89.444 444 444 444 169 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 89.444 444 444 444 169(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
89.444 444 444 444 169(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. First, convert to binary (in base 2) the integer part: 89.
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;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

89(10) =

101 1001(2)


3. Convert to binary (base 2) the fractional part: 0.444 444 444 444 169.

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) 0.444 444 444 444 169 × 2 = 0 + 0.888 888 888 888 338;
  • 2) 0.888 888 888 888 338 × 2 = 1 + 0.777 777 777 776 676;
  • 3) 0.777 777 777 776 676 × 2 = 1 + 0.555 555 555 553 352;
  • 4) 0.555 555 555 553 352 × 2 = 1 + 0.111 111 111 106 704;
  • 5) 0.111 111 111 106 704 × 2 = 0 + 0.222 222 222 213 408;
  • 6) 0.222 222 222 213 408 × 2 = 0 + 0.444 444 444 426 816;
  • 7) 0.444 444 444 426 816 × 2 = 0 + 0.888 888 888 853 632;
  • 8) 0.888 888 888 853 632 × 2 = 1 + 0.777 777 777 707 264;
  • 9) 0.777 777 777 707 264 × 2 = 1 + 0.555 555 555 414 528;
  • 10) 0.555 555 555 414 528 × 2 = 1 + 0.111 111 110 829 056;
  • 11) 0.111 111 110 829 056 × 2 = 0 + 0.222 222 221 658 112;
  • 12) 0.222 222 221 658 112 × 2 = 0 + 0.444 444 443 316 224;
  • 13) 0.444 444 443 316 224 × 2 = 0 + 0.888 888 886 632 448;
  • 14) 0.888 888 886 632 448 × 2 = 1 + 0.777 777 773 264 896;
  • 15) 0.777 777 773 264 896 × 2 = 1 + 0.555 555 546 529 792;
  • 16) 0.555 555 546 529 792 × 2 = 1 + 0.111 111 093 059 584;
  • 17) 0.111 111 093 059 584 × 2 = 0 + 0.222 222 186 119 168;
  • 18) 0.222 222 186 119 168 × 2 = 0 + 0.444 444 372 238 336;
  • 19) 0.444 444 372 238 336 × 2 = 0 + 0.888 888 744 476 672;
  • 20) 0.888 888 744 476 672 × 2 = 1 + 0.777 777 488 953 344;
  • 21) 0.777 777 488 953 344 × 2 = 1 + 0.555 554 977 906 688;
  • 22) 0.555 554 977 906 688 × 2 = 1 + 0.111 109 955 813 376;
  • 23) 0.111 109 955 813 376 × 2 = 0 + 0.222 219 911 626 752;
  • 24) 0.222 219 911 626 752 × 2 = 0 + 0.444 439 823 253 504;
  • 25) 0.444 439 823 253 504 × 2 = 0 + 0.888 879 646 507 008;
  • 26) 0.888 879 646 507 008 × 2 = 1 + 0.777 759 293 014 016;
  • 27) 0.777 759 293 014 016 × 2 = 1 + 0.555 518 586 028 032;
  • 28) 0.555 518 586 028 032 × 2 = 1 + 0.111 037 172 056 064;
  • 29) 0.111 037 172 056 064 × 2 = 0 + 0.222 074 344 112 128;
  • 30) 0.222 074 344 112 128 × 2 = 0 + 0.444 148 688 224 256;
  • 31) 0.444 148 688 224 256 × 2 = 0 + 0.888 297 376 448 512;
  • 32) 0.888 297 376 448 512 × 2 = 1 + 0.776 594 752 897 024;
  • 33) 0.776 594 752 897 024 × 2 = 1 + 0.553 189 505 794 048;
  • 34) 0.553 189 505 794 048 × 2 = 1 + 0.106 379 011 588 096;
  • 35) 0.106 379 011 588 096 × 2 = 0 + 0.212 758 023 176 192;
  • 36) 0.212 758 023 176 192 × 2 = 0 + 0.425 516 046 352 384;
  • 37) 0.425 516 046 352 384 × 2 = 0 + 0.851 032 092 704 768;
  • 38) 0.851 032 092 704 768 × 2 = 1 + 0.702 064 185 409 536;
  • 39) 0.702 064 185 409 536 × 2 = 1 + 0.404 128 370 819 072;
  • 40) 0.404 128 370 819 072 × 2 = 0 + 0.808 256 741 638 144;
  • 41) 0.808 256 741 638 144 × 2 = 1 + 0.616 513 483 276 288;
  • 42) 0.616 513 483 276 288 × 2 = 1 + 0.233 026 966 552 576;
  • 43) 0.233 026 966 552 576 × 2 = 0 + 0.466 053 933 105 152;
  • 44) 0.466 053 933 105 152 × 2 = 0 + 0.932 107 866 210 304;
  • 45) 0.932 107 866 210 304 × 2 = 1 + 0.864 215 732 420 608;
  • 46) 0.864 215 732 420 608 × 2 = 1 + 0.728 431 464 841 216;
  • 47) 0.728 431 464 841 216 × 2 = 1 + 0.456 862 929 682 432;
  • 48) 0.456 862 929 682 432 × 2 = 0 + 0.913 725 859 364 864;
  • 49) 0.913 725 859 364 864 × 2 = 1 + 0.827 451 718 729 728;
  • 50) 0.827 451 718 729 728 × 2 = 1 + 0.654 903 437 459 456;
  • 51) 0.654 903 437 459 456 × 2 = 1 + 0.309 806 874 918 912;
  • 52) 0.309 806 874 918 912 × 2 = 0 + 0.619 613 749 837 824;
  • 53) 0.619 613 749 837 824 × 2 = 1 + 0.239 227 499 675 648;

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).


4. 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:


0.444 444 444 444 169(10) =

0.0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1100 1110 1110 1(2)

5. Positive number before normalization:

89.444 444 444 444 169(10) =

101 1001.0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1100 1110 1110 1(2)

6. 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:


89.444 444 444 444 169(10) =

101 1001.0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1100 1110 1110 1(2) =

101 1001.0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1100 1110 1110 1(2) × 20 =

1.0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011 1011 101(2) × 26


7. 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 0 (a positive number)

Exponent (unadjusted): 6

Mantissa (not normalized):
1.0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011 1011 101


8. 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)


9. 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;

10. 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)


11. 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. 0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011 101 1101 =

0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0101

Mantissa (52 bits) =
0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011


Decimal number 89.444 444 444 444 169 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0110 0101 1100 0111 0001 1100 0111 0001 1100 0111 0001 1011 0011

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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