6.285 714 285 714 285 832 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 832(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
6.285 714 285 714 285 832(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: 0.285 714 285 714 285 832.

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) 0.285 714 285 714 285 832 × 2 = 0 + 0.571 428 571 428 571 664;
  • 2) 0.571 428 571 428 571 664 × 2 = 1 + 0.142 857 142 857 143 328;
  • 3) 0.142 857 142 857 143 328 × 2 = 0 + 0.285 714 285 714 286 656;
  • 4) 0.285 714 285 714 286 656 × 2 = 0 + 0.571 428 571 428 573 312;
  • 5) 0.571 428 571 428 573 312 × 2 = 1 + 0.142 857 142 857 146 624;
  • 6) 0.142 857 142 857 146 624 × 2 = 0 + 0.285 714 285 714 293 248;
  • 7) 0.285 714 285 714 293 248 × 2 = 0 + 0.571 428 571 428 586 496;
  • 8) 0.571 428 571 428 586 496 × 2 = 1 + 0.142 857 142 857 172 992;
  • 9) 0.142 857 142 857 172 992 × 2 = 0 + 0.285 714 285 714 345 984;
  • 10) 0.285 714 285 714 345 984 × 2 = 0 + 0.571 428 571 428 691 968;
  • 11) 0.571 428 571 428 691 968 × 2 = 1 + 0.142 857 142 857 383 936;
  • 12) 0.142 857 142 857 383 936 × 2 = 0 + 0.285 714 285 714 767 872;
  • 13) 0.285 714 285 714 767 872 × 2 = 0 + 0.571 428 571 429 535 744;
  • 14) 0.571 428 571 429 535 744 × 2 = 1 + 0.142 857 142 859 071 488;
  • 15) 0.142 857 142 859 071 488 × 2 = 0 + 0.285 714 285 718 142 976;
  • 16) 0.285 714 285 718 142 976 × 2 = 0 + 0.571 428 571 436 285 952;
  • 17) 0.571 428 571 436 285 952 × 2 = 1 + 0.142 857 142 872 571 904;
  • 18) 0.142 857 142 872 571 904 × 2 = 0 + 0.285 714 285 745 143 808;
  • 19) 0.285 714 285 745 143 808 × 2 = 0 + 0.571 428 571 490 287 616;
  • 20) 0.571 428 571 490 287 616 × 2 = 1 + 0.142 857 142 980 575 232;
  • 21) 0.142 857 142 980 575 232 × 2 = 0 + 0.285 714 285 961 150 464;
  • 22) 0.285 714 285 961 150 464 × 2 = 0 + 0.571 428 571 922 300 928;
  • 23) 0.571 428 571 922 300 928 × 2 = 1 + 0.142 857 143 844 601 856;
  • 24) 0.142 857 143 844 601 856 × 2 = 0 + 0.285 714 287 689 203 712;
  • 25) 0.285 714 287 689 203 712 × 2 = 0 + 0.571 428 575 378 407 424;
  • 26) 0.571 428 575 378 407 424 × 2 = 1 + 0.142 857 150 756 814 848;
  • 27) 0.142 857 150 756 814 848 × 2 = 0 + 0.285 714 301 513 629 696;
  • 28) 0.285 714 301 513 629 696 × 2 = 0 + 0.571 428 603 027 259 392;
  • 29) 0.571 428 603 027 259 392 × 2 = 1 + 0.142 857 206 054 518 784;
  • 30) 0.142 857 206 054 518 784 × 2 = 0 + 0.285 714 412 109 037 568;
  • 31) 0.285 714 412 109 037 568 × 2 = 0 + 0.571 428 824 218 075 136;
  • 32) 0.571 428 824 218 075 136 × 2 = 1 + 0.142 857 648 436 150 272;
  • 33) 0.142 857 648 436 150 272 × 2 = 0 + 0.285 715 296 872 300 544;
  • 34) 0.285 715 296 872 300 544 × 2 = 0 + 0.571 430 593 744 601 088;
  • 35) 0.571 430 593 744 601 088 × 2 = 1 + 0.142 861 187 489 202 176;
  • 36) 0.142 861 187 489 202 176 × 2 = 0 + 0.285 722 374 978 404 352;
  • 37) 0.285 722 374 978 404 352 × 2 = 0 + 0.571 444 749 956 808 704;
  • 38) 0.571 444 749 956 808 704 × 2 = 1 + 0.142 889 499 913 617 408;
  • 39) 0.142 889 499 913 617 408 × 2 = 0 + 0.285 778 999 827 234 816;
  • 40) 0.285 778 999 827 234 816 × 2 = 0 + 0.571 557 999 654 469 632;
  • 41) 0.571 557 999 654 469 632 × 2 = 1 + 0.143 115 999 308 939 264;
  • 42) 0.143 115 999 308 939 264 × 2 = 0 + 0.286 231 998 617 878 528;
  • 43) 0.286 231 998 617 878 528 × 2 = 0 + 0.572 463 997 235 757 056;
  • 44) 0.572 463 997 235 757 056 × 2 = 1 + 0.144 927 994 471 514 112;
  • 45) 0.144 927 994 471 514 112 × 2 = 0 + 0.289 855 988 943 028 224;
  • 46) 0.289 855 988 943 028 224 × 2 = 0 + 0.579 711 977 886 056 448;
  • 47) 0.579 711 977 886 056 448 × 2 = 1 + 0.159 423 955 772 112 896;
  • 48) 0.159 423 955 772 112 896 × 2 = 0 + 0.318 847 911 544 225 792;
  • 49) 0.318 847 911 544 225 792 × 2 = 0 + 0.637 695 823 088 451 584;
  • 50) 0.637 695 823 088 451 584 × 2 = 1 + 0.275 391 646 176 903 168;
  • 51) 0.275 391 646 176 903 168 × 2 = 0 + 0.550 783 292 353 806 336;
  • 52) 0.550 783 292 353 806 336 × 2 = 1 + 0.101 566 584 707 612 672;
  • 53) 0.101 566 584 707 612 672 × 2 = 0 + 0.203 133 169 415 225 344;

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:


0.285 714 285 714 285 832(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0(2)

5. Positive number before normalization:

6.285 714 285 714 285 832(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 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:


6.285 714 285 714 285 832(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 010(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 010


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 010 =

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 6.285 714 285 714 285 832 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

Share

» Convert decimal 6.285 714 285 714 285 84 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 minutes

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