5.143 999 999 999 999 239 719 272 736 692 786 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.143 999 999 999 999 239 719 272 736 692 786 6(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
5.143 999 999 999 999 239 719 272 736 692 786 6(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: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.143 999 999 999 999 239 719 272 736 692 786 6.

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) 0.143 999 999 999 999 239 719 272 736 692 786 6 × 2 = 0 + 0.287 999 999 999 998 479 438 545 473 385 573 2;
  • 2) 0.287 999 999 999 998 479 438 545 473 385 573 2 × 2 = 0 + 0.575 999 999 999 996 958 877 090 946 771 146 4;
  • 3) 0.575 999 999 999 996 958 877 090 946 771 146 4 × 2 = 1 + 0.151 999 999 999 993 917 754 181 893 542 292 8;
  • 4) 0.151 999 999 999 993 917 754 181 893 542 292 8 × 2 = 0 + 0.303 999 999 999 987 835 508 363 787 084 585 6;
  • 5) 0.303 999 999 999 987 835 508 363 787 084 585 6 × 2 = 0 + 0.607 999 999 999 975 671 016 727 574 169 171 2;
  • 6) 0.607 999 999 999 975 671 016 727 574 169 171 2 × 2 = 1 + 0.215 999 999 999 951 342 033 455 148 338 342 4;
  • 7) 0.215 999 999 999 951 342 033 455 148 338 342 4 × 2 = 0 + 0.431 999 999 999 902 684 066 910 296 676 684 8;
  • 8) 0.431 999 999 999 902 684 066 910 296 676 684 8 × 2 = 0 + 0.863 999 999 999 805 368 133 820 593 353 369 6;
  • 9) 0.863 999 999 999 805 368 133 820 593 353 369 6 × 2 = 1 + 0.727 999 999 999 610 736 267 641 186 706 739 2;
  • 10) 0.727 999 999 999 610 736 267 641 186 706 739 2 × 2 = 1 + 0.455 999 999 999 221 472 535 282 373 413 478 4;
  • 11) 0.455 999 999 999 221 472 535 282 373 413 478 4 × 2 = 0 + 0.911 999 999 998 442 945 070 564 746 826 956 8;
  • 12) 0.911 999 999 998 442 945 070 564 746 826 956 8 × 2 = 1 + 0.823 999 999 996 885 890 141 129 493 653 913 6;
  • 13) 0.823 999 999 996 885 890 141 129 493 653 913 6 × 2 = 1 + 0.647 999 999 993 771 780 282 258 987 307 827 2;
  • 14) 0.647 999 999 993 771 780 282 258 987 307 827 2 × 2 = 1 + 0.295 999 999 987 543 560 564 517 974 615 654 4;
  • 15) 0.295 999 999 987 543 560 564 517 974 615 654 4 × 2 = 0 + 0.591 999 999 975 087 121 129 035 949 231 308 8;
  • 16) 0.591 999 999 975 087 121 129 035 949 231 308 8 × 2 = 1 + 0.183 999 999 950 174 242 258 071 898 462 617 6;
  • 17) 0.183 999 999 950 174 242 258 071 898 462 617 6 × 2 = 0 + 0.367 999 999 900 348 484 516 143 796 925 235 2;
  • 18) 0.367 999 999 900 348 484 516 143 796 925 235 2 × 2 = 0 + 0.735 999 999 800 696 969 032 287 593 850 470 4;
  • 19) 0.735 999 999 800 696 969 032 287 593 850 470 4 × 2 = 1 + 0.471 999 999 601 393 938 064 575 187 700 940 8;
  • 20) 0.471 999 999 601 393 938 064 575 187 700 940 8 × 2 = 0 + 0.943 999 999 202 787 876 129 150 375 401 881 6;
  • 21) 0.943 999 999 202 787 876 129 150 375 401 881 6 × 2 = 1 + 0.887 999 998 405 575 752 258 300 750 803 763 2;
  • 22) 0.887 999 998 405 575 752 258 300 750 803 763 2 × 2 = 1 + 0.775 999 996 811 151 504 516 601 501 607 526 4;
  • 23) 0.775 999 996 811 151 504 516 601 501 607 526 4 × 2 = 1 + 0.551 999 993 622 303 009 033 203 003 215 052 8;
  • 24) 0.551 999 993 622 303 009 033 203 003 215 052 8 × 2 = 1 + 0.103 999 987 244 606 018 066 406 006 430 105 6;
  • 25) 0.103 999 987 244 606 018 066 406 006 430 105 6 × 2 = 0 + 0.207 999 974 489 212 036 132 812 012 860 211 2;
  • 26) 0.207 999 974 489 212 036 132 812 012 860 211 2 × 2 = 0 + 0.415 999 948 978 424 072 265 624 025 720 422 4;
  • 27) 0.415 999 948 978 424 072 265 624 025 720 422 4 × 2 = 0 + 0.831 999 897 956 848 144 531 248 051 440 844 8;
  • 28) 0.831 999 897 956 848 144 531 248 051 440 844 8 × 2 = 1 + 0.663 999 795 913 696 289 062 496 102 881 689 6;
  • 29) 0.663 999 795 913 696 289 062 496 102 881 689 6 × 2 = 1 + 0.327 999 591 827 392 578 124 992 205 763 379 2;
  • 30) 0.327 999 591 827 392 578 124 992 205 763 379 2 × 2 = 0 + 0.655 999 183 654 785 156 249 984 411 526 758 4;
