2.225 073 858 507 201 383 090 228 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.225 073 858 507 201 383 090 228 1(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
2.225 073 858 507 201 383 090 228 1(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: 0.225 073 858 507 201 383 090 228 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) 0.225 073 858 507 201 383 090 228 1 × 2 = 0 + 0.450 147 717 014 402 766 180 456 2;
  • 2) 0.450 147 717 014 402 766 180 456 2 × 2 = 0 + 0.900 295 434 028 805 532 360 912 4;
  • 3) 0.900 295 434 028 805 532 360 912 4 × 2 = 1 + 0.800 590 868 057 611 064 721 824 8;
  • 4) 0.800 590 868 057 611 064 721 824 8 × 2 = 1 + 0.601 181 736 115 222 129 443 649 6;
  • 5) 0.601 181 736 115 222 129 443 649 6 × 2 = 1 + 0.202 363 472 230 444 258 887 299 2;
  • 6) 0.202 363 472 230 444 258 887 299 2 × 2 = 0 + 0.404 726 944 460 888 517 774 598 4;
  • 7) 0.404 726 944 460 888 517 774 598 4 × 2 = 0 + 0.809 453 888 921 777 035 549 196 8;
  • 8) 0.809 453 888 921 777 035 549 196 8 × 2 = 1 + 0.618 907 777 843 554 071 098 393 6;
  • 9) 0.618 907 777 843 554 071 098 393 6 × 2 = 1 + 0.237 815 555 687 108 142 196 787 2;
  • 10) 0.237 815 555 687 108 142 196 787 2 × 2 = 0 + 0.475 631 111 374 216 284 393 574 4;
  • 11) 0.475 631 111 374 216 284 393 574 4 × 2 = 0 + 0.951 262 222 748 432 568 787 148 8;
  • 12) 0.951 262 222 748 432 568 787 148 8 × 2 = 1 + 0.902 524 445 496 865 137 574 297 6;
  • 13) 0.902 524 445 496 865 137 574 297 6 × 2 = 1 + 0.805 048 890 993 730 275 148 595 2;
  • 14) 0.805 048 890 993 730 275 148 595 2 × 2 = 1 + 0.610 097 781 987 460 550 297 190 4;
  • 15) 0.610 097 781 987 460 550 297 190 4 × 2 = 1 + 0.220 195 563 974 921 100 594 380 8;
  • 16) 0.220 195 563 974 921 100 594 380 8 × 2 = 0 + 0.440 391 127 949 842 201 188 761 6;
  • 17) 0.440 391 127 949 842 201 188 761 6 × 2 = 0 + 0.880 782 255 899 684 402 377 523 2;
  • 18) 0.880 782 255 899 684 402 377 523 2 × 2 = 1 + 0.761 564 511 799 368 804 755 046 4;
  • 19) 0.761 564 511 799 368 804 755 046 4 × 2 = 1 + 0.523 129 023 598 737 609 510 092 8;
  • 20) 0.523 129 023 598 737 609 510 092 8 × 2 = 1 + 0.046 258 047 197 475 219 020 185 6;
  • 21) 0.046 258 047 197 475 219 020 185 6 × 2 = 0 + 0.092 516 094 394 950 438 040 371 2;
  • 22) 0.092 516 094 394 950 438 040 371 2 × 2 = 0 + 0.185 032 188 789 900 876 080 742 4;
  • 23) 0.185 032 188 789 900 876 080 742 4 × 2 = 0 + 0.370 064 377 579 801 752 161 484 8;
  • 24) 0.370 064 377 579 801 752 161 484 8 × 2 = 0 + 0.740 128 755 159 603 504 322 969 6;
  • 25) 0.740 128 755 159 603 504 322 969 6 × 2 = 1 + 0.480 257 510 319 207 008 645 939 2;
  • 26) 0.480 257 510 319 207 008 645 939 2 × 2 = 0 + 0.960 515 020 638 414 017 291 878 4;
  • 27) 0.960 515 020 638 414 017 291 878 4 × 2 = 1 + 0.921 030 041 276 828 034 583 756 8;
  • 28) 0.921 030 041 276 828 034 583 756 8 × 2 = 1 + 0.842 060 082 553 656 069 167 513 6;
