6.285 714 285 714 285 714 285 713 985 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 714 285 713 985(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
6.285 714 285 714 285 714 285 713 985(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: 0.285 714 285 714 285 714 285 713 985.

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) 0.285 714 285 714 285 714 285 713 985 × 2 = 0 + 0.571 428 571 428 571 428 571 427 97;
  • 2) 0.571 428 571 428 571 428 571 427 97 × 2 = 1 + 0.142 857 142 857 142 857 142 855 94;
  • 3) 0.142 857 142 857 142 857 142 855 94 × 2 = 0 + 0.285 714 285 714 285 714 285 711 88;
  • 4) 0.285 714 285 714 285 714 285 711 88 × 2 = 0 + 0.571 428 571 428 571 428 571 423 76;
  • 5) 0.571 428 571 428 571 428 571 423 76 × 2 = 1 + 0.142 857 142 857 142 857 142 847 52;
  • 6) 0.142 857 142 857 142 857 142 847 52 × 2 = 0 + 0.285 714 285 714 285 714 285 695 04;
  • 7) 0.285 714 285 714 285 714 285 695 04 × 2 = 0 + 0.571 428 571 428 571 428 571 390 08;
  • 8) 0.571 428 571 428 571 428 571 390 08 × 2 = 1 + 0.142 857 142 857 142 857 142 780 16;
  • 9) 0.142 857 142 857 142 857 142 780 16 × 2 = 0 + 0.285 714 285 714 285 714 285 560 32;
  • 10) 0.285 714 285 714 285 714 285 560 32 × 2 = 0 + 0.571 428 571 428 571 428 571 120 64;
  • 11) 0.571 428 571 428 571 428 571 120 64 × 2 = 1 + 0.142 857 142 857 142 857 142 241 28;
  • 12) 0.142 857 142 857 142 857 142 241 28 × 2 = 0 + 0.285 714 285 714 285 714 284 482 56;
  • 13) 0.285 714 285 714 285 714 284 482 56 × 2 = 0 + 0.571 428 571 428 571 428 568 965 12;
  • 14) 0.571 428 571 428 571 428 568 965 12 × 2 = 1 + 0.142 857 142 857 142 857 137 930 24;
  • 15) 0.142 857 142 857 142 857 137 930 24 × 2 = 0 + 0.285 714 285 714 285 714 275 860 48;
  • 16) 0.285 714 285 714 285 714 275 860 48 × 2 = 0 + 0.571 428 571 428 571 428 551 720 96;
  • 17) 0.571 428 571 428 571 428 551 720 96 × 2 = 1 + 0.142 857 142 857 142 857 103 441 92;
  • 18) 0.142 857 142 857 142 857 103 441 92 × 2 = 0 + 0.285 714 285 714 285 714 206 883 84;
  • 19) 0.285 714 285 714 285 714 206 883 84 × 2 = 0 + 0.571 428 571 428 571 428 413 767 68;
  • 20) 0.571 428 571 428 571 428 413 767 68 × 2 = 1 + 0.142 857 142 857 142 856 827 535 36;
  • 21) 0.142 857 142 857 142 856 827 535 36 × 2 = 0 + 0.285 714 285 714 285 713 655 070 72;
  • 22) 0.285 714 285 714 285 713 655 070 72 × 2 = 0 + 0.571 428 571 428 571 427 310 141 44;
  • 23) 0.571 428 571 428 571 427 310 141 44 × 2 = 1 + 0.142 857 142 857 142 854 620 282 88;
  • 24) 0.142 857 142 857 142 854 620 282 88 × 2 = 0 + 0.285 714 285 714 285 709 240 565 76;
  • 25) 0.285 714 285 714 285 709 240 565 76 × 2 = 0 + 0.571 428 571 428 571 418 481 131 52;
  • 26) 0.571 428 571 428 571 418 481 131 52 × 2 = 1 + 0.142 857 142 857 142 836 962 263 04;
  • 27) 0.142 857 142 857 142 836 962 263 04 × 2 = 0 + 0.285 714 285 714 285 673 924 526 08;
