1.123 412 341 234 123 412 341 259 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.123 412 341 234 123 412 341 259(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
1.123 412 341 234 123 412 341 259(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: 0.123 412 341 234 123 412 341 259.

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) 0.123 412 341 234 123 412 341 259 × 2 = 0 + 0.246 824 682 468 246 824 682 518;
  • 2) 0.246 824 682 468 246 824 682 518 × 2 = 0 + 0.493 649 364 936 493 649 365 036;
  • 3) 0.493 649 364 936 493 649 365 036 × 2 = 0 + 0.987 298 729 872 987 298 730 072;
  • 4) 0.987 298 729 872 987 298 730 072 × 2 = 1 + 0.974 597 459 745 974 597 460 144;
  • 5) 0.974 597 459 745 974 597 460 144 × 2 = 1 + 0.949 194 919 491 949 194 920 288;
  • 6) 0.949 194 919 491 949 194 920 288 × 2 = 1 + 0.898 389 838 983 898 389 840 576;
  • 7) 0.898 389 838 983 898 389 840 576 × 2 = 1 + 0.796 779 677 967 796 779 681 152;
  • 8) 0.796 779 677 967 796 779 681 152 × 2 = 1 + 0.593 559 355 935 593 559 362 304;
  • 9) 0.593 559 355 935 593 559 362 304 × 2 = 1 + 0.187 118 711 871 187 118 724 608;
  • 10) 0.187 118 711 871 187 118 724 608 × 2 = 0 + 0.374 237 423 742 374 237 449 216;
  • 11) 0.374 237 423 742 374 237 449 216 × 2 = 0 + 0.748 474 847 484 748 474 898 432;
  • 12) 0.748 474 847 484 748 474 898 432 × 2 = 1 + 0.496 949 694 969 496 949 796 864;
  • 13) 0.496 949 694 969 496 949 796 864 × 2 = 0 + 0.993 899 389 938 993 899 593 728;
  • 14) 0.993 899 389 938 993 899 593 728 × 2 = 1 + 0.987 798 779 877 987 799 187 456;
  • 15) 0.987 798 779 877 987 799 187 456 × 2 = 1 + 0.975 597 559 755 975 598 374 912;
  • 16) 0.975 597 559 755 975 598 374 912 × 2 = 1 + 0.951 195 119 511 951 196 749 824;
  • 17) 0.951 195 119 511 951 196 749 824 × 2 = 1 + 0.902 390 239 023 902 393 499 648;
  • 18) 0.902 390 239 023 902 393 499 648 × 2 = 1 + 0.804 780 478 047 804 786 999 296;
  • 19) 0.804 780 478 047 804 786 999 296 × 2 = 1 + 0.609 560 956 095 609 573 998 592;
  • 20) 0.609 560 956 095 609 573 998 592 × 2 = 1 + 0.219 121 912 191 219 147 997 184;
  • 21) 0.219 121 912 191 219 147 997 184 × 2 = 0 + 0.438 243 824 382 438 295 994 368;
  • 22) 0.438 243 824 382 438 295 994 368 × 2 = 0 + 0.876 487 648 764 876 591 988 736;
  • 23) 0.876 487 648 764 876 591 988 736 × 2 = 1 + 0.752 975 297 529 753 183 977 472;
  • 24) 0.752 975 297 529 753 183 977 472 × 2 = 1 + 0.505 950 595 059 506 367 954 944;
  • 25) 0.505 950 595 059 506 367 954 944 × 2 = 1 + 0.011 901 190 119 012 735 909 888;
  • 26) 0.011 901 190 119 012 735 909 888 × 2 = 0 + 0.023 802 380 238 025 471 819 776;
  • 27) 0.023 802 380 238 025 471 819 776 × 2 = 0 + 0.047 604 760 476 050 943 639 552;
  • 28) 0.047 604 760 476 050 943 639 552 × 2 = 0 + 0.095 209 520 952 101 887 279 104;
  • 29) 0.095 209 520 952 101 887 279 104 × 2 = 0 + 0.190 419 041 904 203 774 558 208;
  • 30) 0.190 419 041 904 203 774 558 208 × 2 = 0 + 0.380 838 083 808 407 549 116 416;
  • 31) 0.380 838 083 808 407 549 116 416 × 2 = 0 + 0.761 676 167 616 815 098 232 832;
  • 32) 0.761 676 167 616 815 098 232 832 × 2 = 1 + 0.523 352 335 233 630 196 465 664;
  • 33) 0.523 352 335 233 630 196 465 664 × 2 = 1 + 0.046 704 670 467 260 392 931 328;
  • 34) 0.046 704 670 467 260 392 931 328 × 2 = 0 + 0.093 409 340 934 520 785 862 656;
  • 35) 0.093 409 340 934 520 785 862 656 × 2 = 0 + 0.186 818 681 869 041 571 725 312;
  • 36) 0.186 818 681 869 041 571 725 312 × 2 = 0 + 0.373 637 363 738 083 143 450 624;
  • 37) 0.373 637 363 738 083 143 450 624 × 2 = 0 + 0.747 274 727 476 166 286 901 248;
  • 38) 0.747 274 727 476 166 286 901 248 × 2 = 1 + 0.494 549 454 952 332 573 802 496;
  • 39) 0.494 549 454 952 332 573 802 496 × 2 = 0 + 0.989 098 909 904 665 147 604 992;
  • 40) 0.989 098 909 904 665 147 604 992 × 2 = 1 + 0.978 197 819 809 330 295 209 984;
  • 41) 0.978 197 819 809 330 295 209 984 × 2 = 1 + 0.956 395 639 618 660 590 419 968;
  • 42) 0.956 395 639 618 660 590 419 968 × 2 = 1 + 0.912 791 279 237 321 180 839 936;
  • 43) 0.912 791 279 237 321 180 839 936 × 2 = 1 + 0.825 582 558 474 642 361 679 872;
  • 44) 0.825 582 558 474 642 361 679 872 × 2 = 1 + 0.651 165 116 949 284 723 359 744;
  • 45) 0.651 165 116 949 284 723 359 744 × 2 = 1 + 0.302 330 233 898 569 446 719 488;
  • 46) 0.302 330 233 898 569 446 719 488 × 2 = 0 + 0.604 660 467 797 138 893 438 976;
  • 47) 0.604 660 467 797 138 893 438 976 × 2 = 1 + 0.209 320 935 594 277 786 877 952;
  • 48) 0.209 320 935 594 277 786 877 952 × 2 = 0 + 0.418 641 871 188 555 573 755 904;
  • 49) 0.418 641 871 188 555 573 755 904 × 2 = 0 + 0.837 283 742 377 111 147 511 808;
  • 50) 0.837 283 742 377 111 147 511 808 × 2 = 1 + 0.674 567 484 754 222 295 023 616;
  • 51) 0.674 567 484 754 222 295 023 616 × 2 = 1 + 0.349 134 969 508 444 590 047 232;
  • 52) 0.349 134 969 508 444 590 047 232 × 2 = 0 + 0.698 269 939 016 889 180 094 464;
  • 53) 0.698 269 939 016 889 180 094 464 × 2 = 1 + 0.396 539 878 033 778 360 188 928;

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:


0.123 412 341 234 123 412 341 259(10) =

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

5. Positive number before normalization:

1.123 412 341 234 123 412 341 259(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:


1.123 412 341 234 123 412 341 259(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 1.123 412 341 234 123 412 341 259 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