0.002 204 718 637 485 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.002 204 718 637 485(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
0.002 204 718 637 485(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.002 204 718 637 485.

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) 0.002 204 718 637 485 × 2 = 0 + 0.004 409 437 274 97;
  • 2) 0.004 409 437 274 97 × 2 = 0 + 0.008 818 874 549 94;
  • 3) 0.008 818 874 549 94 × 2 = 0 + 0.017 637 749 099 88;
  • 4) 0.017 637 749 099 88 × 2 = 0 + 0.035 275 498 199 76;
  • 5) 0.035 275 498 199 76 × 2 = 0 + 0.070 550 996 399 52;
  • 6) 0.070 550 996 399 52 × 2 = 0 + 0.141 101 992 799 04;
  • 7) 0.141 101 992 799 04 × 2 = 0 + 0.282 203 985 598 08;
  • 8) 0.282 203 985 598 08 × 2 = 0 + 0.564 407 971 196 16;
  • 9) 0.564 407 971 196 16 × 2 = 1 + 0.128 815 942 392 32;
  • 10) 0.128 815 942 392 32 × 2 = 0 + 0.257 631 884 784 64;
  • 11) 0.257 631 884 784 64 × 2 = 0 + 0.515 263 769 569 28;
  • 12) 0.515 263 769 569 28 × 2 = 1 + 0.030 527 539 138 56;
  • 13) 0.030 527 539 138 56 × 2 = 0 + 0.061 055 078 277 12;
  • 14) 0.061 055 078 277 12 × 2 = 0 + 0.122 110 156 554 24;
  • 15) 0.122 110 156 554 24 × 2 = 0 + 0.244 220 313 108 48;
  • 16) 0.244 220 313 108 48 × 2 = 0 + 0.488 440 626 216 96;
  • 17) 0.488 440 626 216 96 × 2 = 0 + 0.976 881 252 433 92;
  • 18) 0.976 881 252 433 92 × 2 = 1 + 0.953 762 504 867 84;
  • 19) 0.953 762 504 867 84 × 2 = 1 + 0.907 525 009 735 68;
  • 20) 0.907 525 009 735 68 × 2 = 1 + 0.815 050 019 471 36;
  • 21) 0.815 050 019 471 36 × 2 = 1 + 0.630 100 038 942 72;
  • 22) 0.630 100 038 942 72 × 2 = 1 + 0.260 200 077 885 44;
  • 23) 0.260 200 077 885 44 × 2 = 0 + 0.520 400 155 770 88;
  • 24) 0.520 400 155 770 88 × 2 = 1 + 0.040 800 311 541 76;
  • 25) 0.040 800 311 541 76 × 2 = 0 + 0.081 600 623 083 52;
  • 26) 0.081 600 623 083 52 × 2 = 0 + 0.163 201 246 167 04;
  • 27) 0.163 201 246 167 04 × 2 = 0 + 0.326 402 492 334 08;
  • 28) 0.326 402 492 334 08 × 2 = 0 + 0.652 804 984 668 16;
  • 29) 0.652 804 984 668 16 × 2 = 1 + 0.305 609 969 336 32;
  • 30) 0.305 609 969 336 32 × 2 = 0 + 0.611 219 938 672 64;
  • 31) 0.611 219 938 672 64 × 2 = 1 + 0.222 439 877 345 28;
  • 32) 0.222 439 877 345 28 × 2 = 0 + 0.444 879 754 690 56;
  • 33) 0.444 879 754 690 56 × 2 = 0 + 0.889 759 509 381 12;
  • 34) 0.889 759 509 381 12 × 2 = 1 + 0.779 519 018 762 24;
  • 35) 0.779 519 018 762 24 × 2 = 1 + 0.559 038 037 524 48;
  • 36) 0.559 038 037 524 48 × 2 = 1 + 0.118 076 075 048 96;
  • 37) 0.118 076 075 048 96 × 2 = 0 + 0.236 152 150 097 92;
  • 38) 0.236 152 150 097 92 × 2 = 0 + 0.472 304 300 195 84;
  • 39) 0.472 304 300 195 84 × 2 = 0 + 0.944 608 600 391 68;
  • 40) 0.944 608 600 391 68 × 2 = 1 + 0.889 217 200 783 36;
  • 41) 0.889 217 200 783 36 × 2 = 1 + 0.778 434 401 566 72;
  • 42) 0.778 434 401 566 72 × 2 = 1 + 0.556 868 803 133 44;
  • 43) 0.556 868 803 133 44 × 2 = 1 + 0.113 737 606 266 88;
  • 44) 0.113 737 606 266 88 × 2 = 0 + 0.227 475 212 533 76;
  • 45) 0.227 475 212 533 76 × 2 = 0 + 0.454 950 425 067 52;
  • 46) 0.454 950 425 067 52 × 2 = 0 + 0.909 900 850 135 04;
  • 47) 0.909 900 850 135 04 × 2 = 1 + 0.819 801 700 270 08;
  • 48) 0.819 801 700 270 08 × 2 = 1 + 0.639 603 400 540 16;
  • 49) 0.639 603 400 540 16 × 2 = 1 + 0.279 206 801 080 32;
  • 50) 0.279 206 801 080 32 × 2 = 0 + 0.558 413 602 160 64;
  • 51) 0.558 413 602 160 64 × 2 = 1 + 0.116 827 204 321 28;
  • 52) 0.116 827 204 321 28 × 2 = 0 + 0.233 654 408 642 56;
  • 53) 0.233 654 408 642 56 × 2 = 0 + 0.467 308 817 285 12;
  • 54) 0.467 308 817 285 12 × 2 = 0 + 0.934 617 634 570 24;
  • 55) 0.934 617 634 570 24 × 2 = 1 + 0.869 235 269 140 48;
  • 56) 0.869 235 269 140 48 × 2 = 1 + 0.738 470 538 280 96;
  • 57) 0.738 470 538 280 96 × 2 = 1 + 0.476 941 076 561 92;
  • 58) 0.476 941 076 561 92 × 2 = 0 + 0.953 882 153 123 84;
  • 59) 0.953 882 153 123 84 × 2 = 1 + 0.907 764 306 247 68;
  • 60) 0.907 764 306 247 68 × 2 = 1 + 0.815 528 612 495 36;
  • 61) 0.815 528 612 495 36 × 2 = 1 + 0.631 057 224 990 72;

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:


0.002 204 718 637 485(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1110 0011 1010 0011 1011 1(2)

5. Positive number before normalization:

0.002 204 718 637 485(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1110 0011 1010 0011 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.002 204 718 637 485(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1110 0011 1010 0011 1011 1(2) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1110 0011 1010 0011 1011 1(2) × 20 =

1.0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111(2) × 2-9


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): -9

Mantissa (not normalized):
1.0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-9 + 2(11-1) - 1 =

(-9 + 1 023)(10) =

1 014(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;
  • 1 014 ÷ 2 = 507 + 0;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1014(10) =

011 1111 0110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111 =

0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111


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) =
011 1111 0110

Mantissa (52 bits) =
0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111


Decimal number 0.002 204 718 637 485 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0110 - 0010 0000 1111 1010 0001 0100 1110 0011 1100 0111 0100 0111 0111

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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