-84.129 999 999 999 926 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -84.129 999 999 999 926(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
-84.129 999 999 999 926(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:

|-84.129 999 999 999 926| = 84.129 999 999 999 926


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: 0.129 999 999 999 926.

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) 0.129 999 999 999 926 × 2 = 0 + 0.259 999 999 999 852;
  • 2) 0.259 999 999 999 852 × 2 = 0 + 0.519 999 999 999 704;
  • 3) 0.519 999 999 999 704 × 2 = 1 + 0.039 999 999 999 408;
  • 4) 0.039 999 999 999 408 × 2 = 0 + 0.079 999 999 998 816;
  • 5) 0.079 999 999 998 816 × 2 = 0 + 0.159 999 999 997 632;
  • 6) 0.159 999 999 997 632 × 2 = 0 + 0.319 999 999 995 264;
  • 7) 0.319 999 999 995 264 × 2 = 0 + 0.639 999 999 990 528;
  • 8) 0.639 999 999 990 528 × 2 = 1 + 0.279 999 999 981 056;
  • 9) 0.279 999 999 981 056 × 2 = 0 + 0.559 999 999 962 112;
  • 10) 0.559 999 999 962 112 × 2 = 1 + 0.119 999 999 924 224;
  • 11) 0.119 999 999 924 224 × 2 = 0 + 0.239 999 999 848 448;
  • 12) 0.239 999 999 848 448 × 2 = 0 + 0.479 999 999 696 896;
  • 13) 0.479 999 999 696 896 × 2 = 0 + 0.959 999 999 393 792;
  • 14) 0.959 999 999 393 792 × 2 = 1 + 0.919 999 998 787 584;
  • 15) 0.919 999 998 787 584 × 2 = 1 + 0.839 999 997 575 168;
  • 16) 0.839 999 997 575 168 × 2 = 1 + 0.679 999 995 150 336;
  • 17) 0.679 999 995 150 336 × 2 = 1 + 0.359 999 990 300 672;
  • 18) 0.359 999 990 300 672 × 2 = 0 + 0.719 999 980 601 344;
  • 19) 0.719 999 980 601 344 × 2 = 1 + 0.439 999 961 202 688;
  • 20) 0.439 999 961 202 688 × 2 = 0 + 0.879 999 922 405 376;
  • 21) 0.879 999 922 405 376 × 2 = 1 + 0.759 999 844 810 752;
  • 22) 0.759 999 844 810 752 × 2 = 1 + 0.519 999 689 621 504;
  • 23) 0.519 999 689 621 504 × 2 = 1 + 0.039 999 379 243 008;
  • 24) 0.039 999 379 243 008 × 2 = 0 + 0.079 998 758 486 016;
  • 25) 0.079 998 758 486 016 × 2 = 0 + 0.159 997 516 972 032;
  • 26) 0.159 997 516 972 032 × 2 = 0 + 0.319 995 033 944 064;
  • 27) 0.319 995 033 944 064 × 2 = 0 + 0.639 990 067 888 128;
  • 28) 0.639 990 067 888 128 × 2 = 1 + 0.279 980 135 776 256;
  • 29) 0.279 980 135 776 256 × 2 = 0 + 0.559 960 271 552 512;
  • 30) 0.559 960 271 552 512 × 2 = 1 + 0.119 920 543 105 024;
  • 31) 0.119 920 543 105 024 × 2 = 0 + 0.239 841 086 210 048;
  • 32) 0.239 841 086 210 048 × 2 = 0 + 0.479 682 172 420 096;
  • 33) 0.479 682 172 420 096 × 2 = 0 + 0.959 364 344 840 192;
  • 34) 0.959 364 344 840 192 × 2 = 1 + 0.918 728 689 680 384;
  • 35) 0.918 728 689 680 384 × 2 = 1 + 0.837 457 379 360 768;
  • 36) 0.837 457 379 360 768 × 2 = 1 + 0.674 914 758 721 536;
  • 37) 0.674 914 758 721 536 × 2 = 1 + 0.349 829 517 443 072;
  • 38) 0.349 829 517 443 072 × 2 = 0 + 0.699 659 034 886 144;
  • 39) 0.699 659 034 886 144 × 2 = 1 + 0.399 318 069 772 288;
  • 40) 0.399 318 069 772 288 × 2 = 0 + 0.798 636 139 544 576;
  • 41) 0.798 636 139 544 576 × 2 = 1 + 0.597 272 279 089 152;
  • 42) 0.597 272 279 089 152 × 2 = 1 + 0.194 544 558 178 304;
  • 43) 0.194 544 558 178 304 × 2 = 0 + 0.389 089 116 356 608;
  • 44) 0.389 089 116 356 608 × 2 = 0 + 0.778 178 232 713 216;
  • 45) 0.778 178 232 713 216 × 2 = 1 + 0.556 356 465 426 432;
  • 46) 0.556 356 465 426 432 × 2 = 1 + 0.112 712 930 852 864;
  • 47) 0.112 712 930 852 864 × 2 = 0 + 0.225 425 861 705 728;
  • 48) 0.225 425 861 705 728 × 2 = 0 + 0.450 851 723 411 456;
  • 49) 0.450 851 723 411 456 × 2 = 0 + 0.901 703 446 822 912;
  • 50) 0.901 703 446 822 912 × 2 = 1 + 0.803 406 893 645 824;
  • 51) 0.803 406 893 645 824 × 2 = 1 + 0.606 813 787 291 648;
  • 52) 0.606 813 787 291 648 × 2 = 1 + 0.213 627 574 583 296;
  • 53) 0.213 627 574 583 296 × 2 = 0 + 0.427 255 149 166 592;

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:


0.129 999 999 999 926(10) =

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

6. Positive number before normalization:

84.129 999 999 999 926(10) =

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


84.129 999 999 999 926(10) =

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

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

1.0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0011 0001 110(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 0011 0001 110


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 0011 000 1110 =

0101 0000 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0011


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 0011


Decimal number -84.129 999 999 999 926 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 0011

» Convert decimal -84.129 999 999 999 956 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