6.285 714 285 714 285 539 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 539(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
6.285 714 285 714 285 539(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: 0.285 714 285 714 285 539.

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) 0.285 714 285 714 285 539 × 2 = 0 + 0.571 428 571 428 571 078;
  • 2) 0.571 428 571 428 571 078 × 2 = 1 + 0.142 857 142 857 142 156;
  • 3) 0.142 857 142 857 142 156 × 2 = 0 + 0.285 714 285 714 284 312;
  • 4) 0.285 714 285 714 284 312 × 2 = 0 + 0.571 428 571 428 568 624;
  • 5) 0.571 428 571 428 568 624 × 2 = 1 + 0.142 857 142 857 137 248;
  • 6) 0.142 857 142 857 137 248 × 2 = 0 + 0.285 714 285 714 274 496;
  • 7) 0.285 714 285 714 274 496 × 2 = 0 + 0.571 428 571 428 548 992;
  • 8) 0.571 428 571 428 548 992 × 2 = 1 + 0.142 857 142 857 097 984;
  • 9) 0.142 857 142 857 097 984 × 2 = 0 + 0.285 714 285 714 195 968;
  • 10) 0.285 714 285 714 195 968 × 2 = 0 + 0.571 428 571 428 391 936;
  • 11) 0.571 428 571 428 391 936 × 2 = 1 + 0.142 857 142 856 783 872;
  • 12) 0.142 857 142 856 783 872 × 2 = 0 + 0.285 714 285 713 567 744;
  • 13) 0.285 714 285 713 567 744 × 2 = 0 + 0.571 428 571 427 135 488;
  • 14) 0.571 428 571 427 135 488 × 2 = 1 + 0.142 857 142 854 270 976;
  • 15) 0.142 857 142 854 270 976 × 2 = 0 + 0.285 714 285 708 541 952;
  • 16) 0.285 714 285 708 541 952 × 2 = 0 + 0.571 428 571 417 083 904;
  • 17) 0.571 428 571 417 083 904 × 2 = 1 + 0.142 857 142 834 167 808;
  • 18) 0.142 857 142 834 167 808 × 2 = 0 + 0.285 714 285 668 335 616;
  • 19) 0.285 714 285 668 335 616 × 2 = 0 + 0.571 428 571 336 671 232;
  • 20) 0.571 428 571 336 671 232 × 2 = 1 + 0.142 857 142 673 342 464;
  • 21) 0.142 857 142 673 342 464 × 2 = 0 + 0.285 714 285 346 684 928;
  • 22) 0.285 714 285 346 684 928 × 2 = 0 + 0.571 428 570 693 369 856;
  • 23) 0.571 428 570 693 369 856 × 2 = 1 + 0.142 857 141 386 739 712;
  • 24) 0.142 857 141 386 739 712 × 2 = 0 + 0.285 714 282 773 479 424;
  • 25) 0.285 714 282 773 479 424 × 2 = 0 + 0.571 428 565 546 958 848;
  • 26) 0.571 428 565 546 958 848 × 2 = 1 + 0.142 857 131 093 917 696;
  • 27) 0.142 857 131 093 917 696 × 2 = 0 + 0.285 714 262 187 835 392;
  • 28) 0.285 714 262 187 835 392 × 2 = 0 + 0.571 428 524 375 670 784;
  • 29) 0.571 428 524 375 670 784 × 2 = 1 + 0.142 857 048 751 341 568;
  • 30) 0.142 857 048 751 341 568 × 2 = 0 + 0.285 714 097 502 683 136;
  • 31) 0.285 714 097 502 683 136 × 2 = 0 + 0.571 428 195 005 366 272;
  • 32) 0.571 428 195 005 366 272 × 2 = 1 + 0.142 856 390 010 732 544;
  • 33) 0.142 856 390 010 732 544 × 2 = 0 + 0.285 712 780 021 465 088;
  • 34) 0.285 712 780 021 465 088 × 2 = 0 + 0.571 425 560 042 930 176;
  • 35) 0.571 425 560 042 930 176 × 2 = 1 + 0.142 851 120 085 860 352;
  • 36) 0.142 851 120 085 860 352 × 2 = 0 + 0.285 702 240 171 720 704;
  • 37) 0.285 702 240 171 720 704 × 2 = 0 + 0.571 404 480 343 441 408;
  • 38) 0.571 404 480 343 441 408 × 2 = 1 + 0.142 808 960 686 882 816;
  • 39) 0.142 808 960 686 882 816 × 2 = 0 + 0.285 617 921 373 765 632;
  • 40) 0.285 617 921 373 765 632 × 2 = 0 + 0.571 235 842 747 531 264;
  • 41) 0.571 235 842 747 531 264 × 2 = 1 + 0.142 471 685 495 062 528;
  • 42) 0.142 471 685 495 062 528 × 2 = 0 + 0.284 943 370 990 125 056;
  • 43) 0.284 943 370 990 125 056 × 2 = 0 + 0.569 886 741 980 250 112;
  • 44) 0.569 886 741 980 250 112 × 2 = 1 + 0.139 773 483 960 500 224;
  • 45) 0.139 773 483 960 500 224 × 2 = 0 + 0.279 546 967 921 000 448;
  • 46) 0.279 546 967 921 000 448 × 2 = 0 + 0.559 093 935 842 000 896;
  • 47) 0.559 093 935 842 000 896 × 2 = 1 + 0.118 187 871 684 001 792;
  • 48) 0.118 187 871 684 001 792 × 2 = 0 + 0.236 375 743 368 003 584;
  • 49) 0.236 375 743 368 003 584 × 2 = 0 + 0.472 751 486 736 007 168;
  • 50) 0.472 751 486 736 007 168 × 2 = 0 + 0.945 502 973 472 014 336;
  • 51) 0.945 502 973 472 014 336 × 2 = 1 + 0.891 005 946 944 028 672;
  • 52) 0.891 005 946 944 028 672 × 2 = 1 + 0.782 011 893 888 057 344;
  • 53) 0.782 011 893 888 057 344 × 2 = 1 + 0.564 023 787 776 114 688;

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:


0.285 714 285 714 285 539(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1(2)

5. Positive number before normalization:

6.285 714 285 714 285 539(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 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:


6.285 714 285 714 285 539(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 111(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 1000 111


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 1000 111 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000


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 1000


Decimal number 6.285 714 285 714 285 539 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 1000

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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