3.123 145 321 678 989 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.123 145 321 678 989 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
3.123 145 321 678 989 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: 0.123 145 321 678 989 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) 0.123 145 321 678 989 5 × 2 = 0 + 0.246 290 643 357 979;
  • 2) 0.246 290 643 357 979 × 2 = 0 + 0.492 581 286 715 958;
  • 3) 0.492 581 286 715 958 × 2 = 0 + 0.985 162 573 431 916;
  • 4) 0.985 162 573 431 916 × 2 = 1 + 0.970 325 146 863 832;
  • 5) 0.970 325 146 863 832 × 2 = 1 + 0.940 650 293 727 664;
  • 6) 0.940 650 293 727 664 × 2 = 1 + 0.881 300 587 455 328;
  • 7) 0.881 300 587 455 328 × 2 = 1 + 0.762 601 174 910 656;
  • 8) 0.762 601 174 910 656 × 2 = 1 + 0.525 202 349 821 312;
  • 9) 0.525 202 349 821 312 × 2 = 1 + 0.050 404 699 642 624;
  • 10) 0.050 404 699 642 624 × 2 = 0 + 0.100 809 399 285 248;
  • 11) 0.100 809 399 285 248 × 2 = 0 + 0.201 618 798 570 496;
  • 12) 0.201 618 798 570 496 × 2 = 0 + 0.403 237 597 140 992;
  • 13) 0.403 237 597 140 992 × 2 = 0 + 0.806 475 194 281 984;
  • 14) 0.806 475 194 281 984 × 2 = 1 + 0.612 950 388 563 968;
  • 15) 0.612 950 388 563 968 × 2 = 1 + 0.225 900 777 127 936;
  • 16) 0.225 900 777 127 936 × 2 = 0 + 0.451 801 554 255 872;
  • 17) 0.451 801 554 255 872 × 2 = 0 + 0.903 603 108 511 744;
  • 18) 0.903 603 108 511 744 × 2 = 1 + 0.807 206 217 023 488;
  • 19) 0.807 206 217 023 488 × 2 = 1 + 0.614 412 434 046 976;
  • 20) 0.614 412 434 046 976 × 2 = 1 + 0.228 824 868 093 952;
  • 21) 0.228 824 868 093 952 × 2 = 0 + 0.457 649 736 187 904;
  • 22) 0.457 649 736 187 904 × 2 = 0 + 0.915 299 472 375 808;
  • 23) 0.915 299 472 375 808 × 2 = 1 + 0.830 598 944 751 616;
  • 24) 0.830 598 944 751 616 × 2 = 1 + 0.661 197 889 503 232;
  • 25) 0.661 197 889 503 232 × 2 = 1 + 0.322 395 779 006 464;
  • 26) 0.322 395 779 006 464 × 2 = 0 + 0.644 791 558 012 928;
  • 27) 0.644 791 558 012 928 × 2 = 1 + 0.289 583 116 025 856;
  • 28) 0.289 583 116 025 856 × 2 = 0 + 0.579 166 232 051 712;
  • 29) 0.579 166 232 051 712 × 2 = 1 + 0.158 332 464 103 424;
  • 30) 0.158 332 464 103 424 × 2 = 0 + 0.316 664 928 206 848;
  • 31) 0.316 664 928 206 848 × 2 = 0 + 0.633 329 856 413 696;
  • 32) 0.633 329 856 413 696 × 2 = 1 + 0.266 659 712 827 392;
  • 33) 0.266 659 712 827 392 × 2 = 0 + 0.533 319 425 654 784;
  • 34) 0.533 319 425 654 784 × 2 = 1 + 0.066 638 851 309 568;
  • 35) 0.066 638 851 309 568 × 2 = 0 + 0.133 277 702 619 136;
  • 36) 0.133 277 702 619 136 × 2 = 0 + 0.266 555 405 238 272;
  • 37) 0.266 555 405 238 272 × 2 = 0 + 0.533 110 810 476 544;
  • 38) 0.533 110 810 476 544 × 2 = 1 + 0.066 221 620 953 088;
  • 39) 0.066 221 620 953 088 × 2 = 0 + 0.132 443 241 906 176;
  • 40) 0.132 443 241 906 176 × 2 = 0 + 0.264 886 483 812 352;
  • 41) 0.264 886 483 812 352 × 2 = 0 + 0.529 772 967 624 704;
  • 42) 0.529 772 967 624 704 × 2 = 1 + 0.059 545 935 249 408;
  • 43) 0.059 545 935 249 408 × 2 = 0 + 0.119 091 870 498 816;
  • 44) 0.119 091 870 498 816 × 2 = 0 + 0.238 183 740 997 632;
  • 45) 0.238 183 740 997 632 × 2 = 0 + 0.476 367 481 995 264;
  • 46) 0.476 367 481 995 264 × 2 = 0 + 0.952 734 963 990 528;
  • 47) 0.952 734 963 990 528 × 2 = 1 + 0.905 469 927 981 056;
  • 48) 0.905 469 927 981 056 × 2 = 1 + 0.810 939 855 962 112;
  • 49) 0.810 939 855 962 112 × 2 = 1 + 0.621 879 711 924 224;
  • 50) 0.621 879 711 924 224 × 2 = 1 + 0.243 759 423 848 448;
  • 51) 0.243 759 423 848 448 × 2 = 0 + 0.487 518 847 696 896;
  • 52) 0.487 518 847 696 896 × 2 = 0 + 0.975 037 695 393 792;
  • 53) 0.975 037 695 393 792 × 2 = 1 + 0.950 075 390 787 584;

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:


0.123 145 321 678 989 5(10) =

0.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0011 1100 1(2)

5. Positive number before normalization:

3.123 145 321 678 989 5(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0011 1100 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:


3.123 145 321 678 989 5(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0011 1100 1(2) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0011 1100 1(2) × 20 =

1.1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0001 1110 01(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 0001 1110 01


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 0001 1110 01 =

1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0001 1110


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 0001 1110


Decimal number 3.123 145 321 678 989 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 0001 1110

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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