-199.219 999 999 999 998 863 221 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -199.219 999 999 999 998 863 221(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
-199.219 999 999 999 998 863 221(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:

|-199.219 999 999 999 998 863 221| = 199.219 999 999 999 998 863 221


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: 0.219 999 999 999 998 863 221.

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) 0.219 999 999 999 998 863 221 × 2 = 0 + 0.439 999 999 999 997 726 442;
  • 2) 0.439 999 999 999 997 726 442 × 2 = 0 + 0.879 999 999 999 995 452 884;
  • 3) 0.879 999 999 999 995 452 884 × 2 = 1 + 0.759 999 999 999 990 905 768;
  • 4) 0.759 999 999 999 990 905 768 × 2 = 1 + 0.519 999 999 999 981 811 536;
  • 5) 0.519 999 999 999 981 811 536 × 2 = 1 + 0.039 999 999 999 963 623 072;
  • 6) 0.039 999 999 999 963 623 072 × 2 = 0 + 0.079 999 999 999 927 246 144;
  • 7) 0.079 999 999 999 927 246 144 × 2 = 0 + 0.159 999 999 999 854 492 288;
  • 8) 0.159 999 999 999 854 492 288 × 2 = 0 + 0.319 999 999 999 708 984 576;
  • 9) 0.319 999 999 999 708 984 576 × 2 = 0 + 0.639 999 999 999 417 969 152;
  • 10) 0.639 999 999 999 417 969 152 × 2 = 1 + 0.279 999 999 998 835 938 304;
  • 11) 0.279 999 999 998 835 938 304 × 2 = 0 + 0.559 999 999 997 671 876 608;
  • 12) 0.559 999 999 997 671 876 608 × 2 = 1 + 0.119 999 999 995 343 753 216;
  • 13) 0.119 999 999 995 343 753 216 × 2 = 0 + 0.239 999 999 990 687 506 432;
  • 14) 0.239 999 999 990 687 506 432 × 2 = 0 + 0.479 999 999 981 375 012 864;
  • 15) 0.479 999 999 981 375 012 864 × 2 = 0 + 0.959 999 999 962 750 025 728;
  • 16) 0.959 999 999 962 750 025 728 × 2 = 1 + 0.919 999 999 925 500 051 456;
  • 17) 0.919 999 999 925 500 051 456 × 2 = 1 + 0.839 999 999 851 000 102 912;
  • 18) 0.839 999 999 851 000 102 912 × 2 = 1 + 0.679 999 999 702 000 205 824;
  • 19) 0.679 999 999 702 000 205 824 × 2 = 1 + 0.359 999 999 404 000 411 648;
  • 20) 0.359 999 999 404 000 411 648 × 2 = 0 + 0.719 999 998 808 000 823 296;
  • 21) 0.719 999 998 808 000 823 296 × 2 = 1 + 0.439 999 997 616 001 646 592;
  • 22) 0.439 999 997 616 001 646 592 × 2 = 0 + 0.879 999 995 232 003 293 184;
  • 23) 0.879 999 995 232 003 293 184 × 2 = 1 + 0.759 999 990 464 006 586 368;
  • 24) 0.759 999 990 464 006 586 368 × 2 = 1 + 0.519 999 980 928 013 172 736;
  • 25) 0.519 999 980 928 013 172 736 × 2 = 1 + 0.039 999 961 856 026 345 472;
  • 26) 0.039 999 961 856 026 345 472 × 2 = 0 + 0.079 999 923 712 052 690 944;
  • 27) 0.079 999 923 712 052 690 944 × 2 = 0 + 0.159 999 847 424 105 381 888;
  • 28) 0.159 999 847 424 105 381 888 × 2 = 0 + 0.319 999 694 848 210 763 776;
  • 29) 0.319 999 694 848 210 763 776 × 2 = 0 + 0.639 999 389 696 421 527 552;
  • 30) 0.639 999 389 696 421 527 552 × 2 = 1 + 0.279 998 779 392 843 055 104;
  • 31) 0.279 998 779 392 843 055 104 × 2 = 0 + 0.559 997 558 785 686 110 208;
  • 32) 0.559 997 558 785 686 110 208 × 2 = 1 + 0.119 995 117 571 372 220 416;
  • 33) 0.119 995 117 571 372 220 416 × 2 = 0 + 0.239 990 235 142 744 440 832;
  • 34) 0.239 990 235 142 744 440 832 × 2 = 0 + 0.479 980 470 285 488 881 664;
  • 35) 0.479 980 470 285 488 881 664 × 2 = 0 + 0.959 960 940 570 977 763 328;
  • 36) 0.959 960 940 570 977 763 328 × 2 = 1 + 0.919 921 881 141 955 526 656;
  • 37) 0.919 921 881 141 955 526 656 × 2 = 1 + 0.839 843 762 283 911 053 312;
  • 38) 0.839 843 762 283 911 053 312 × 2 = 1 + 0.679 687 524 567 822 106 624;
  • 39) 0.679 687 524 567 822 106 624 × 2 = 1 + 0.359 375 049 135 644 213 248;
  • 40) 0.359 375 049 135 644 213 248 × 2 = 0 + 0.718 750 098 271 288 426 496;
  • 41) 0.718 750 098 271 288 426 496 × 2 = 1 + 0.437 500 196 542 576 852 992;
  • 42) 0.437 500 196 542 576 852 992 × 2 = 0 + 0.875 000 393 085 153 705 984;
  • 43) 0.875 000 393 085 153 705 984 × 2 = 1 + 0.750 000 786 170 307 411 968;
  • 44) 0.750 000 786 170 307 411 968 × 2 = 1 + 0.500 001 572 340 614 823 936;
  • 45) 0.500 001 572 340 614 823 936 × 2 = 1 + 0.000 003 144 681 229 647 872;
  • 46) 0.000 003 144 681 229 647 872 × 2 = 0 + 0.000 006 289 362 459 295 744;
  • 47) 0.000 006 289 362 459 295 744 × 2 = 0 + 0.000 012 578 724 918 591 488;
  • 48) 0.000 012 578 724 918 591 488 × 2 = 0 + 0.000 025 157 449 837 182 976;
  • 49) 0.000 025 157 449 837 182 976 × 2 = 0 + 0.000 050 314 899 674 365 952;
  • 50) 0.000 050 314 899 674 365 952 × 2 = 0 + 0.000 100 629 799 348 731 904;
  • 51) 0.000 100 629 799 348 731 904 × 2 = 0 + 0.000 201 259 598 697 463 808;
  • 52) 0.000 201 259 598 697 463 808 × 2 = 0 + 0.000 402 519 197 394 927 616;
  • 53) 0.000 402 519 197 394 927 616 × 2 = 0 + 0.000 805 038 394 789 855 232;

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:


0.219 999 999 999 998 863 221(10) =

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

6. Positive number before normalization:

199.219 999 999 999 998 863 221(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:


199.219 999 999 999 998 863 221(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 -199.219 999 999 999 998 863 221 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

» Convert decimal -199.219 999 999 999 998 863 288 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