  • 29) 0.842 060 082 553 656 069 167 513 6 × 2 = 1 + 0.684 120 165 107 312 138 335 027 2;
  • 30) 0.684 120 165 107 312 138 335 027 2 × 2 = 1 + 0.368 240 330 214 624 276 670 054 4;
  • 31) 0.368 240 330 214 624 276 670 054 4 × 2 = 0 + 0.736 480 660 429 248 553 340 108 8;
  • 32) 0.736 480 660 429 248 553 340 108 8 × 2 = 1 + 0.472 961 320 858 497 106 680 217 6;
  • 33) 0.472 961 320 858 497 106 680 217 6 × 2 = 0 + 0.945 922 641 716 994 213 360 435 2;
  • 34) 0.945 922 641 716 994 213 360 435 2 × 2 = 1 + 0.891 845 283 433 988 426 720 870 4;
  • 35) 0.891 845 283 433 988 426 720 870 4 × 2 = 1 + 0.783 690 566 867 976 853 441 740 8;
  • 36) 0.783 690 566 867 976 853 441 740 8 × 2 = 1 + 0.567 381 133 735 953 706 883 481 6;
  • 37) 0.567 381 133 735 953 706 883 481 6 × 2 = 1 + 0.134 762 267 471 907 413 766 963 2;
  • 38) 0.134 762 267 471 907 413 766 963 2 × 2 = 0 + 0.269 524 534 943 814 827 533 926 4;
  • 39) 0.269 524 534 943 814 827 533 926 4 × 2 = 0 + 0.539 049 069 887 629 655 067 852 8;
  • 40) 0.539 049 069 887 629 655 067 852 8 × 2 = 1 + 0.078 098 139 775 259 310 135 705 6;
  • 41) 0.078 098 139 775 259 310 135 705 6 × 2 = 0 + 0.156 196 279 550 518 620 271 411 2;
  • 42) 0.156 196 279 550 518 620 271 411 2 × 2 = 0 + 0.312 392 559 101 037 240 542 822 4;
  • 43) 0.312 392 559 101 037 240 542 822 4 × 2 = 0 + 0.624 785 118 202 074 481 085 644 8;
  • 44) 0.624 785 118 202 074 481 085 644 8 × 2 = 1 + 0.249 570 236 404 148 962 171 289 6;
  • 45) 0.249 570 236 404 148 962 171 289 6 × 2 = 0 + 0.499 140 472 808 297 924 342 579 2;
  • 46) 0.499 140 472 808 297 924 342 579 2 × 2 = 0 + 0.998 280 945 616 595 848 685 158 4;
  • 47) 0.998 280 945 616 595 848 685 158 4 × 2 = 1 + 0.996 561 891 233 191 697 370 316 8;
  • 48) 0.996 561 891 233 191 697 370 316 8 × 2 = 1 + 0.993 123 782 466 383 394 740 633 6;
  • 49) 0.993 123 782 466 383 394 740 633 6 × 2 = 1 + 0.986 247 564 932 766 789 481 267 2;
  • 50) 0.986 247 564 932 766 789 481 267 2 × 2 = 1 + 0.972 495 129 865 533 578 962 534 4;
  • 51) 0.972 495 129 865 533 578 962 534 4 × 2 = 1 + 0.944 990 259 731 067 157 925 068 8;
  • 52) 0.944 990 259 731 067 157 925 068 8 × 2 = 1 + 0.889 980 519 462 134 315 850 137 6;
  • 53) 0.889 980 519 462 134 315 850 137 6 × 2 = 1 + 0.779 961 038 924 268 631 700 275 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:


0.225 073 858 507 201 383 090 228 1(10) =

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

5. Positive number before normalization:

2.225 073 858 507 201 383 090 228 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:


2.225 073 858 507 201 383 090 228 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


Decimal number 2.225 073 858 507 201 383 090 228 1 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

» Convert decimal 2.225 073 858 507 201 383 090 228 9 to 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