2.714 285 714 285 714 285 714 906 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 285 714 285 714 906(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
2.714 285 714 285 714 285 714 906(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.714 285 714 285 714 285 714 906.

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) 0.714 285 714 285 714 285 714 906 × 2 = 1 + 0.428 571 428 571 428 571 429 812;
  • 2) 0.428 571 428 571 428 571 429 812 × 2 = 0 + 0.857 142 857 142 857 142 859 624;
  • 3) 0.857 142 857 142 857 142 859 624 × 2 = 1 + 0.714 285 714 285 714 285 719 248;
  • 4) 0.714 285 714 285 714 285 719 248 × 2 = 1 + 0.428 571 428 571 428 571 438 496;
  • 5) 0.428 571 428 571 428 571 438 496 × 2 = 0 + 0.857 142 857 142 857 142 876 992;
  • 6) 0.857 142 857 142 857 142 876 992 × 2 = 1 + 0.714 285 714 285 714 285 753 984;
  • 7) 0.714 285 714 285 714 285 753 984 × 2 = 1 + 0.428 571 428 571 428 571 507 968;
  • 8) 0.428 571 428 571 428 571 507 968 × 2 = 0 + 0.857 142 857 142 857 143 015 936;
  • 9) 0.857 142 857 142 857 143 015 936 × 2 = 1 + 0.714 285 714 285 714 286 031 872;
  • 10) 0.714 285 714 285 714 286 031 872 × 2 = 1 + 0.428 571 428 571 428 572 063 744;
  • 11) 0.428 571 428 571 428 572 063 744 × 2 = 0 + 0.857 142 857 142 857 144 127 488;
  • 12) 0.857 142 857 142 857 144 127 488 × 2 = 1 + 0.714 285 714 285 714 288 254 976;
  • 13) 0.714 285 714 285 714 288 254 976 × 2 = 1 + 0.428 571 428 571 428 576 509 952;
  • 14) 0.428 571 428 571 428 576 509 952 × 2 = 0 + 0.857 142 857 142 857 153 019 904;
  • 15) 0.857 142 857 142 857 153 019 904 × 2 = 1 + 0.714 285 714 285 714 306 039 808;
  • 16) 0.714 285 714 285 714 306 039 808 × 2 = 1 + 0.428 571 428 571 428 612 079 616;
  • 17) 0.428 571 428 571 428 612 079 616 × 2 = 0 + 0.857 142 857 142 857 224 159 232;
  • 18) 0.857 142 857 142 857 224 159 232 × 2 = 1 + 0.714 285 714 285 714 448 318 464;
  • 19) 0.714 285 714 285 714 448 318 464 × 2 = 1 + 0.428 571 428 571 428 896 636 928;
  • 20) 0.428 571 428 571 428 896 636 928 × 2 = 0 + 0.857 142 857 142 857 793 273 856;
  • 21) 0.857 142 857 142 857 793 273 856 × 2 = 1 + 0.714 285 714 285 715 586 547 712;
  • 22) 0.714 285 714 285 715 586 547 712 × 2 = 1 + 0.428 571 428 571 431 173 095 424;
  • 23) 0.428 571 428 571 431 173 095 424 × 2 = 0 + 0.857 142 857 142 862 346 190 848;
  • 24) 0.857 142 857 142 862 346 190 848 × 2 = 1 + 0.714 285 714 285 724 692 381 696;
  • 25) 0.714 285 714 285 724 692 381 696 × 2 = 1 + 0.428 571 428 571 449 384 763 392;
  • 26) 0.428 571 428 571 449 384 763 392 × 2 = 0 + 0.857 142 857 142 898 769 526 784;
  • 27) 0.857 142 857 142 898 769 526 784 × 2 = 1 + 0.714 285 714 285 797 539 053 568;
  • 28) 0.714 285 714 285 797 539 053 568 × 2 = 1 + 0.428 571 428 571 595 078 107 136;
  • 29) 0.428 571 428 571 595 078 107 136 × 2 = 0 + 0.857 142 857 143 190 156 214 272;
  • 30) 0.857 142 857 143 190 156 214 272 × 2 = 1 + 0.714 285 714 286 380 312 428 544;
  • 31) 0.714 285 714 286 380 312 428 544 × 2 = 1 + 0.428 571 428 572 760 624 857 088;
  • 32) 0.428 571 428 572 760 624 857 088 × 2 = 0 + 0.857 142 857 145 521 249 714 176;
  • 33) 0.857 142 857 145 521 249 714 176 × 2 = 1 + 0.714 285 714 291 042 499 428 352;
  • 34) 0.714 285 714 291 042 499 428 352 × 2 = 1 + 0.428 571 428 582 084 998 856 704;
  • 35) 0.428 571 428 582 084 998 856 704 × 2 = 0 + 0.857 142 857 164 169 997 713 408;
  • 36) 0.857 142 857 164 169 997 713 408 × 2 = 1 + 0.714 285 714 328 339 995 426 816;
  • 37) 0.714 285 714 328 339 995 426 816 × 2 = 1 + 0.428 571 428 656 679 990 853 632;
  • 38) 0.428 571 428 656 679 990 853 632 × 2 = 0 + 0.857 142 857 313 359 981 707 264;
  • 39) 0.857 142 857 313 359 981 707 264 × 2 = 1 + 0.714 285 714 626 719 963 414 528;
  • 40) 0.714 285 714 626 719 963 414 528 × 2 = 1 + 0.428 571 429 253 439 926 829 056;
  • 41) 0.428 571 429 253 439 926 829 056 × 2 = 0 + 0.857 142 858 506 879 853 658 112;
  • 42) 0.857 142 858 506 879 853 658 112 × 2 = 1 + 0.714 285 717 013 759 707 316 224;
  • 43) 0.714 285 717 013 759 707 316 224 × 2 = 1 + 0.428 571 434 027 519 414 632 448;
  • 44) 0.428 571 434 027 519 414 632 448 × 2 = 0 + 0.857 142 868 055 038 829 264 896;
  • 45) 0.857 142 868 055 038 829 264 896 × 2 = 1 + 0.714 285 736 110 077 658 529 792;
  • 46) 0.714 285 736 110 077 658 529 792 × 2 = 1 + 0.428 571 472 220 155 317 059 584;
  • 47) 0.428 571 472 220 155 317 059 584 × 2 = 0 + 0.857 142 944 440 310 634 119 168;
  • 48) 0.857 142 944 440 310 634 119 168 × 2 = 1 + 0.714 285 888 880 621 268 238 336;
  • 49) 0.714 285 888 880 621 268 238 336 × 2 = 1 + 0.428 571 777 761 242 536 476 672;
  • 50) 0.428 571 777 761 242 536 476 672 × 2 = 0 + 0.857 143 555 522 485 072 953 344;
  • 51) 0.857 143 555 522 485 072 953 344 × 2 = 1 + 0.714 287 111 044 970 145 906 688;
  • 52) 0.714 287 111 044 970 145 906 688 × 2 = 1 + 0.428 574 222 089 940 291 813 376;
  • 53) 0.428 574 222 089 940 291 813 376 × 2 = 0 + 0.857 148 444 179 880 583 626 752;

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).


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:


0.714 285 714 285 714 285 714 906(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2)

5. Positive number before normalization:

2.714 285 714 285 714 285 714 906(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(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:


2.714 285 714 285 714 285 714 906(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10(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):
1.0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10


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. 0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10 =

0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 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) =
100 0000 0000

Mantissa (52 bits) =
0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101


Decimal number 2.714 285 714 285 714 285 714 906 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 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