1.123 412 341 234 123 412 341 232 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.123 412 341 234 123 412 341 232 92(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
1.123 412 341 234 123 412 341 232 92(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.123 412 341 234 123 412 341 232 92.

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) 0.123 412 341 234 123 412 341 232 92 × 2 = 0 + 0.246 824 682 468 246 824 682 465 84;
  • 2) 0.246 824 682 468 246 824 682 465 84 × 2 = 0 + 0.493 649 364 936 493 649 364 931 68;
  • 3) 0.493 649 364 936 493 649 364 931 68 × 2 = 0 + 0.987 298 729 872 987 298 729 863 36;
  • 4) 0.987 298 729 872 987 298 729 863 36 × 2 = 1 + 0.974 597 459 745 974 597 459 726 72;
  • 5) 0.974 597 459 745 974 597 459 726 72 × 2 = 1 + 0.949 194 919 491 949 194 919 453 44;
  • 6) 0.949 194 919 491 949 194 919 453 44 × 2 = 1 + 0.898 389 838 983 898 389 838 906 88;
  • 7) 0.898 389 838 983 898 389 838 906 88 × 2 = 1 + 0.796 779 677 967 796 779 677 813 76;
  • 8) 0.796 779 677 967 796 779 677 813 76 × 2 = 1 + 0.593 559 355 935 593 559 355 627 52;
  • 9) 0.593 559 355 935 593 559 355 627 52 × 2 = 1 + 0.187 118 711 871 187 118 711 255 04;
  • 10) 0.187 118 711 871 187 118 711 255 04 × 2 = 0 + 0.374 237 423 742 374 237 422 510 08;
  • 11) 0.374 237 423 742 374 237 422 510 08 × 2 = 0 + 0.748 474 847 484 748 474 845 020 16;
  • 12) 0.748 474 847 484 748 474 845 020 16 × 2 = 1 + 0.496 949 694 969 496 949 690 040 32;
  • 13) 0.496 949 694 969 496 949 690 040 32 × 2 = 0 + 0.993 899 389 938 993 899 380 080 64;
  • 14) 0.993 899 389 938 993 899 380 080 64 × 2 = 1 + 0.987 798 779 877 987 798 760 161 28;
  • 15) 0.987 798 779 877 987 798 760 161 28 × 2 = 1 + 0.975 597 559 755 975 597 520 322 56;
  • 16) 0.975 597 559 755 975 597 520 322 56 × 2 = 1 + 0.951 195 119 511 951 195 040 645 12;
  • 17) 0.951 195 119 511 951 195 040 645 12 × 2 = 1 + 0.902 390 239 023 902 390 081 290 24;
  • 18) 0.902 390 239 023 902 390 081 290 24 × 2 = 1 + 0.804 780 478 047 804 780 162 580 48;
  • 19) 0.804 780 478 047 804 780 162 580 48 × 2 = 1 + 0.609 560 956 095 609 560 325 160 96;
  • 20) 0.609 560 956 095 609 560 325 160 96 × 2 = 1 + 0.219 121 912 191 219 120 650 321 92;
  • 21) 0.219 121 912 191 219 120 650 321 92 × 2 = 0 + 0.438 243 824 382 438 241 300 643 84;
  • 22) 0.438 243 824 382 438 241 300 643 84 × 2 = 0 + 0.876 487 648 764 876 482 601 287 68;
  • 23) 0.876 487 648 764 876 482 601 287 68 × 2 = 1 + 0.752 975 297 529 752 965 202 575 36;
  • 24) 0.752 975 297 529 752 965 202 575 36 × 2 = 1 + 0.505 950 595 059 505 930 405 150 72;
  • 25) 0.505 950 595 059 505 930 405 150 72 × 2 = 1 + 0.011 901 190 119 011 860 810 301 44;
  • 26) 0.011 901 190 119 011 860 810 301 44 × 2 = 0 + 0.023 802 380 238 023 721 620 602 88;
  • 27) 0.023 802 380 238 023 721 620 602 88 × 2 = 0 + 0.047 604 760 476 047 443 241 205 76;
  • 28) 0.047 604 760 476 047 443 241 205 76 × 2 = 0 + 0.095 209 520 952 094 886 482 411 52;
  • 29) 0.095 209 520 952 094 886 482 411 52 × 2 = 0 + 0.190 419 041 904 189 772 964 823 04;
  • 30) 0.190 419 041 904 189 772 964 823 04 × 2 = 0 + 0.380 838 083 808 379 545 929 646 08;
