6.285 714 285 714 285 688 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 688 6(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
6.285 714 285 714 285 688 6(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: 0.285 714 285 714 285 688 6.

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) 0.285 714 285 714 285 688 6 × 2 = 0 + 0.571 428 571 428 571 377 2;
  • 2) 0.571 428 571 428 571 377 2 × 2 = 1 + 0.142 857 142 857 142 754 4;
  • 3) 0.142 857 142 857 142 754 4 × 2 = 0 + 0.285 714 285 714 285 508 8;
  • 4) 0.285 714 285 714 285 508 8 × 2 = 0 + 0.571 428 571 428 571 017 6;
  • 5) 0.571 428 571 428 571 017 6 × 2 = 1 + 0.142 857 142 857 142 035 2;
  • 6) 0.142 857 142 857 142 035 2 × 2 = 0 + 0.285 714 285 714 284 070 4;
  • 7) 0.285 714 285 714 284 070 4 × 2 = 0 + 0.571 428 571 428 568 140 8;
  • 8) 0.571 428 571 428 568 140 8 × 2 = 1 + 0.142 857 142 857 136 281 6;
  • 9) 0.142 857 142 857 136 281 6 × 2 = 0 + 0.285 714 285 714 272 563 2;
  • 10) 0.285 714 285 714 272 563 2 × 2 = 0 + 0.571 428 571 428 545 126 4;
  • 11) 0.571 428 571 428 545 126 4 × 2 = 1 + 0.142 857 142 857 090 252 8;
  • 12) 0.142 857 142 857 090 252 8 × 2 = 0 + 0.285 714 285 714 180 505 6;
  • 13) 0.285 714 285 714 180 505 6 × 2 = 0 + 0.571 428 571 428 361 011 2;
  • 14) 0.571 428 571 428 361 011 2 × 2 = 1 + 0.142 857 142 856 722 022 4;
  • 15) 0.142 857 142 856 722 022 4 × 2 = 0 + 0.285 714 285 713 444 044 8;
  • 16) 0.285 714 285 713 444 044 8 × 2 = 0 + 0.571 428 571 426 888 089 6;
  • 17) 0.571 428 571 426 888 089 6 × 2 = 1 + 0.142 857 142 853 776 179 2;
  • 18) 0.142 857 142 853 776 179 2 × 2 = 0 + 0.285 714 285 707 552 358 4;
  • 19) 0.285 714 285 707 552 358 4 × 2 = 0 + 0.571 428 571 415 104 716 8;
  • 20) 0.571 428 571 415 104 716 8 × 2 = 1 + 0.142 857 142 830 209 433 6;
  • 21) 0.142 857 142 830 209 433 6 × 2 = 0 + 0.285 714 285 660 418 867 2;
  • 22) 0.285 714 285 660 418 867 2 × 2 = 0 + 0.571 428 571 320 837 734 4;
  • 23) 0.571 428 571 320 837 734 4 × 2 = 1 + 0.142 857 142 641 675 468 8;
  • 24) 0.142 857 142 641 675 468 8 × 2 = 0 + 0.285 714 285 283 350 937 6;
  • 25) 0.285 714 285 283 350 937 6 × 2 = 0 + 0.571 428 570 566 701 875 2;
  • 26) 0.571 428 570 566 701 875 2 × 2 = 1 + 0.142 857 141 133 403 750 4;
  • 27) 0.142 857 141 133 403 750 4 × 2 = 0 + 0.285 714 282 266 807 500 8;
  • 28) 0.285 714 282 266 807 500 8 × 2 = 0 + 0.571 428 564 533 615 001 6;
  • 29) 0.571 428 564 533 615 001 6 × 2 = 1 + 0.142 857 129 067 230 003 2;
  • 30) 0.142 857 129 067 230 003 2 × 2 = 0 + 0.285 714 258 134 460 006 4;
  • 31) 0.285 714 258 134 460 006 4 × 2 = 0 + 0.571 428 516 268 920 012 8;
  • 32) 0.571 428 516 268 920 012 8 × 2 = 1 + 0.142 857 032 537 840 025 6;
  • 33) 0.142 857 032 537 840 025 6 × 2 = 0 + 0.285 714 065 075 680 051 2;
  • 34) 0.285 714 065 075 680 051 2 × 2 = 0 + 0.571 428 130 151 360 102 4;
  • 35) 0.571 428 130 151 360 102 4 × 2 = 1 + 0.142 856 260 302 720 204 8;
  • 36) 0.142 856 260 302 720 204 8 × 2 = 0 + 0.285 712 520 605 440 409 6;
  • 37) 0.285 712 520 605 440 409 6 × 2 = 0 + 0.571 425 041 210 880 819 2;
  • 38) 0.571 425 041 210 880 819 2 × 2 = 1 + 0.142 850 082 421 761 638 4;
  • 39) 0.142 850 082 421 761 638 4 × 2 = 0 + 0.285 700 164 843 523 276 8;
  • 40) 0.285 700 164 843 523 276 8 × 2 = 0 + 0.571 400 329 687 046 553 6;
  • 41) 0.571 400 329 687 046 553 6 × 2 = 1 + 0.142 800 659 374 093 107 2;
  • 42) 0.142 800 659 374 093 107 2 × 2 = 0 + 0.285 601 318 748 186 214 4;
  • 43) 0.285 601 318 748 186 214 4 × 2 = 0 + 0.571 202 637 496 372 428 8;
  • 44) 0.571 202 637 496 372 428 8 × 2 = 1 + 0.142 405 274 992 744 857 6;
  • 45) 0.142 405 274 992 744 857 6 × 2 = 0 + 0.284 810 549 985 489 715 2;
  • 46) 0.284 810 549 985 489 715 2 × 2 = 0 + 0.569 621 099 970 979 430 4;
  • 47) 0.569 621 099 970 979 430 4 × 2 = 1 + 0.139 242 199 941 958 860 8;
  • 48) 0.139 242 199 941 958 860 8 × 2 = 0 + 0.278 484 399 883 917 721 6;
  • 49) 0.278 484 399 883 917 721 6 × 2 = 0 + 0.556 968 799 767 835 443 2;
  • 50) 0.556 968 799 767 835 443 2 × 2 = 1 + 0.113 937 599 535 670 886 4;
  • 51) 0.113 937 599 535 670 886 4 × 2 = 0 + 0.227 875 199 071 341 772 8;
  • 52) 0.227 875 199 071 341 772 8 × 2 = 0 + 0.455 750 398 142 683 545 6;
  • 53) 0.455 750 398 142 683 545 6 × 2 = 0 + 0.911 500 796 285 367 091 2;

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:


0.285 714 285 714 285 688 6(10) =

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

5. Positive number before normalization:

6.285 714 285 714 285 688 6(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:


6.285 714 285 714 285 688 6(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 6.285 714 285 714 285 688 6 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