Decimal to 64 Bit IEEE 754 Binary: Convert Number 2.225 073 858 507 201 383 1 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 2.225 073 858 507 201 383 1(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.225 073 858 507 201 383 1.

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) 0.225 073 858 507 201 383 1 × 2 = 0 + 0.450 147 717 014 402 766 2;
  • 2) 0.450 147 717 014 402 766 2 × 2 = 0 + 0.900 295 434 028 805 532 4;
  • 3) 0.900 295 434 028 805 532 4 × 2 = 1 + 0.800 590 868 057 611 064 8;
  • 4) 0.800 590 868 057 611 064 8 × 2 = 1 + 0.601 181 736 115 222 129 6;
  • 5) 0.601 181 736 115 222 129 6 × 2 = 1 + 0.202 363 472 230 444 259 2;
  • 6) 0.202 363 472 230 444 259 2 × 2 = 0 + 0.404 726 944 460 888 518 4;
  • 7) 0.404 726 944 460 888 518 4 × 2 = 0 + 0.809 453 888 921 777 036 8;
  • 8) 0.809 453 888 921 777 036 8 × 2 = 1 + 0.618 907 777 843 554 073 6;
  • 9) 0.618 907 777 843 554 073 6 × 2 = 1 + 0.237 815 555 687 108 147 2;
  • 10) 0.237 815 555 687 108 147 2 × 2 = 0 + 0.475 631 111 374 216 294 4;
  • 11) 0.475 631 111 374 216 294 4 × 2 = 0 + 0.951 262 222 748 432 588 8;
  • 12) 0.951 262 222 748 432 588 8 × 2 = 1 + 0.902 524 445 496 865 177 6;
  • 13) 0.902 524 445 496 865 177 6 × 2 = 1 + 0.805 048 890 993 730 355 2;
  • 14) 0.805 048 890 993 730 355 2 × 2 = 1 + 0.610 097 781 987 460 710 4;
  • 15) 0.610 097 781 987 460 710 4 × 2 = 1 + 0.220 195 563 974 921 420 8;
  • 16) 0.220 195 563 974 921 420 8 × 2 = 0 + 0.440 391 127 949 842 841 6;
  • 17) 0.440 391 127 949 842 841 6 × 2 = 0 + 0.880 782 255 899 685 683 2;
  • 18) 0.880 782 255 899 685 683 2 × 2 = 1 + 0.761 564 511 799 371 366 4;
  • 19) 0.761 564 511 799 371 366 4 × 2 = 1 + 0.523 129 023 598 742 732 8;
  • 20) 0.523 129 023 598 742 732 8 × 2 = 1 + 0.046 258 047 197 485 465 6;
  • 21) 0.046 258 047 197 485 465 6 × 2 = 0 + 0.092 516 094 394 970 931 2;
  • 22) 0.092 516 094 394 970 931 2 × 2 = 0 + 0.185 032 188 789 941 862 4;
  • 23) 0.185 032 188 789 941 862 4 × 2 = 0 + 0.370 064 377 579 883 724 8;
  • 24) 0.370 064 377 579 883 724 8 × 2 = 0 + 0.740 128 755 159 767 449 6;
  • 25) 0.740 128 755 159 767 449 6 × 2 = 1 + 0.480 257 510 319 534 899 2;
  • 26) 0.480 257 510 319 534 899 2 × 2 = 0 + 0.960 515 020 639 069 798 4;
  • 27) 0.960 515 020 639 069 798 4 × 2 = 1 + 0.921 030 041 278 139 596 8;
  • 28) 0.921 030 041 278 139 596 8 × 2 = 1 + 0.842 060 082 556 279 193 6;
  • 29) 0.842 060 082 556 279 193 6 × 2 = 1 + 0.684 120 165 112 558 387 2;
  • 30) 0.684 120 165 112 558 387 2 × 2 = 1 + 0.368 240 330 225 116 774 4;
  • 31) 0.368 240 330 225 116 774 4 × 2 = 0 + 0.736 480 660 450 233 548 8;
  • 32) 0.736 480 660 450 233 548 8 × 2 = 1 + 0.472 961 320 900 467 097 6;
  • 33) 0.472 961 320 900 467 097 6 × 2 = 0 + 0.945 922 641 800 934 195 2;
  • 34) 0.945 922 641 800 934 195 2 × 2 = 1 + 0.891 845 283 601 868 390 4;
  • 35) 0.891 845 283 601 868 390 4 × 2 = 1 + 0.783 690 567 203 736 780 8;
  • 36) 0.783 690 567 203 736 780 8 × 2 = 1 + 0.567 381 134 407 473 561 6;
  • 37) 0.567 381 134 407 473 561 6 × 2 = 1 + 0.134 762 268 814 947 123 2;
  • 38) 0.134 762 268 814 947 123 2 × 2 = 0 + 0.269 524 537 629 894 246 4;
  • 39) 0.269 524 537 629 894 246 4 × 2 = 0 + 0.539 049 075 259 788 492 8;
  • 40) 0.539 049 075 259 788 492 8 × 2 = 1 + 0.078 098 150 519 576 985 6;
  • 41) 0.078 098 150 519 576 985 6 × 2 = 0 + 0.156 196 301 039 153 971 2;
  • 42) 0.156 196 301 039 153 971 2 × 2 = 0 + 0.312 392 602 078 307 942 4;
  • 43) 0.312 392 602 078 307 942 4 × 2 = 0 + 0.624 785 204 156 615 884 8;
  • 44) 0.624 785 204 156 615 884 8 × 2 = 1 + 0.249 570 408 313 231 769 6;
  • 45) 0.249 570 408 313 231 769 6 × 2 = 0 + 0.499 140 816 626 463 539 2;
  • 46) 0.499 140 816 626 463 539 2 × 2 = 0 + 0.998 281 633 252 927 078 4;
  • 47) 0.998 281 633 252 927 078 4 × 2 = 1 + 0.996 563 266 505 854 156 8;
  • 48) 0.996 563 266 505 854 156 8 × 2 = 1 + 0.993 126 533 011 708 313 6;
  • 49) 0.993 126 533 011 708 313 6 × 2 = 1 + 0.986 253 066 023 416 627 2;
  • 50) 0.986 253 066 023 416 627 2 × 2 = 1 + 0.972 506 132 046 833 254 4;
  • 51) 0.972 506 132 046 833 254 4 × 2 = 1 + 0.945 012 264 093 666 508 8;
  • 52) 0.945 012 264 093 666 508 8 × 2 = 1 + 0.890 024 528 187 333 017 6;
  • 53) 0.890 024 528 187 333 017 6 × 2 = 1 + 0.780 049 056 374 666 035 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:


0.225 073 858 507 201 383 1(10) =

0.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2)

5. Positive number before normalization:

2.225 073 858 507 201 383 1(10) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.225 073 858 507 201 383 1(10) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2) × 20 =

1.0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11(2) × 21


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): 1

Mantissa (not normalized):
1.0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11 =

0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111


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 0000

Mantissa (52 bits) =
0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111


The base ten decimal number 2.225 073 858 507 201 383 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111

» Decimal number in base ten 2.225 073 858 507 201 378 3 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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