6.285 714 285 714 285 714 285 714 016 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 714 285 714 016(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
6.285 714 285 714 285 714 285 714 016(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: 0.285 714 285 714 285 714 285 714 016.

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) 0.285 714 285 714 285 714 285 714 016 × 2 = 0 + 0.571 428 571 428 571 428 571 428 032;
  • 2) 0.571 428 571 428 571 428 571 428 032 × 2 = 1 + 0.142 857 142 857 142 857 142 856 064;
  • 3) 0.142 857 142 857 142 857 142 856 064 × 2 = 0 + 0.285 714 285 714 285 714 285 712 128;
  • 4) 0.285 714 285 714 285 714 285 712 128 × 2 = 0 + 0.571 428 571 428 571 428 571 424 256;
  • 5) 0.571 428 571 428 571 428 571 424 256 × 2 = 1 + 0.142 857 142 857 142 857 142 848 512;
  • 6) 0.142 857 142 857 142 857 142 848 512 × 2 = 0 + 0.285 714 285 714 285 714 285 697 024;
  • 7) 0.285 714 285 714 285 714 285 697 024 × 2 = 0 + 0.571 428 571 428 571 428 571 394 048;
  • 8) 0.571 428 571 428 571 428 571 394 048 × 2 = 1 + 0.142 857 142 857 142 857 142 788 096;
  • 9) 0.142 857 142 857 142 857 142 788 096 × 2 = 0 + 0.285 714 285 714 285 714 285 576 192;
  • 10) 0.285 714 285 714 285 714 285 576 192 × 2 = 0 + 0.571 428 571 428 571 428 571 152 384;
  • 11) 0.571 428 571 428 571 428 571 152 384 × 2 = 1 + 0.142 857 142 857 142 857 142 304 768;
  • 12) 0.142 857 142 857 142 857 142 304 768 × 2 = 0 + 0.285 714 285 714 285 714 284 609 536;
  • 13) 0.285 714 285 714 285 714 284 609 536 × 2 = 0 + 0.571 428 571 428 571 428 569 219 072;
  • 14) 0.571 428 571 428 571 428 569 219 072 × 2 = 1 + 0.142 857 142 857 142 857 138 438 144;
  • 15) 0.142 857 142 857 142 857 138 438 144 × 2 = 0 + 0.285 714 285 714 285 714 276 876 288;
  • 16) 0.285 714 285 714 285 714 276 876 288 × 2 = 0 + 0.571 428 571 428 571 428 553 752 576;
  • 17) 0.571 428 571 428 571 428 553 752 576 × 2 = 1 + 0.142 857 142 857 142 857 107 505 152;
  • 18) 0.142 857 142 857 142 857 107 505 152 × 2 = 0 + 0.285 714 285 714 285 714 215 010 304;
  • 19) 0.285 714 285 714 285 714 215 010 304 × 2 = 0 + 0.571 428 571 428 571 428 430 020 608;
  • 20) 0.571 428 571 428 571 428 430 020 608 × 2 = 1 + 0.142 857 142 857 142 856 860 041 216;
  • 21) 0.142 857 142 857 142 856 860 041 216 × 2 = 0 + 0.285 714 285 714 285 713 720 082 432;
  • 22) 0.285 714 285 714 285 713 720 082 432 × 2 = 0 + 0.571 428 571 428 571 427 440 164 864;
  • 23) 0.571 428 571 428 571 427 440 164 864 × 2 = 1 + 0.142 857 142 857 142 854 880 329 728;
  • 24) 0.142 857 142 857 142 854 880 329 728 × 2 = 0 + 0.285 714 285 714 285 709 760 659 456;
  • 25) 0.285 714 285 714 285 709 760 659 456 × 2 = 0 + 0.571 428 571 428 571 419 521 318 912;
  • 26) 0.571 428 571 428 571 419 521 318 912 × 2 = 1 + 0.142 857 142 857 142 839 042 637 824;
  • 27) 0.142 857 142 857 142 839 042 637 824 × 2 = 0 + 0.285 714 285 714 285 678 085 275 648;
  • 28) 0.285 714 285 714 285 678 085 275 648 × 2 = 0 + 0.571 428 571 428 571 356 170 551 296;
