-199.219 999 999 999 998 863 131 622 783 841 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -199.219 999 999 999 998 863 131 622 783 841 86(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
-199.219 999 999 999 998 863 131 622 783 841 86(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. Start with the positive version of the number:

|-199.219 999 999 999 998 863 131 622 783 841 86| = 199.219 999 999 999 998 863 131 622 783 841 86


2. First, convert to binary (in base 2) the integer part: 199.
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;
  • 199 ÷ 2 = 99 + 1;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. 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.

199(10) =

1100 0111(2)


4. Convert to binary (base 2) the fractional part: 0.219 999 999 999 998 863 131 622 783 841 86.

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) 0.219 999 999 999 998 863 131 622 783 841 86 × 2 = 0 + 0.439 999 999 999 997 726 263 245 567 683 72;
  • 2) 0.439 999 999 999 997 726 263 245 567 683 72 × 2 = 0 + 0.879 999 999 999 995 452 526 491 135 367 44;
  • 3) 0.879 999 999 999 995 452 526 491 135 367 44 × 2 = 1 + 0.759 999 999 999 990 905 052 982 270 734 88;
  • 4) 0.759 999 999 999 990 905 052 982 270 734 88 × 2 = 1 + 0.519 999 999 999 981 810 105 964 541 469 76;
  • 5) 0.519 999 999 999 981 810 105 964 541 469 76 × 2 = 1 + 0.039 999 999 999 963 620 211 929 082 939 52;
  • 6) 0.039 999 999 999 963 620 211 929 082 939 52 × 2 = 0 + 0.079 999 999 999 927 240 423 858 165 879 04;
  • 7) 0.079 999 999 999 927 240 423 858 165 879 04 × 2 = 0 + 0.159 999 999 999 854 480 847 716 331 758 08;
  • 8) 0.159 999 999 999 854 480 847 716 331 758 08 × 2 = 0 + 0.319 999 999 999 708 961 695 432 663 516 16;
  • 9) 0.319 999 999 999 708 961 695 432 663 516 16 × 2 = 0 + 0.639 999 999 999 417 923 390 865 327 032 32;
  • 10) 0.639 999 999 999 417 923 390 865 327 032 32 × 2 = 1 + 0.279 999 999 998 835 846 781 730 654 064 64;
  • 11) 0.279 999 999 998 835 846 781 730 654 064 64 × 2 = 0 + 0.559 999 999 997 671 693 563 461 308 129 28;
  • 12) 0.559 999 999 997 671 693 563 461 308 129 28 × 2 = 1 + 0.119 999 999 995 343 387 126 922 616 258 56;
  • 13) 0.119 999 999 995 343 387 126 922 616 258 56 × 2 = 0 + 0.239 999 999 990 686 774 253 845 232 517 12;
  • 14) 0.239 999 999 990 686 774 253 845 232 517 12 × 2 = 0 + 0.479 999 999 981 373 548 507 690 465 034 24;
  • 15) 0.479 999 999 981 373 548 507 690 465 034 24 × 2 = 0 + 0.959 999 999 962 747 097 015 380 930 068 48;
  • 16) 0.959 999 999 962 747 097 015 380 930 068 48 × 2 = 1 + 0.919 999 999 925 494 194 030 761 860 136 96;
  • 17) 0.919 999 999 925 494 194 030 761 860 136 96 × 2 = 1 + 0.839 999 999 850 988 388 061 523 720 273 92;
  • 18) 0.839 999 999 850 988 388 061 523 720 273 92 × 2 = 1 + 0.679 999 999 701 976 776 123 047 440 547 84;
  • 19) 0.679 999 999 701 976 776 123 047 440 547 84 × 2 = 1 + 0.359 999 999 403 953 552 246 094 881 095 68;
  • 20) 0.359 999 999 403 953 552 246 094 881 095 68 × 2 = 0 + 0.719 999 998 807 907 104 492 189 762 191 36;
  • 21) 0.719 999 998 807 907 104 492 189 762 191 36 × 2 = 1 + 0.439 999 997 615 814 208 984 379 524 382 72;
  • 22) 0.439 999 997 615 814 208 984 379 524 382 72 × 2 = 0 + 0.879 999 995 231 628 417 968 759 048 765 44;
  • 23) 0.879 999 995 231 628 417 968 759 048 765 44 × 2 = 1 + 0.759 999 990 463 256 835 937 518 097 530 88;
  • 24) 0.759 999 990 463 256 835 937 518 097 530 88 × 2 = 1 + 0.519 999 980 926 513 671 875 036 195 061 76;
  • 25) 0.519 999 980 926 513 671 875 036 195 061 76 × 2 = 1 + 0.039 999 961 853 027 343 750 072 390 123 52;
  • 26) 0.039 999 961 853 027 343 750 072 390 123 52 × 2 = 0 + 0.079 999 923 706 054 687 500 144 780 247 04;
  • 27) 0.079 999 923 706 054 687 500 144 780 247 04 × 2 = 0 + 0.159 999 847 412 109 375 000 289 560 494 08;
  • 28) 0.159 999 847 412 109 375 000 289 560 494 08 × 2 = 0 + 0.319 999 694 824 218 750 000 579 120 988 16;
  • 29) 0.319 999 694 824 218 750 000 579 120 988 16 × 2 = 0 + 0.639 999 389 648 437 500 001 158 241 976 32;
  • 30) 0.639 999 389 648 437 500 001 158 241 976 32 × 2 = 1 + 0.279 998 779 296 875 000 002 316 483 952 64;
