.123 145 321 678 998 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.123 145 321 678 998 5(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
3.123 145 321 678 998 5(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.123 145 321 678 998 5.

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) 0.123 145 321 678 998 5 × 2 = 0 + 0.246 290 643 357 997;
  • 2) 0.246 290 643 357 997 × 2 = 0 + 0.492 581 286 715 994;
  • 3) 0.492 581 286 715 994 × 2 = 0 + 0.985 162 573 431 988;
  • 4) 0.985 162 573 431 988 × 2 = 1 + 0.970 325 146 863 976;
  • 5) 0.970 325 146 863 976 × 2 = 1 + 0.940 650 293 727 952;
  • 6) 0.940 650 293 727 952 × 2 = 1 + 0.881 300 587 455 904;
  • 7) 0.881 300 587 455 904 × 2 = 1 + 0.762 601 174 911 808;
  • 8) 0.762 601 174 911 808 × 2 = 1 + 0.525 202 349 823 616;
  • 9) 0.525 202 349 823 616 × 2 = 1 + 0.050 404 699 647 232;
  • 10) 0.050 404 699 647 232 × 2 = 0 + 0.100 809 399 294 464;
  • 11) 0.100 809 399 294 464 × 2 = 0 + 0.201 618 798 588 928;
  • 12) 0.201 618 798 588 928 × 2 = 0 + 0.403 237 597 177 856;
  • 13) 0.403 237 597 177 856 × 2 = 0 + 0.806 475 194 355 712;
  • 14) 0.806 475 194 355 712 × 2 = 1 + 0.612 950 388 711 424;
  • 15) 0.612 950 388 711 424 × 2 = 1 + 0.225 900 777 422 848;
  • 16) 0.225 900 777 422 848 × 2 = 0 + 0.451 801 554 845 696;
  • 17) 0.451 801 554 845 696 × 2 = 0 + 0.903 603 109 691 392;
  • 18) 0.903 603 109 691 392 × 2 = 1 + 0.807 206 219 382 784;
  • 19) 0.807 206 219 382 784 × 2 = 1 + 0.614 412 438 765 568;
  • 20) 0.614 412 438 765 568 × 2 = 1 + 0.228 824 877 531 136;
  • 21) 0.228 824 877 531 136 × 2 = 0 + 0.457 649 755 062 272;
  • 22) 0.457 649 755 062 272 × 2 = 0 + 0.915 299 510 124 544;
  • 23) 0.915 299 510 124 544 × 2 = 1 + 0.830 599 020 249 088;
  • 24) 0.830 599 020 249 088 × 2 = 1 + 0.661 198 040 498 176;
  • 25) 0.661 198 040 498 176 × 2 = 1 + 0.322 396 080 996 352;
  • 26) 0.322 396 080 996 352 × 2 = 0 + 0.644 792 161 992 704;
  • 27) 0.644 792 161 992 704 × 2 = 1 + 0.289 584 323 985 408;
  • 28) 0.289 584 323 985 408 × 2 = 0 + 0.579 168 647 970 816;
  • 29) 0.579 168 647 970 816 × 2 = 1 + 0.158 337 295 941 632;
  • 30) 0.158 337 295 941 632 × 2 = 0 + 0.316 674 591 883 264;
  • 31) 0.316 674 591 883 264 × 2 = 0 + 0.633 349 183 766 528;
  • 32) 0.633 349 183 766 528 × 2 = 1 + 0.266 698 367 533 056;
  • 33) 0.266 698 367 533 056 × 2 = 0 + 0.533 396 735 066 112;
  • 34) 0.533 396 735 066 112 × 2 = 1 + 0.066 793 470 132 224;
  • 35) 0.066 793 470 132 224 × 2 = 0 + 0.133 586 940 264 448;
  • 36) 0.133 586 940 264 448 × 2 = 0 + 0.267 173 880 528 896;
  • 37) 0.267 173 880 528 896 × 2 = 0 + 0.534 347 761 057 792;
  • 38) 0.534 347 761 057 792 × 2 = 1 + 0.068 695 522 115 584;
  • 39) 0.068 695 522 115 584 × 2 = 0 + 0.137 391 044 231 168;
  • 40) 0.137 391 044 231 168 × 2 = 0 + 0.274 782 088 462 336;
  • 41) 0.274 782 088 462 336 × 2 = 0 + 0.549 564 176 924 672;
  • 42) 0.549 564 176 924 672 × 2 = 1 + 0.099 128 353 849 344;
  • 43) 0.099 128 353 849 344 × 2 = 0 + 0.198 256 707 698 688;
  • 44) 0.198 256 707 698 688 × 2 = 0 + 0.396 513 415 397 376;
  • 45) 0.396 513 415 397 376 × 2 = 0 + 0.793 026 830 794 752;
  • 46) 0.793 026 830 794 752 × 2 = 1 + 0.586 053 661 589 504;
  • 47) 0.586 053 661 589 504 × 2 = 1 + 0.172 107 323 179 008;
  • 48) 0.172 107 323 179 008 × 2 = 0 + 0.344 214 646 358 016;
  • 49) 0.344 214 646 358 016 × 2 = 0 + 0.688 429 292 716 032;
  • 50) 0.688 429 292 716 032 × 2 = 1 + 0.376 858 585 432 064;
  • 51) 0.376 858 585 432 064 × 2 = 0 + 0.753 717 170 864 128;
  • 52) 0.753 717 170 864 128 × 2 = 1 + 0.507 434 341 728 256;
  • 53) 0.507 434 341 728 256 × 2 = 1 + 0.014 868 683 456 512;

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:


0.123 145 321 678 998 5(10) =

0.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0110 0101 1(2)

5. Positive number before normalization:

3.123 145 321 678 998 5(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0110 0101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.123 145 321 678 998 5(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0110 0101 1(2) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0110 0101 1(2) × 20 =

1.1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010 11(2) × 21


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): 1

Mantissa (not normalized):
1.1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010 11 =

1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010


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) =
100 0000 0000

Mantissa (52 bits) =
1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010


Decimal number 3.123 145 321 678 998 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0011 0010

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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