  • 28) 0.285 714 285 714 285 673 924 526 08 × 2 = 0 + 0.571 428 571 428 571 347 849 052 16;
  • 29) 0.571 428 571 428 571 347 849 052 16 × 2 = 1 + 0.142 857 142 857 142 695 698 104 32;
  • 30) 0.142 857 142 857 142 695 698 104 32 × 2 = 0 + 0.285 714 285 714 285 391 396 208 64;
  • 31) 0.285 714 285 714 285 391 396 208 64 × 2 = 0 + 0.571 428 571 428 570 782 792 417 28;
  • 32) 0.571 428 571 428 570 782 792 417 28 × 2 = 1 + 0.142 857 142 857 141 565 584 834 56;
  • 33) 0.142 857 142 857 141 565 584 834 56 × 2 = 0 + 0.285 714 285 714 283 131 169 669 12;
  • 34) 0.285 714 285 714 283 131 169 669 12 × 2 = 0 + 0.571 428 571 428 566 262 339 338 24;
  • 35) 0.571 428 571 428 566 262 339 338 24 × 2 = 1 + 0.142 857 142 857 132 524 678 676 48;
  • 36) 0.142 857 142 857 132 524 678 676 48 × 2 = 0 + 0.285 714 285 714 265 049 357 352 96;
  • 37) 0.285 714 285 714 265 049 357 352 96 × 2 = 0 + 0.571 428 571 428 530 098 714 705 92;
  • 38) 0.571 428 571 428 530 098 714 705 92 × 2 = 1 + 0.142 857 142 857 060 197 429 411 84;
  • 39) 0.142 857 142 857 060 197 429 411 84 × 2 = 0 + 0.285 714 285 714 120 394 858 823 68;
  • 40) 0.285 714 285 714 120 394 858 823 68 × 2 = 0 + 0.571 428 571 428 240 789 717 647 36;
  • 41) 0.571 428 571 428 240 789 717 647 36 × 2 = 1 + 0.142 857 142 856 481 579 435 294 72;
  • 42) 0.142 857 142 856 481 579 435 294 72 × 2 = 0 + 0.285 714 285 712 963 158 870 589 44;
  • 43) 0.285 714 285 712 963 158 870 589 44 × 2 = 0 + 0.571 428 571 425 926 317 741 178 88;
  • 44) 0.571 428 571 425 926 317 741 178 88 × 2 = 1 + 0.142 857 142 851 852 635 482 357 76;
  • 45) 0.142 857 142 851 852 635 482 357 76 × 2 = 0 + 0.285 714 285 703 705 270 964 715 52;
  • 46) 0.285 714 285 703 705 270 964 715 52 × 2 = 0 + 0.571 428 571 407 410 541 929 431 04;
  • 47) 0.571 428 571 407 410 541 929 431 04 × 2 = 1 + 0.142 857 142 814 821 083 858 862 08;
  • 48) 0.142 857 142 814 821 083 858 862 08 × 2 = 0 + 0.285 714 285 629 642 167 717 724 16;
  • 49) 0.285 714 285 629 642 167 717 724 16 × 2 = 0 + 0.571 428 571 259 284 335 435 448 32;
  • 50) 0.571 428 571 259 284 335 435 448 32 × 2 = 1 + 0.142 857 142 518 568 670 870 896 64;
  • 51) 0.142 857 142 518 568 670 870 896 64 × 2 = 0 + 0.285 714 285 037 137 341 741 793 28;
  • 52) 0.285 714 285 037 137 341 741 793 28 × 2 = 0 + 0.571 428 570 074 274 683 483 586 56;
  • 53) 0.571 428 570 074 274 683 483 586 56 × 2 = 1 + 0.142 857 140 148 549 366 967 173 12;

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:


0.285 714 285 714 285 714 285 713 985(10) =

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

5. Positive number before normalization:

6.285 714 285 714 285 714 285 713 985(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:


6.285 714 285 714 285 714 285 713 985(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 6.285 714 285 714 285 714 285 713 985 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