  • 31) 0.655 999 183 654 785 156 249 984 411 526 758 4 × 2 = 1 + 0.311 998 367 309 570 312 499 968 823 053 516 8;
  • 32) 0.311 998 367 309 570 312 499 968 823 053 516 8 × 2 = 0 + 0.623 996 734 619 140 624 999 937 646 107 033 6;
  • 33) 0.623 996 734 619 140 624 999 937 646 107 033 6 × 2 = 1 + 0.247 993 469 238 281 249 999 875 292 214 067 2;
  • 34) 0.247 993 469 238 281 249 999 875 292 214 067 2 × 2 = 0 + 0.495 986 938 476 562 499 999 750 584 428 134 4;
  • 35) 0.495 986 938 476 562 499 999 750 584 428 134 4 × 2 = 0 + 0.991 973 876 953 124 999 999 501 168 856 268 8;
  • 36) 0.991 973 876 953 124 999 999 501 168 856 268 8 × 2 = 1 + 0.983 947 753 906 249 999 999 002 337 712 537 6;
  • 37) 0.983 947 753 906 249 999 999 002 337 712 537 6 × 2 = 1 + 0.967 895 507 812 499 999 998 004 675 425 075 2;
  • 38) 0.967 895 507 812 499 999 998 004 675 425 075 2 × 2 = 1 + 0.935 791 015 624 999 999 996 009 350 850 150 4;
  • 39) 0.935 791 015 624 999 999 996 009 350 850 150 4 × 2 = 1 + 0.871 582 031 249 999 999 992 018 701 700 300 8;
  • 40) 0.871 582 031 249 999 999 992 018 701 700 300 8 × 2 = 1 + 0.743 164 062 499 999 999 984 037 403 400 601 6;
  • 41) 0.743 164 062 499 999 999 984 037 403 400 601 6 × 2 = 1 + 0.486 328 124 999 999 999 968 074 806 801 203 2;
  • 42) 0.486 328 124 999 999 999 968 074 806 801 203 2 × 2 = 0 + 0.972 656 249 999 999 999 936 149 613 602 406 4;
  • 43) 0.972 656 249 999 999 999 936 149 613 602 406 4 × 2 = 1 + 0.945 312 499 999 999 999 872 299 227 204 812 8;
  • 44) 0.945 312 499 999 999 999 872 299 227 204 812 8 × 2 = 1 + 0.890 624 999 999 999 999 744 598 454 409 625 6;
  • 45) 0.890 624 999 999 999 999 744 598 454 409 625 6 × 2 = 1 + 0.781 249 999 999 999 999 489 196 908 819 251 2;
  • 46) 0.781 249 999 999 999 999 489 196 908 819 251 2 × 2 = 1 + 0.562 499 999 999 999 998 978 393 817 638 502 4;
  • 47) 0.562 499 999 999 999 998 978 393 817 638 502 4 × 2 = 1 + 0.124 999 999 999 999 997 956 787 635 277 004 8;
  • 48) 0.124 999 999 999 999 997 956 787 635 277 004 8 × 2 = 0 + 0.249 999 999 999 999 995 913 575 270 554 009 6;
  • 49) 0.249 999 999 999 999 995 913 575 270 554 009 6 × 2 = 0 + 0.499 999 999 999 999 991 827 150 541 108 019 2;
  • 50) 0.499 999 999 999 999 991 827 150 541 108 019 2 × 2 = 0 + 0.999 999 999 999 999 983 654 301 082 216 038 4;
  • 51) 0.999 999 999 999 999 983 654 301 082 216 038 4 × 2 = 1 + 0.999 999 999 999 999 967 308 602 164 432 076 8;
  • 52) 0.999 999 999 999 999 967 308 602 164 432 076 8 × 2 = 1 + 0.999 999 999 999 999 934 617 204 328 864 153 6;
  • 53) 0.999 999 999 999 999 934 617 204 328 864 153 6 × 2 = 1 + 0.999 999 999 999 999 869 234 408 657 728 307 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:


0.143 999 999 999 999 239 719 272 736 692 786 6(10) =

0.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2)

5. Positive number before normalization:

5.143 999 999 999 999 239 719 272 736 692 786 6(10) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.143 999 999 999 999 239 719 272 736 692 786 6(10) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2) × 20 =

1.0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111(2) × 22


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): 2

Mantissa (not normalized):
1.0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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. 0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111 =

0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000


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 0001

Mantissa (52 bits) =
0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000


Decimal number 5.143 999 999 999 999 239 719 272 736 692 786 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000

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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