  • 31) 0.380 838 083 808 379 545 929 646 08 × 2 = 0 + 0.761 676 167 616 759 091 859 292 16;
  • 32) 0.761 676 167 616 759 091 859 292 16 × 2 = 1 + 0.523 352 335 233 518 183 718 584 32;
  • 33) 0.523 352 335 233 518 183 718 584 32 × 2 = 1 + 0.046 704 670 467 036 367 437 168 64;
  • 34) 0.046 704 670 467 036 367 437 168 64 × 2 = 0 + 0.093 409 340 934 072 734 874 337 28;
  • 35) 0.093 409 340 934 072 734 874 337 28 × 2 = 0 + 0.186 818 681 868 145 469 748 674 56;
  • 36) 0.186 818 681 868 145 469 748 674 56 × 2 = 0 + 0.373 637 363 736 290 939 497 349 12;
  • 37) 0.373 637 363 736 290 939 497 349 12 × 2 = 0 + 0.747 274 727 472 581 878 994 698 24;
  • 38) 0.747 274 727 472 581 878 994 698 24 × 2 = 1 + 0.494 549 454 945 163 757 989 396 48;
  • 39) 0.494 549 454 945 163 757 989 396 48 × 2 = 0 + 0.989 098 909 890 327 515 978 792 96;
  • 40) 0.989 098 909 890 327 515 978 792 96 × 2 = 1 + 0.978 197 819 780 655 031 957 585 92;
  • 41) 0.978 197 819 780 655 031 957 585 92 × 2 = 1 + 0.956 395 639 561 310 063 915 171 84;
  • 42) 0.956 395 639 561 310 063 915 171 84 × 2 = 1 + 0.912 791 279 122 620 127 830 343 68;
  • 43) 0.912 791 279 122 620 127 830 343 68 × 2 = 1 + 0.825 582 558 245 240 255 660 687 36;
  • 44) 0.825 582 558 245 240 255 660 687 36 × 2 = 1 + 0.651 165 116 490 480 511 321 374 72;
  • 45) 0.651 165 116 490 480 511 321 374 72 × 2 = 1 + 0.302 330 232 980 961 022 642 749 44;
  • 46) 0.302 330 232 980 961 022 642 749 44 × 2 = 0 + 0.604 660 465 961 922 045 285 498 88;
  • 47) 0.604 660 465 961 922 045 285 498 88 × 2 = 1 + 0.209 320 931 923 844 090 570 997 76;
  • 48) 0.209 320 931 923 844 090 570 997 76 × 2 = 0 + 0.418 641 863 847 688 181 141 995 52;
  • 49) 0.418 641 863 847 688 181 141 995 52 × 2 = 0 + 0.837 283 727 695 376 362 283 991 04;
  • 50) 0.837 283 727 695 376 362 283 991 04 × 2 = 1 + 0.674 567 455 390 752 724 567 982 08;
  • 51) 0.674 567 455 390 752 724 567 982 08 × 2 = 1 + 0.349 134 910 781 505 449 135 964 16;
  • 52) 0.349 134 910 781 505 449 135 964 16 × 2 = 0 + 0.698 269 821 563 010 898 271 928 32;
  • 53) 0.698 269 821 563 010 898 271 928 32 × 2 = 1 + 0.396 539 643 126 021 796 543 856 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:


0.123 412 341 234 123 412 341 232 92(10) =

0.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2)

5. Positive number before normalization:

1.123 412 341 234 123 412 341 232 92(10) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.123 412 341 234 123 412 341 232 92(10) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2) × 20


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): 0

Mantissa (not normalized):
1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1023(10) =

011 1111 1111(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 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1 =

0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110


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 1111

Mantissa (52 bits) =
0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110


Decimal number 1.123 412 341 234 123 412 341 232 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110

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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