  • 29) 0.571 428 571 428 571 356 170 551 296 × 2 = 1 + 0.142 857 142 857 142 712 341 102 592;
  • 30) 0.142 857 142 857 142 712 341 102 592 × 2 = 0 + 0.285 714 285 714 285 424 682 205 184;
  • 31) 0.285 714 285 714 285 424 682 205 184 × 2 = 0 + 0.571 428 571 428 570 849 364 410 368;
  • 32) 0.571 428 571 428 570 849 364 410 368 × 2 = 1 + 0.142 857 142 857 141 698 728 820 736;
  • 33) 0.142 857 142 857 141 698 728 820 736 × 2 = 0 + 0.285 714 285 714 283 397 457 641 472;
  • 34) 0.285 714 285 714 283 397 457 641 472 × 2 = 0 + 0.571 428 571 428 566 794 915 282 944;
  • 35) 0.571 428 571 428 566 794 915 282 944 × 2 = 1 + 0.142 857 142 857 133 589 830 565 888;
  • 36) 0.142 857 142 857 133 589 830 565 888 × 2 = 0 + 0.285 714 285 714 267 179 661 131 776;
  • 37) 0.285 714 285 714 267 179 661 131 776 × 2 = 0 + 0.571 428 571 428 534 359 322 263 552;
  • 38) 0.571 428 571 428 534 359 322 263 552 × 2 = 1 + 0.142 857 142 857 068 718 644 527 104;
  • 39) 0.142 857 142 857 068 718 644 527 104 × 2 = 0 + 0.285 714 285 714 137 437 289 054 208;
  • 40) 0.285 714 285 714 137 437 289 054 208 × 2 = 0 + 0.571 428 571 428 274 874 578 108 416;
  • 41) 0.571 428 571 428 274 874 578 108 416 × 2 = 1 + 0.142 857 142 856 549 749 156 216 832;
  • 42) 0.142 857 142 856 549 749 156 216 832 × 2 = 0 + 0.285 714 285 713 099 498 312 433 664;
  • 43) 0.285 714 285 713 099 498 312 433 664 × 2 = 0 + 0.571 428 571 426 198 996 624 867 328;
  • 44) 0.571 428 571 426 198 996 624 867 328 × 2 = 1 + 0.142 857 142 852 397 993 249 734 656;
  • 45) 0.142 857 142 852 397 993 249 734 656 × 2 = 0 + 0.285 714 285 704 795 986 499 469 312;
  • 46) 0.285 714 285 704 795 986 499 469 312 × 2 = 0 + 0.571 428 571 409 591 972 998 938 624;
  • 47) 0.571 428 571 409 591 972 998 938 624 × 2 = 1 + 0.142 857 142 819 183 945 997 877 248;
  • 48) 0.142 857 142 819 183 945 997 877 248 × 2 = 0 + 0.285 714 285 638 367 891 995 754 496;
  • 49) 0.285 714 285 638 367 891 995 754 496 × 2 = 0 + 0.571 428 571 276 735 783 991 508 992;
  • 50) 0.571 428 571 276 735 783 991 508 992 × 2 = 1 + 0.142 857 142 553 471 567 983 017 984;
  • 51) 0.142 857 142 553 471 567 983 017 984 × 2 = 0 + 0.285 714 285 106 943 135 966 035 968;
  • 52) 0.285 714 285 106 943 135 966 035 968 × 2 = 0 + 0.571 428 570 213 886 271 932 071 936;
  • 53) 0.571 428 570 213 886 271 932 071 936 × 2 = 1 + 0.142 857 140 427 772 543 864 143 872;

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:


0.285 714 285 714 285 714 285 714 016(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2)

5. Positive number before normalization:

6.285 714 285 714 285 714 285 714 016(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(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:


6.285 714 285 714 285 714 285 714 016(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001(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 001


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 001 =

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 6.285 714 285 714 285 714 285 714 016 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