-84.129 999 999 999 941 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -84.129 999 999 999 941(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
-84.129 999 999 999 941(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:

|-84.129 999 999 999 941| = 84.129 999 999 999 941


2. First, convert to binary (in base 2) the integer part: 84.
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;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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.

84(10) =

101 0100(2)


4. Convert to binary (base 2) the fractional part: 0.129 999 999 999 941.

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) 0.129 999 999 999 941 × 2 = 0 + 0.259 999 999 999 882;
  • 2) 0.259 999 999 999 882 × 2 = 0 + 0.519 999 999 999 764;
  • 3) 0.519 999 999 999 764 × 2 = 1 + 0.039 999 999 999 528;
  • 4) 0.039 999 999 999 528 × 2 = 0 + 0.079 999 999 999 056;
  • 5) 0.079 999 999 999 056 × 2 = 0 + 0.159 999 999 998 112;
  • 6) 0.159 999 999 998 112 × 2 = 0 + 0.319 999 999 996 224;
  • 7) 0.319 999 999 996 224 × 2 = 0 + 0.639 999 999 992 448;
  • 8) 0.639 999 999 992 448 × 2 = 1 + 0.279 999 999 984 896;
  • 9) 0.279 999 999 984 896 × 2 = 0 + 0.559 999 999 969 792;
  • 10) 0.559 999 999 969 792 × 2 = 1 + 0.119 999 999 939 584;
  • 11) 0.119 999 999 939 584 × 2 = 0 + 0.239 999 999 879 168;
  • 12) 0.239 999 999 879 168 × 2 = 0 + 0.479 999 999 758 336;
  • 13) 0.479 999 999 758 336 × 2 = 0 + 0.959 999 999 516 672;
  • 14) 0.959 999 999 516 672 × 2 = 1 + 0.919 999 999 033 344;
  • 15) 0.919 999 999 033 344 × 2 = 1 + 0.839 999 998 066 688;
  • 16) 0.839 999 998 066 688 × 2 = 1 + 0.679 999 996 133 376;
  • 17) 0.679 999 996 133 376 × 2 = 1 + 0.359 999 992 266 752;
  • 18) 0.359 999 992 266 752 × 2 = 0 + 0.719 999 984 533 504;
  • 19) 0.719 999 984 533 504 × 2 = 1 + 0.439 999 969 067 008;
  • 20) 0.439 999 969 067 008 × 2 = 0 + 0.879 999 938 134 016;
  • 21) 0.879 999 938 134 016 × 2 = 1 + 0.759 999 876 268 032;
  • 22) 0.759 999 876 268 032 × 2 = 1 + 0.519 999 752 536 064;
  • 23) 0.519 999 752 536 064 × 2 = 1 + 0.039 999 505 072 128;
  • 24) 0.039 999 505 072 128 × 2 = 0 + 0.079 999 010 144 256;
  • 25) 0.079 999 010 144 256 × 2 = 0 + 0.159 998 020 288 512;
  • 26) 0.159 998 020 288 512 × 2 = 0 + 0.319 996 040 577 024;
  • 27) 0.319 996 040 577 024 × 2 = 0 + 0.639 992 081 154 048;
  • 28) 0.639 992 081 154 048 × 2 = 1 + 0.279 984 162 308 096;
  • 29) 0.279 984 162 308 096 × 2 = 0 + 0.559 968 324 616 192;
  • 30) 0.559 968 324 616 192 × 2 = 1 + 0.119 936 649 232 384;
  • 31) 0.119 936 649 232 384 × 2 = 0 + 0.239 873 298 464 768;
  • 32) 0.239 873 298 464 768 × 2 = 0 + 0.479 746 596 929 536;
  • 33) 0.479 746 596 929 536 × 2 = 0 + 0.959 493 193 859 072;
  • 34) 0.959 493 193 859 072 × 2 = 1 + 0.918 986 387 718 144;
  • 35) 0.918 986 387 718 144 × 2 = 1 + 0.837 972 775 436 288;
  • 36) 0.837 972 775 436 288 × 2 = 1 + 0.675 945 550 872 576;
  • 37) 0.675 945 550 872 576 × 2 = 1 + 0.351 891 101 745 152;
  • 38) 0.351 891 101 745 152 × 2 = 0 + 0.703 782 203 490 304;
  • 39) 0.703 782 203 490 304 × 2 = 1 + 0.407 564 406 980 608;
  • 40) 0.407 564 406 980 608 × 2 = 0 + 0.815 128 813 961 216;
  • 41) 0.815 128 813 961 216 × 2 = 1 + 0.630 257 627 922 432;
  • 42) 0.630 257 627 922 432 × 2 = 1 + 0.260 515 255 844 864;
  • 43) 0.260 515 255 844 864 × 2 = 0 + 0.521 030 511 689 728;
  • 44) 0.521 030 511 689 728 × 2 = 1 + 0.042 061 023 379 456;
  • 45) 0.042 061 023 379 456 × 2 = 0 + 0.084 122 046 758 912;
  • 46) 0.084 122 046 758 912 × 2 = 0 + 0.168 244 093 517 824;
  • 47) 0.168 244 093 517 824 × 2 = 0 + 0.336 488 187 035 648;
  • 48) 0.336 488 187 035 648 × 2 = 0 + 0.672 976 374 071 296;
  • 49) 0.672 976 374 071 296 × 2 = 1 + 0.345 952 748 142 592;
  • 50) 0.345 952 748 142 592 × 2 = 0 + 0.691 905 496 285 184;
  • 51) 0.691 905 496 285 184 × 2 = 1 + 0.383 810 992 570 368;
  • 52) 0.383 810 992 570 368 × 2 = 0 + 0.767 621 985 140 736;
  • 53) 0.767 621 985 140 736 × 2 = 1 + 0.535 243 970 281 472;

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:


0.129 999 999 999 941(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 0000 1010 1(2)

6. Positive number before normalization:

84.129 999 999 999 941(10) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 0000 1010 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


84.129 999 999 999 941(10) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 0000 1010 1(2) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 0000 1010 1(2) × 20 =

1.0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100 0010 101(2) × 26


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): 6

Mantissa (not normalized):
1.0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100 0010 101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 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) =

1029(10) =

100 0000 0101(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. 0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100 001 0101 =

0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100


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 0101

Mantissa (52 bits) =
0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100


Decimal number -84.129 999 999 999 941 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0100

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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