6.285 714 285 714 285 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 23(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 23(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 23.

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 23 × 2 = 0 + 0.571 428 571 428 570 46;
  • 2) 0.571 428 571 428 570 46 × 2 = 1 + 0.142 857 142 857 140 92;
  • 3) 0.142 857 142 857 140 92 × 2 = 0 + 0.285 714 285 714 281 84;
  • 4) 0.285 714 285 714 281 84 × 2 = 0 + 0.571 428 571 428 563 68;
  • 5) 0.571 428 571 428 563 68 × 2 = 1 + 0.142 857 142 857 127 36;
  • 6) 0.142 857 142 857 127 36 × 2 = 0 + 0.285 714 285 714 254 72;
  • 7) 0.285 714 285 714 254 72 × 2 = 0 + 0.571 428 571 428 509 44;
  • 8) 0.571 428 571 428 509 44 × 2 = 1 + 0.142 857 142 857 018 88;
  • 9) 0.142 857 142 857 018 88 × 2 = 0 + 0.285 714 285 714 037 76;
  • 10) 0.285 714 285 714 037 76 × 2 = 0 + 0.571 428 571 428 075 52;
  • 11) 0.571 428 571 428 075 52 × 2 = 1 + 0.142 857 142 856 151 04;
  • 12) 0.142 857 142 856 151 04 × 2 = 0 + 0.285 714 285 712 302 08;
  • 13) 0.285 714 285 712 302 08 × 2 = 0 + 0.571 428 571 424 604 16;
  • 14) 0.571 428 571 424 604 16 × 2 = 1 + 0.142 857 142 849 208 32;
  • 15) 0.142 857 142 849 208 32 × 2 = 0 + 0.285 714 285 698 416 64;
  • 16) 0.285 714 285 698 416 64 × 2 = 0 + 0.571 428 571 396 833 28;
  • 17) 0.571 428 571 396 833 28 × 2 = 1 + 0.142 857 142 793 666 56;
  • 18) 0.142 857 142 793 666 56 × 2 = 0 + 0.285 714 285 587 333 12;
  • 19) 0.285 714 285 587 333 12 × 2 = 0 + 0.571 428 571 174 666 24;
  • 20) 0.571 428 571 174 666 24 × 2 = 1 + 0.142 857 142 349 332 48;
  • 21) 0.142 857 142 349 332 48 × 2 = 0 + 0.285 714 284 698 664 96;
  • 22) 0.285 714 284 698 664 96 × 2 = 0 + 0.571 428 569 397 329 92;
  • 23) 0.571 428 569 397 329 92 × 2 = 1 + 0.142 857 138 794 659 84;
  • 24) 0.142 857 138 794 659 84 × 2 = 0 + 0.285 714 277 589 319 68;
  • 25) 0.285 714 277 589 319 68 × 2 = 0 + 0.571 428 555 178 639 36;
  • 26) 0.571 428 555 178 639 36 × 2 = 1 + 0.142 857 110 357 278 72;
  • 27) 0.142 857 110 357 278 72 × 2 = 0 + 0.285 714 220 714 557 44;
  • 28) 0.285 714 220 714 557 44 × 2 = 0 + 0.571 428 441 429 114 88;
  • 29) 0.571 428 441 429 114 88 × 2 = 1 + 0.142 856 882 858 229 76;
  • 30) 0.142 856 882 858 229 76 × 2 = 0 + 0.285 713 765 716 459 52;
  • 31) 0.285 713 765 716 459 52 × 2 = 0 + 0.571 427 531 432 919 04;
  • 32) 0.571 427 531 432 919 04 × 2 = 1 + 0.142 855 062 865 838 08;
  • 33) 0.142 855 062 865 838 08 × 2 = 0 + 0.285 710 125 731 676 16;
  • 34) 0.285 710 125 731 676 16 × 2 = 0 + 0.571 420 251 463 352 32;
  • 35) 0.571 420 251 463 352 32 × 2 = 1 + 0.142 840 502 926 704 64;
  • 36) 0.142 840 502 926 704 64 × 2 = 0 + 0.285 681 005 853 409 28;
  • 37) 0.285 681 005 853 409 28 × 2 = 0 + 0.571 362 011 706 818 56;
  • 38) 0.571 362 011 706 818 56 × 2 = 1 + 0.142 724 023 413 637 12;
  • 39) 0.142 724 023 413 637 12 × 2 = 0 + 0.285 448 046 827 274 24;
  • 40) 0.285 448 046 827 274 24 × 2 = 0 + 0.570 896 093 654 548 48;
  • 41) 0.570 896 093 654 548 48 × 2 = 1 + 0.141 792 187 309 096 96;
  • 42) 0.141 792 187 309 096 96 × 2 = 0 + 0.283 584 374 618 193 92;
  • 43) 0.283 584 374 618 193 92 × 2 = 0 + 0.567 168 749 236 387 84;
  • 44) 0.567 168 749 236 387 84 × 2 = 1 + 0.134 337 498 472 775 68;
  • 45) 0.134 337 498 472 775 68 × 2 = 0 + 0.268 674 996 945 551 36;
  • 46) 0.268 674 996 945 551 36 × 2 = 0 + 0.537 349 993 891 102 72;
  • 47) 0.537 349 993 891 102 72 × 2 = 1 + 0.074 699 987 782 205 44;
  • 48) 0.074 699 987 782 205 44 × 2 = 0 + 0.149 399 975 564 410 88;
  • 49) 0.149 399 975 564 410 88 × 2 = 0 + 0.298 799 951 128 821 76;
  • 50) 0.298 799 951 128 821 76 × 2 = 0 + 0.597 599 902 257 643 52;
  • 51) 0.597 599 902 257 643 52 × 2 = 1 + 0.195 199 804 515 287 04;
  • 52) 0.195 199 804 515 287 04 × 2 = 0 + 0.390 399 609 030 574 08;
  • 53) 0.390 399 609 030 574 08 × 2 = 0 + 0.780 799 218 061 148 16;

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 23(10) =

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

5. Positive number before normalization:

6.285 714 285 714 285 23(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0010 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 23(10) =

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

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

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 100(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 100


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 100 =

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 23 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