  • 31) 0.279 998 779 296 875 000 002 316 483 952 64 × 2 = 0 + 0.559 997 558 593 750 000 004 632 967 905 28;
  • 32) 0.559 997 558 593 750 000 004 632 967 905 28 × 2 = 1 + 0.119 995 117 187 500 000 009 265 935 810 56;
  • 33) 0.119 995 117 187 500 000 009 265 935 810 56 × 2 = 0 + 0.239 990 234 375 000 000 018 531 871 621 12;
  • 34) 0.239 990 234 375 000 000 018 531 871 621 12 × 2 = 0 + 0.479 980 468 750 000 000 037 063 743 242 24;
  • 35) 0.479 980 468 750 000 000 037 063 743 242 24 × 2 = 0 + 0.959 960 937 500 000 000 074 127 486 484 48;
  • 36) 0.959 960 937 500 000 000 074 127 486 484 48 × 2 = 1 + 0.919 921 875 000 000 000 148 254 972 968 96;
  • 37) 0.919 921 875 000 000 000 148 254 972 968 96 × 2 = 1 + 0.839 843 750 000 000 000 296 509 945 937 92;
  • 38) 0.839 843 750 000 000 000 296 509 945 937 92 × 2 = 1 + 0.679 687 500 000 000 000 593 019 891 875 84;
  • 39) 0.679 687 500 000 000 000 593 019 891 875 84 × 2 = 1 + 0.359 375 000 000 000 001 186 039 783 751 68;
  • 40) 0.359 375 000 000 000 001 186 039 783 751 68 × 2 = 0 + 0.718 750 000 000 000 002 372 079 567 503 36;
  • 41) 0.718 750 000 000 000 002 372 079 567 503 36 × 2 = 1 + 0.437 500 000 000 000 004 744 159 135 006 72;
  • 42) 0.437 500 000 000 000 004 744 159 135 006 72 × 2 = 0 + 0.875 000 000 000 000 009 488 318 270 013 44;
  • 43) 0.875 000 000 000 000 009 488 318 270 013 44 × 2 = 1 + 0.750 000 000 000 000 018 976 636 540 026 88;
  • 44) 0.750 000 000 000 000 018 976 636 540 026 88 × 2 = 1 + 0.500 000 000 000 000 037 953 273 080 053 76;
  • 45) 0.500 000 000 000 000 037 953 273 080 053 76 × 2 = 1 + 0.000 000 000 000 000 075 906 546 160 107 52;
  • 46) 0.000 000 000 000 000 075 906 546 160 107 52 × 2 = 0 + 0.000 000 000 000 000 151 813 092 320 215 04;
  • 47) 0.000 000 000 000 000 151 813 092 320 215 04 × 2 = 0 + 0.000 000 000 000 000 303 626 184 640 430 08;
  • 48) 0.000 000 000 000 000 303 626 184 640 430 08 × 2 = 0 + 0.000 000 000 000 000 607 252 369 280 860 16;
  • 49) 0.000 000 000 000 000 607 252 369 280 860 16 × 2 = 0 + 0.000 000 000 000 001 214 504 738 561 720 32;
  • 50) 0.000 000 000 000 001 214 504 738 561 720 32 × 2 = 0 + 0.000 000 000 000 002 429 009 477 123 440 64;
  • 51) 0.000 000 000 000 002 429 009 477 123 440 64 × 2 = 0 + 0.000 000 000 000 004 858 018 954 246 881 28;
  • 52) 0.000 000 000 000 004 858 018 954 246 881 28 × 2 = 0 + 0.000 000 000 000 009 716 037 908 493 762 56;
  • 53) 0.000 000 000 000 009 716 037 908 493 762 56 × 2 = 0 + 0.000 000 000 000 019 432 075 816 987 525 12;

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).


5. 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:


0.219 999 999 999 998 863 131 622 783 841 86(10) =

0.0011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0(2)

6. Positive number before normalization:

199.219 999 999 999 998 863 131 622 783 841 86(10) =

1100 0111.0011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


199.219 999 999 999 998 863 131 622 783 841 86(10) =

1100 0111.0011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0(2) =

1100 0111.0011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0(2) × 20 =

1.1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000(2) × 27


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): 7

Mantissa (not normalized):
1.1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(10)


10. 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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1030(10) =

100 0000 0110(2)


12. 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. 1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 =

1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0110

Mantissa (52 bits) =
1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111


Decimal number -199.219 999 999 999 998 863 131 622 783 841 86 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0110 - 1000 1110 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111

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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