2.225 073 858 507 201 382 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.225 073 858 507 201 382 42(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
2.225 073 858 507 201 382 42(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: 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: 0.225 073 858 507 201 382 42.

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) 0.225 073 858 507 201 382 42 × 2 = 0 + 0.450 147 717 014 402 764 84;
  • 2) 0.450 147 717 014 402 764 84 × 2 = 0 + 0.900 295 434 028 805 529 68;
  • 3) 0.900 295 434 028 805 529 68 × 2 = 1 + 0.800 590 868 057 611 059 36;
  • 4) 0.800 590 868 057 611 059 36 × 2 = 1 + 0.601 181 736 115 222 118 72;
  • 5) 0.601 181 736 115 222 118 72 × 2 = 1 + 0.202 363 472 230 444 237 44;
  • 6) 0.202 363 472 230 444 237 44 × 2 = 0 + 0.404 726 944 460 888 474 88;
  • 7) 0.404 726 944 460 888 474 88 × 2 = 0 + 0.809 453 888 921 776 949 76;
  • 8) 0.809 453 888 921 776 949 76 × 2 = 1 + 0.618 907 777 843 553 899 52;
  • 9) 0.618 907 777 843 553 899 52 × 2 = 1 + 0.237 815 555 687 107 799 04;
  • 10) 0.237 815 555 687 107 799 04 × 2 = 0 + 0.475 631 111 374 215 598 08;
  • 11) 0.475 631 111 374 215 598 08 × 2 = 0 + 0.951 262 222 748 431 196 16;
  • 12) 0.951 262 222 748 431 196 16 × 2 = 1 + 0.902 524 445 496 862 392 32;
  • 13) 0.902 524 445 496 862 392 32 × 2 = 1 + 0.805 048 890 993 724 784 64;
  • 14) 0.805 048 890 993 724 784 64 × 2 = 1 + 0.610 097 781 987 449 569 28;
  • 15) 0.610 097 781 987 449 569 28 × 2 = 1 + 0.220 195 563 974 899 138 56;
  • 16) 0.220 195 563 974 899 138 56 × 2 = 0 + 0.440 391 127 949 798 277 12;
  • 17) 0.440 391 127 949 798 277 12 × 2 = 0 + 0.880 782 255 899 596 554 24;
  • 18) 0.880 782 255 899 596 554 24 × 2 = 1 + 0.761 564 511 799 193 108 48;
  • 19) 0.761 564 511 799 193 108 48 × 2 = 1 + 0.523 129 023 598 386 216 96;
  • 20) 0.523 129 023 598 386 216 96 × 2 = 1 + 0.046 258 047 196 772 433 92;
  • 21) 0.046 258 047 196 772 433 92 × 2 = 0 + 0.092 516 094 393 544 867 84;
  • 22) 0.092 516 094 393 544 867 84 × 2 = 0 + 0.185 032 188 787 089 735 68;
  • 23) 0.185 032 188 787 089 735 68 × 2 = 0 + 0.370 064 377 574 179 471 36;
  • 24) 0.370 064 377 574 179 471 36 × 2 = 0 + 0.740 128 755 148 358 942 72;
  • 25) 0.740 128 755 148 358 942 72 × 2 = 1 + 0.480 257 510 296 717 885 44;
  • 26) 0.480 257 510 296 717 885 44 × 2 = 0 + 0.960 515 020 593 435 770 88;
  • 27) 0.960 515 020 593 435 770 88 × 2 = 1 + 0.921 030 041 186 871 541 76;
  • 28) 0.921 030 041 186 871 541 76 × 2 = 1 + 0.842 060 082 373 743 083 52;
  • 29) 0.842 060 082 373 743 083 52 × 2 = 1 + 0.684 120 164 747 486 167 04;
  • 30) 0.684 120 164 747 486 167 04 × 2 = 1 + 0.368 240 329 494 972 334 08;
  • 31) 0.368 240 329 494 972 334 08 × 2 = 0 + 0.736 480 658 989 944 668 16;
  • 32) 0.736 480 658 989 944 668 16 × 2 = 1 + 0.472 961 317 979 889 336 32;
  • 33) 0.472 961 317 979 889 336 32 × 2 = 0 + 0.945 922 635 959 778 672 64;
  • 34) 0.945 922 635 959 778 672 64 × 2 = 1 + 0.891 845 271 919 557 345 28;
  • 35) 0.891 845 271 919 557 345 28 × 2 = 1 + 0.783 690 543 839 114 690 56;
  • 36) 0.783 690 543 839 114 690 56 × 2 = 1 + 0.567 381 087 678 229 381 12;
  • 37) 0.567 381 087 678 229 381 12 × 2 = 1 + 0.134 762 175 356 458 762 24;
  • 38) 0.134 762 175 356 458 762 24 × 2 = 0 + 0.269 524 350 712 917 524 48;
  • 39) 0.269 524 350 712 917 524 48 × 2 = 0 + 0.539 048 701 425 835 048 96;
  • 40) 0.539 048 701 425 835 048 96 × 2 = 1 + 0.078 097 402 851 670 097 92;
  • 41) 0.078 097 402 851 670 097 92 × 2 = 0 + 0.156 194 805 703 340 195 84;
  • 42) 0.156 194 805 703 340 195 84 × 2 = 0 + 0.312 389 611 406 680 391 68;
  • 43) 0.312 389 611 406 680 391 68 × 2 = 0 + 0.624 779 222 813 360 783 36;
  • 44) 0.624 779 222 813 360 783 36 × 2 = 1 + 0.249 558 445 626 721 566 72;
  • 45) 0.249 558 445 626 721 566 72 × 2 = 0 + 0.499 116 891 253 443 133 44;
  • 46) 0.499 116 891 253 443 133 44 × 2 = 0 + 0.998 233 782 506 886 266 88;
  • 47) 0.998 233 782 506 886 266 88 × 2 = 1 + 0.996 467 565 013 772 533 76;
  • 48) 0.996 467 565 013 772 533 76 × 2 = 1 + 0.992 935 130 027 545 067 52;
  • 49) 0.992 935 130 027 545 067 52 × 2 = 1 + 0.985 870 260 055 090 135 04;
  • 50) 0.985 870 260 055 090 135 04 × 2 = 1 + 0.971 740 520 110 180 270 08;
  • 51) 0.971 740 520 110 180 270 08 × 2 = 1 + 0.943 481 040 220 360 540 16;
  • 52) 0.943 481 040 220 360 540 16 × 2 = 1 + 0.886 962 080 440 721 080 32;
  • 53) 0.886 962 080 440 721 080 32 × 2 = 1 + 0.773 924 160 881 442 160 64;

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:


0.225 073 858 507 201 382 42(10) =

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

5. Positive number before normalization:

2.225 073 858 507 201 382 42(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:


2.225 073 858 507 201 382 42(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


Decimal number 2.225 073 858 507 201 382 42 converted to 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

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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