2.714 285 714 285 714 276 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 285 714 276 4(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
2.714 285 714 285 714 276 4(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: 0.714 285 714 285 714 276 4.

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) 0.714 285 714 285 714 276 4 × 2 = 1 + 0.428 571 428 571 428 552 8;
  • 2) 0.428 571 428 571 428 552 8 × 2 = 0 + 0.857 142 857 142 857 105 6;
  • 3) 0.857 142 857 142 857 105 6 × 2 = 1 + 0.714 285 714 285 714 211 2;
  • 4) 0.714 285 714 285 714 211 2 × 2 = 1 + 0.428 571 428 571 428 422 4;
  • 5) 0.428 571 428 571 428 422 4 × 2 = 0 + 0.857 142 857 142 856 844 8;
  • 6) 0.857 142 857 142 856 844 8 × 2 = 1 + 0.714 285 714 285 713 689 6;
  • 7) 0.714 285 714 285 713 689 6 × 2 = 1 + 0.428 571 428 571 427 379 2;
  • 8) 0.428 571 428 571 427 379 2 × 2 = 0 + 0.857 142 857 142 854 758 4;
  • 9) 0.857 142 857 142 854 758 4 × 2 = 1 + 0.714 285 714 285 709 516 8;
  • 10) 0.714 285 714 285 709 516 8 × 2 = 1 + 0.428 571 428 571 419 033 6;
  • 11) 0.428 571 428 571 419 033 6 × 2 = 0 + 0.857 142 857 142 838 067 2;
  • 12) 0.857 142 857 142 838 067 2 × 2 = 1 + 0.714 285 714 285 676 134 4;
  • 13) 0.714 285 714 285 676 134 4 × 2 = 1 + 0.428 571 428 571 352 268 8;
  • 14) 0.428 571 428 571 352 268 8 × 2 = 0 + 0.857 142 857 142 704 537 6;
  • 15) 0.857 142 857 142 704 537 6 × 2 = 1 + 0.714 285 714 285 409 075 2;
  • 16) 0.714 285 714 285 409 075 2 × 2 = 1 + 0.428 571 428 570 818 150 4;
  • 17) 0.428 571 428 570 818 150 4 × 2 = 0 + 0.857 142 857 141 636 300 8;
  • 18) 0.857 142 857 141 636 300 8 × 2 = 1 + 0.714 285 714 283 272 601 6;
  • 19) 0.714 285 714 283 272 601 6 × 2 = 1 + 0.428 571 428 566 545 203 2;
  • 20) 0.428 571 428 566 545 203 2 × 2 = 0 + 0.857 142 857 133 090 406 4;
  • 21) 0.857 142 857 133 090 406 4 × 2 = 1 + 0.714 285 714 266 180 812 8;
  • 22) 0.714 285 714 266 180 812 8 × 2 = 1 + 0.428 571 428 532 361 625 6;
  • 23) 0.428 571 428 532 361 625 6 × 2 = 0 + 0.857 142 857 064 723 251 2;
  • 24) 0.857 142 857 064 723 251 2 × 2 = 1 + 0.714 285 714 129 446 502 4;
  • 25) 0.714 285 714 129 446 502 4 × 2 = 1 + 0.428 571 428 258 893 004 8;
  • 26) 0.428 571 428 258 893 004 8 × 2 = 0 + 0.857 142 856 517 786 009 6;
  • 27) 0.857 142 856 517 786 009 6 × 2 = 1 + 0.714 285 713 035 572 019 2;
  • 28) 0.714 285 713 035 572 019 2 × 2 = 1 + 0.428 571 426 071 144 038 4;
  • 29) 0.428 571 426 071 144 038 4 × 2 = 0 + 0.857 142 852 142 288 076 8;
  • 30) 0.857 142 852 142 288 076 8 × 2 = 1 + 0.714 285 704 284 576 153 6;
  • 31) 0.714 285 704 284 576 153 6 × 2 = 1 + 0.428 571 408 569 152 307 2;
  • 32) 0.428 571 408 569 152 307 2 × 2 = 0 + 0.857 142 817 138 304 614 4;
  • 33) 0.857 142 817 138 304 614 4 × 2 = 1 + 0.714 285 634 276 609 228 8;
  • 34) 0.714 285 634 276 609 228 8 × 2 = 1 + 0.428 571 268 553 218 457 6;
  • 35) 0.428 571 268 553 218 457 6 × 2 = 0 + 0.857 142 537 106 436 915 2;
  • 36) 0.857 142 537 106 436 915 2 × 2 = 1 + 0.714 285 074 212 873 830 4;
  • 37) 0.714 285 074 212 873 830 4 × 2 = 1 + 0.428 570 148 425 747 660 8;
  • 38) 0.428 570 148 425 747 660 8 × 2 = 0 + 0.857 140 296 851 495 321 6;
  • 39) 0.857 140 296 851 495 321 6 × 2 = 1 + 0.714 280 593 702 990 643 2;
  • 40) 0.714 280 593 702 990 643 2 × 2 = 1 + 0.428 561 187 405 981 286 4;
  • 41) 0.428 561 187 405 981 286 4 × 2 = 0 + 0.857 122 374 811 962 572 8;
  • 42) 0.857 122 374 811 962 572 8 × 2 = 1 + 0.714 244 749 623 925 145 6;
  • 43) 0.714 244 749 623 925 145 6 × 2 = 1 + 0.428 489 499 247 850 291 2;
  • 44) 0.428 489 499 247 850 291 2 × 2 = 0 + 0.856 978 998 495 700 582 4;
  • 45) 0.856 978 998 495 700 582 4 × 2 = 1 + 0.713 957 996 991 401 164 8;
  • 46) 0.713 957 996 991 401 164 8 × 2 = 1 + 0.427 915 993 982 802 329 6;
  • 47) 0.427 915 993 982 802 329 6 × 2 = 0 + 0.855 831 987 965 604 659 2;
  • 48) 0.855 831 987 965 604 659 2 × 2 = 1 + 0.711 663 975 931 209 318 4;
  • 49) 0.711 663 975 931 209 318 4 × 2 = 1 + 0.423 327 951 862 418 636 8;
  • 50) 0.423 327 951 862 418 636 8 × 2 = 0 + 0.846 655 903 724 837 273 6;
  • 51) 0.846 655 903 724 837 273 6 × 2 = 1 + 0.693 311 807 449 674 547 2;
  • 52) 0.693 311 807 449 674 547 2 × 2 = 1 + 0.386 623 614 899 349 094 4;
  • 53) 0.386 623 614 899 349 094 4 × 2 = 0 + 0.773 247 229 798 698 188 8;

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:


0.714 285 714 285 714 276 4(10) =

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

5. Positive number before normalization:

2.714 285 714 285 714 276 4(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:


2.714 285 714 285 714 276 4(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 2.714 285 714 285 714 276 4 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

» Convert decimal 2.714 285 714 285 714 273 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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