-84.129 999 999 999 811 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -84.129 999 999 999 811(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
-84.129 999 999 999 811(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:

|-84.129 999 999 999 811| = 84.129 999 999 999 811


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: 0.129 999 999 999 811.

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) 0.129 999 999 999 811 × 2 = 0 + 0.259 999 999 999 622;
  • 2) 0.259 999 999 999 622 × 2 = 0 + 0.519 999 999 999 244;
  • 3) 0.519 999 999 999 244 × 2 = 1 + 0.039 999 999 998 488;
  • 4) 0.039 999 999 998 488 × 2 = 0 + 0.079 999 999 996 976;
  • 5) 0.079 999 999 996 976 × 2 = 0 + 0.159 999 999 993 952;
  • 6) 0.159 999 999 993 952 × 2 = 0 + 0.319 999 999 987 904;
  • 7) 0.319 999 999 987 904 × 2 = 0 + 0.639 999 999 975 808;
  • 8) 0.639 999 999 975 808 × 2 = 1 + 0.279 999 999 951 616;
  • 9) 0.279 999 999 951 616 × 2 = 0 + 0.559 999 999 903 232;
  • 10) 0.559 999 999 903 232 × 2 = 1 + 0.119 999 999 806 464;
  • 11) 0.119 999 999 806 464 × 2 = 0 + 0.239 999 999 612 928;
  • 12) 0.239 999 999 612 928 × 2 = 0 + 0.479 999 999 225 856;
  • 13) 0.479 999 999 225 856 × 2 = 0 + 0.959 999 998 451 712;
  • 14) 0.959 999 998 451 712 × 2 = 1 + 0.919 999 996 903 424;
  • 15) 0.919 999 996 903 424 × 2 = 1 + 0.839 999 993 806 848;
  • 16) 0.839 999 993 806 848 × 2 = 1 + 0.679 999 987 613 696;
  • 17) 0.679 999 987 613 696 × 2 = 1 + 0.359 999 975 227 392;
  • 18) 0.359 999 975 227 392 × 2 = 0 + 0.719 999 950 454 784;
  • 19) 0.719 999 950 454 784 × 2 = 1 + 0.439 999 900 909 568;
  • 20) 0.439 999 900 909 568 × 2 = 0 + 0.879 999 801 819 136;
  • 21) 0.879 999 801 819 136 × 2 = 1 + 0.759 999 603 638 272;
  • 22) 0.759 999 603 638 272 × 2 = 1 + 0.519 999 207 276 544;
  • 23) 0.519 999 207 276 544 × 2 = 1 + 0.039 998 414 553 088;
  • 24) 0.039 998 414 553 088 × 2 = 0 + 0.079 996 829 106 176;
  • 25) 0.079 996 829 106 176 × 2 = 0 + 0.159 993 658 212 352;
  • 26) 0.159 993 658 212 352 × 2 = 0 + 0.319 987 316 424 704;
  • 27) 0.319 987 316 424 704 × 2 = 0 + 0.639 974 632 849 408;
  • 28) 0.639 974 632 849 408 × 2 = 1 + 0.279 949 265 698 816;
  • 29) 0.279 949 265 698 816 × 2 = 0 + 0.559 898 531 397 632;
  • 30) 0.559 898 531 397 632 × 2 = 1 + 0.119 797 062 795 264;
  • 31) 0.119 797 062 795 264 × 2 = 0 + 0.239 594 125 590 528;
  • 32) 0.239 594 125 590 528 × 2 = 0 + 0.479 188 251 181 056;
  • 33) 0.479 188 251 181 056 × 2 = 0 + 0.958 376 502 362 112;
  • 34) 0.958 376 502 362 112 × 2 = 1 + 0.916 753 004 724 224;
  • 35) 0.916 753 004 724 224 × 2 = 1 + 0.833 506 009 448 448;
  • 36) 0.833 506 009 448 448 × 2 = 1 + 0.667 012 018 896 896;
  • 37) 0.667 012 018 896 896 × 2 = 1 + 0.334 024 037 793 792;
  • 38) 0.334 024 037 793 792 × 2 = 0 + 0.668 048 075 587 584;
  • 39) 0.668 048 075 587 584 × 2 = 1 + 0.336 096 151 175 168;
  • 40) 0.336 096 151 175 168 × 2 = 0 + 0.672 192 302 350 336;
  • 41) 0.672 192 302 350 336 × 2 = 1 + 0.344 384 604 700 672;
  • 42) 0.344 384 604 700 672 × 2 = 0 + 0.688 769 209 401 344;
  • 43) 0.688 769 209 401 344 × 2 = 1 + 0.377 538 418 802 688;
  • 44) 0.377 538 418 802 688 × 2 = 0 + 0.755 076 837 605 376;
  • 45) 0.755 076 837 605 376 × 2 = 1 + 0.510 153 675 210 752;
  • 46) 0.510 153 675 210 752 × 2 = 1 + 0.020 307 350 421 504;
  • 47) 0.020 307 350 421 504 × 2 = 0 + 0.040 614 700 843 008;
  • 48) 0.040 614 700 843 008 × 2 = 0 + 0.081 229 401 686 016;
  • 49) 0.081 229 401 686 016 × 2 = 0 + 0.162 458 803 372 032;
  • 50) 0.162 458 803 372 032 × 2 = 0 + 0.324 917 606 744 064;
  • 51) 0.324 917 606 744 064 × 2 = 0 + 0.649 835 213 488 128;
  • 52) 0.649 835 213 488 128 × 2 = 1 + 0.299 670 426 976 256;
  • 53) 0.299 670 426 976 256 × 2 = 0 + 0.599 340 853 952 512;

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:


0.129 999 999 999 811(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 1100 0001 0(2)

6. Positive number before normalization:

84.129 999 999 999 811(10) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 1100 0001 0(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:


84.129 999 999 999 811(10) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 1100 0001 0(2) =

101 0100.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 1100 0001 0(2) × 20 =

1.0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1011 0000 010(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 1010 1011 0000 010


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 1010 1011 000 0010 =

0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1011


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 1010 1011


Decimal number -84.129 999 999 999 811 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 1010 1011

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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