1.732 050 808 227 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.732 050 808 227(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
1.732 050 808 227(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.732 050 808 227.

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) 0.732 050 808 227 × 2 = 1 + 0.464 101 616 454;
  • 2) 0.464 101 616 454 × 2 = 0 + 0.928 203 232 908;
  • 3) 0.928 203 232 908 × 2 = 1 + 0.856 406 465 816;
  • 4) 0.856 406 465 816 × 2 = 1 + 0.712 812 931 632;
  • 5) 0.712 812 931 632 × 2 = 1 + 0.425 625 863 264;
  • 6) 0.425 625 863 264 × 2 = 0 + 0.851 251 726 528;
  • 7) 0.851 251 726 528 × 2 = 1 + 0.702 503 453 056;
  • 8) 0.702 503 453 056 × 2 = 1 + 0.405 006 906 112;
  • 9) 0.405 006 906 112 × 2 = 0 + 0.810 013 812 224;
  • 10) 0.810 013 812 224 × 2 = 1 + 0.620 027 624 448;
  • 11) 0.620 027 624 448 × 2 = 1 + 0.240 055 248 896;
  • 12) 0.240 055 248 896 × 2 = 0 + 0.480 110 497 792;
  • 13) 0.480 110 497 792 × 2 = 0 + 0.960 220 995 584;
  • 14) 0.960 220 995 584 × 2 = 1 + 0.920 441 991 168;
  • 15) 0.920 441 991 168 × 2 = 1 + 0.840 883 982 336;
  • 16) 0.840 883 982 336 × 2 = 1 + 0.681 767 964 672;
  • 17) 0.681 767 964 672 × 2 = 1 + 0.363 535 929 344;
  • 18) 0.363 535 929 344 × 2 = 0 + 0.727 071 858 688;
  • 19) 0.727 071 858 688 × 2 = 1 + 0.454 143 717 376;
  • 20) 0.454 143 717 376 × 2 = 0 + 0.908 287 434 752;
  • 21) 0.908 287 434 752 × 2 = 1 + 0.816 574 869 504;
  • 22) 0.816 574 869 504 × 2 = 1 + 0.633 149 739 008;
  • 23) 0.633 149 739 008 × 2 = 1 + 0.266 299 478 016;
  • 24) 0.266 299 478 016 × 2 = 0 + 0.532 598 956 032;
  • 25) 0.532 598 956 032 × 2 = 1 + 0.065 197 912 064;
  • 26) 0.065 197 912 064 × 2 = 0 + 0.130 395 824 128;
  • 27) 0.130 395 824 128 × 2 = 0 + 0.260 791 648 256;
  • 28) 0.260 791 648 256 × 2 = 0 + 0.521 583 296 512;
  • 29) 0.521 583 296 512 × 2 = 1 + 0.043 166 593 024;
  • 30) 0.043 166 593 024 × 2 = 0 + 0.086 333 186 048;
  • 31) 0.086 333 186 048 × 2 = 0 + 0.172 666 372 096;
  • 32) 0.172 666 372 096 × 2 = 0 + 0.345 332 744 192;
  • 33) 0.345 332 744 192 × 2 = 0 + 0.690 665 488 384;
  • 34) 0.690 665 488 384 × 2 = 1 + 0.381 330 976 768;
  • 35) 0.381 330 976 768 × 2 = 0 + 0.762 661 953 536;
  • 36) 0.762 661 953 536 × 2 = 1 + 0.525 323 907 072;
  • 37) 0.525 323 907 072 × 2 = 1 + 0.050 647 814 144;
  • 38) 0.050 647 814 144 × 2 = 0 + 0.101 295 628 288;
  • 39) 0.101 295 628 288 × 2 = 0 + 0.202 591 256 576;
  • 40) 0.202 591 256 576 × 2 = 0 + 0.405 182 513 152;
  • 41) 0.405 182 513 152 × 2 = 0 + 0.810 365 026 304;
  • 42) 0.810 365 026 304 × 2 = 1 + 0.620 730 052 608;
  • 43) 0.620 730 052 608 × 2 = 1 + 0.241 460 105 216;
  • 44) 0.241 460 105 216 × 2 = 0 + 0.482 920 210 432;
  • 45) 0.482 920 210 432 × 2 = 0 + 0.965 840 420 864;
  • 46) 0.965 840 420 864 × 2 = 1 + 0.931 680 841 728;
  • 47) 0.931 680 841 728 × 2 = 1 + 0.863 361 683 456;
  • 48) 0.863 361 683 456 × 2 = 1 + 0.726 723 366 912;
  • 49) 0.726 723 366 912 × 2 = 1 + 0.453 446 733 824;
  • 50) 0.453 446 733 824 × 2 = 0 + 0.906 893 467 648;
  • 51) 0.906 893 467 648 × 2 = 1 + 0.813 786 935 296;
  • 52) 0.813 786 935 296 × 2 = 1 + 0.627 573 870 592;
  • 53) 0.627 573 870 592 × 2 = 1 + 0.255 147 741 184;

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:


0.732 050 808 227(10) =

0.1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1(2)

5. Positive number before normalization:

1.732 050 808 227(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.732 050 808 227(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1(2) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1(2) × 20


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): 0

Mantissa (not normalized):
1.1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011 1 =

1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011


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) =
011 1111 1111

Mantissa (52 bits) =
1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011


Decimal number 1.732 050 808 227 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1011 0110 0111 1010 1110 1000 1000 0101 1000 0110 0111 1011

» Convert decimal 1.732 050 808 183 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