6.285 714 285 714 285 693 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 693 9(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
6.285 714 285 714 285 693 9(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: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.285 714 285 714 285 693 9.

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) 0.285 714 285 714 285 693 9 × 2 = 0 + 0.571 428 571 428 571 387 8;
  • 2) 0.571 428 571 428 571 387 8 × 2 = 1 + 0.142 857 142 857 142 775 6;
  • 3) 0.142 857 142 857 142 775 6 × 2 = 0 + 0.285 714 285 714 285 551 2;
  • 4) 0.285 714 285 714 285 551 2 × 2 = 0 + 0.571 428 571 428 571 102 4;
  • 5) 0.571 428 571 428 571 102 4 × 2 = 1 + 0.142 857 142 857 142 204 8;
  • 6) 0.142 857 142 857 142 204 8 × 2 = 0 + 0.285 714 285 714 284 409 6;
  • 7) 0.285 714 285 714 284 409 6 × 2 = 0 + 0.571 428 571 428 568 819 2;
  • 8) 0.571 428 571 428 568 819 2 × 2 = 1 + 0.142 857 142 857 137 638 4;
  • 9) 0.142 857 142 857 137 638 4 × 2 = 0 + 0.285 714 285 714 275 276 8;
  • 10) 0.285 714 285 714 275 276 8 × 2 = 0 + 0.571 428 571 428 550 553 6;
  • 11) 0.571 428 571 428 550 553 6 × 2 = 1 + 0.142 857 142 857 101 107 2;
  • 12) 0.142 857 142 857 101 107 2 × 2 = 0 + 0.285 714 285 714 202 214 4;
  • 13) 0.285 714 285 714 202 214 4 × 2 = 0 + 0.571 428 571 428 404 428 8;
  • 14) 0.571 428 571 428 404 428 8 × 2 = 1 + 0.142 857 142 856 808 857 6;
  • 15) 0.142 857 142 856 808 857 6 × 2 = 0 + 0.285 714 285 713 617 715 2;
  • 16) 0.285 714 285 713 617 715 2 × 2 = 0 + 0.571 428 571 427 235 430 4;
  • 17) 0.571 428 571 427 235 430 4 × 2 = 1 + 0.142 857 142 854 470 860 8;
  • 18) 0.142 857 142 854 470 860 8 × 2 = 0 + 0.285 714 285 708 941 721 6;
  • 19) 0.285 714 285 708 941 721 6 × 2 = 0 + 0.571 428 571 417 883 443 2;
  • 20) 0.571 428 571 417 883 443 2 × 2 = 1 + 0.142 857 142 835 766 886 4;
  • 21) 0.142 857 142 835 766 886 4 × 2 = 0 + 0.285 714 285 671 533 772 8;
  • 22) 0.285 714 285 671 533 772 8 × 2 = 0 + 0.571 428 571 343 067 545 6;
  • 23) 0.571 428 571 343 067 545 6 × 2 = 1 + 0.142 857 142 686 135 091 2;
  • 24) 0.142 857 142 686 135 091 2 × 2 = 0 + 0.285 714 285 372 270 182 4;
  • 25) 0.285 714 285 372 270 182 4 × 2 = 0 + 0.571 428 570 744 540 364 8;
  • 26) 0.571 428 570 744 540 364 8 × 2 = 1 + 0.142 857 141 489 080 729 6;
  • 27) 0.142 857 141 489 080 729 6 × 2 = 0 + 0.285 714 282 978 161 459 2;
  • 28) 0.285 714 282 978 161 459 2 × 2 = 0 + 0.571 428 565 956 322 918 4;
  • 29) 0.571 428 565 956 322 918 4 × 2 = 1 + 0.142 857 131 912 645 836 8;
  • 30) 0.142 857 131 912 645 836 8 × 2 = 0 + 0.285 714 263 825 291 673 6;
  • 31) 0.285 714 263 825 291 673 6 × 2 = 0 + 0.571 428 527 650 583 347 2;
  • 32) 0.571 428 527 650 583 347 2 × 2 = 1 + 0.142 857 055 301 166 694 4;
  • 33) 0.142 857 055 301 166 694 4 × 2 = 0 + 0.285 714 110 602 333 388 8;
  • 34) 0.285 714 110 602 333 388 8 × 2 = 0 + 0.571 428 221 204 666 777 6;
  • 35) 0.571 428 221 204 666 777 6 × 2 = 1 + 0.142 856 442 409 333 555 2;
  • 36) 0.142 856 442 409 333 555 2 × 2 = 0 + 0.285 712 884 818 667 110 4;
  • 37) 0.285 712 884 818 667 110 4 × 2 = 0 + 0.571 425 769 637 334 220 8;
  • 38) 0.571 425 769 637 334 220 8 × 2 = 1 + 0.142 851 539 274 668 441 6;
  • 39) 0.142 851 539 274 668 441 6 × 2 = 0 + 0.285 703 078 549 336 883 2;
  • 40) 0.285 703 078 549 336 883 2 × 2 = 0 + 0.571 406 157 098 673 766 4;
  • 41) 0.571 406 157 098 673 766 4 × 2 = 1 + 0.142 812 314 197 347 532 8;
  • 42) 0.142 812 314 197 347 532 8 × 2 = 0 + 0.285 624 628 394 695 065 6;
  • 43) 0.285 624 628 394 695 065 6 × 2 = 0 + 0.571 249 256 789 390 131 2;
  • 44) 0.571 249 256 789 390 131 2 × 2 = 1 + 0.142 498 513 578 780 262 4;
  • 45) 0.142 498 513 578 780 262 4 × 2 = 0 + 0.284 997 027 157 560 524 8;
  • 46) 0.284 997 027 157 560 524 8 × 2 = 0 + 0.569 994 054 315 121 049 6;
  • 47) 0.569 994 054 315 121 049 6 × 2 = 1 + 0.139 988 108 630 242 099 2;
  • 48) 0.139 988 108 630 242 099 2 × 2 = 0 + 0.279 976 217 260 484 198 4;
  • 49) 0.279 976 217 260 484 198 4 × 2 = 0 + 0.559 952 434 520 968 396 8;
  • 50) 0.559 952 434 520 968 396 8 × 2 = 1 + 0.119 904 869 041 936 793 6;
  • 51) 0.119 904 869 041 936 793 6 × 2 = 0 + 0.239 809 738 083 873 587 2;
  • 52) 0.239 809 738 083 873 587 2 × 2 = 0 + 0.479 619 476 167 747 174 4;
  • 53) 0.479 619 476 167 747 174 4 × 2 = 0 + 0.959 238 952 335 494 348 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:


0.285 714 285 714 285 693 9(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2)

5. Positive number before normalization:

6.285 714 285 714 285 693 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.285 714 285 714 285 693 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000(2) × 22


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): 2

Mantissa (not normalized):
1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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. 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


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 0001

Mantissa (52 bits) =
1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


Decimal number 6.285 714 285 714 285 693 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001

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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