6.285 714 285 714 284 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 284 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
6.285 714 285 714 284 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: 0.285 714 285 714 284 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) 0.285 714 285 714 284 9 × 2 = 0 + 0.571 428 571 428 569 8;
  • 2) 0.571 428 571 428 569 8 × 2 = 1 + 0.142 857 142 857 139 6;
  • 3) 0.142 857 142 857 139 6 × 2 = 0 + 0.285 714 285 714 279 2;
  • 4) 0.285 714 285 714 279 2 × 2 = 0 + 0.571 428 571 428 558 4;
  • 5) 0.571 428 571 428 558 4 × 2 = 1 + 0.142 857 142 857 116 8;
  • 6) 0.142 857 142 857 116 8 × 2 = 0 + 0.285 714 285 714 233 6;
  • 7) 0.285 714 285 714 233 6 × 2 = 0 + 0.571 428 571 428 467 2;
  • 8) 0.571 428 571 428 467 2 × 2 = 1 + 0.142 857 142 856 934 4;
  • 9) 0.142 857 142 856 934 4 × 2 = 0 + 0.285 714 285 713 868 8;
  • 10) 0.285 714 285 713 868 8 × 2 = 0 + 0.571 428 571 427 737 6;
  • 11) 0.571 428 571 427 737 6 × 2 = 1 + 0.142 857 142 855 475 2;
  • 12) 0.142 857 142 855 475 2 × 2 = 0 + 0.285 714 285 710 950 4;
  • 13) 0.285 714 285 710 950 4 × 2 = 0 + 0.571 428 571 421 900 8;
  • 14) 0.571 428 571 421 900 8 × 2 = 1 + 0.142 857 142 843 801 6;
  • 15) 0.142 857 142 843 801 6 × 2 = 0 + 0.285 714 285 687 603 2;
  • 16) 0.285 714 285 687 603 2 × 2 = 0 + 0.571 428 571 375 206 4;
  • 17) 0.571 428 571 375 206 4 × 2 = 1 + 0.142 857 142 750 412 8;
  • 18) 0.142 857 142 750 412 8 × 2 = 0 + 0.285 714 285 500 825 6;
  • 19) 0.285 714 285 500 825 6 × 2 = 0 + 0.571 428 571 001 651 2;
  • 20) 0.571 428 571 001 651 2 × 2 = 1 + 0.142 857 142 003 302 4;
  • 21) 0.142 857 142 003 302 4 × 2 = 0 + 0.285 714 284 006 604 8;
  • 22) 0.285 714 284 006 604 8 × 2 = 0 + 0.571 428 568 013 209 6;
  • 23) 0.571 428 568 013 209 6 × 2 = 1 + 0.142 857 136 026 419 2;
  • 24) 0.142 857 136 026 419 2 × 2 = 0 + 0.285 714 272 052 838 4;
  • 25) 0.285 714 272 052 838 4 × 2 = 0 + 0.571 428 544 105 676 8;
  • 26) 0.571 428 544 105 676 8 × 2 = 1 + 0.142 857 088 211 353 6;
  • 27) 0.142 857 088 211 353 6 × 2 = 0 + 0.285 714 176 422 707 2;
  • 28) 0.285 714 176 422 707 2 × 2 = 0 + 0.571 428 352 845 414 4;
  • 29) 0.571 428 352 845 414 4 × 2 = 1 + 0.142 856 705 690 828 8;
  • 30) 0.142 856 705 690 828 8 × 2 = 0 + 0.285 713 411 381 657 6;
  • 31) 0.285 713 411 381 657 6 × 2 = 0 + 0.571 426 822 763 315 2;
  • 32) 0.571 426 822 763 315 2 × 2 = 1 + 0.142 853 645 526 630 4;
  • 33) 0.142 853 645 526 630 4 × 2 = 0 + 0.285 707 291 053 260 8;
  • 34) 0.285 707 291 053 260 8 × 2 = 0 + 0.571 414 582 106 521 6;
  • 35) 0.571 414 582 106 521 6 × 2 = 1 + 0.142 829 164 213 043 2;
  • 36) 0.142 829 164 213 043 2 × 2 = 0 + 0.285 658 328 426 086 4;
  • 37) 0.285 658 328 426 086 4 × 2 = 0 + 0.571 316 656 852 172 8;
  • 38) 0.571 316 656 852 172 8 × 2 = 1 + 0.142 633 313 704 345 6;
  • 39) 0.142 633 313 704 345 6 × 2 = 0 + 0.285 266 627 408 691 2;
  • 40) 0.285 266 627 408 691 2 × 2 = 0 + 0.570 533 254 817 382 4;
  • 41) 0.570 533 254 817 382 4 × 2 = 1 + 0.141 066 509 634 764 8;
  • 42) 0.141 066 509 634 764 8 × 2 = 0 + 0.282 133 019 269 529 6;
  • 43) 0.282 133 019 269 529 6 × 2 = 0 + 0.564 266 038 539 059 2;
  • 44) 0.564 266 038 539 059 2 × 2 = 1 + 0.128 532 077 078 118 4;
  • 45) 0.128 532 077 078 118 4 × 2 = 0 + 0.257 064 154 156 236 8;
  • 46) 0.257 064 154 156 236 8 × 2 = 0 + 0.514 128 308 312 473 6;
  • 47) 0.514 128 308 312 473 6 × 2 = 1 + 0.028 256 616 624 947 2;
  • 48) 0.028 256 616 624 947 2 × 2 = 0 + 0.056 513 233 249 894 4;
  • 49) 0.056 513 233 249 894 4 × 2 = 0 + 0.113 026 466 499 788 8;
  • 50) 0.113 026 466 499 788 8 × 2 = 0 + 0.226 052 932 999 577 6;
  • 51) 0.226 052 932 999 577 6 × 2 = 0 + 0.452 105 865 999 155 2;
  • 52) 0.452 105 865 999 155 2 × 2 = 0 + 0.904 211 731 998 310 4;
  • 53) 0.904 211 731 998 310 4 × 2 = 1 + 0.808 423 463 996 620 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:


0.285 714 285 714 284 9(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0000 1(2)

5. Positive number before normalization:

6.285 714 285 714 284 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0000 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:


6.285 714 285 714 284 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0000 1(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0000 1(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 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 1000 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 1000 001 =

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 6.285 714 285 714 284 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 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