64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.010 110 113 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.010 110 113(10) converted and written in 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: 0.010 110 113.

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) 0.010 110 113 × 2 = 0 + 0.020 220 226;
  • 2) 0.020 220 226 × 2 = 0 + 0.040 440 452;
  • 3) 0.040 440 452 × 2 = 0 + 0.080 880 904;
  • 4) 0.080 880 904 × 2 = 0 + 0.161 761 808;
  • 5) 0.161 761 808 × 2 = 0 + 0.323 523 616;
  • 6) 0.323 523 616 × 2 = 0 + 0.647 047 232;
  • 7) 0.647 047 232 × 2 = 1 + 0.294 094 464;
  • 8) 0.294 094 464 × 2 = 0 + 0.588 188 928;
  • 9) 0.588 188 928 × 2 = 1 + 0.176 377 856;
  • 10) 0.176 377 856 × 2 = 0 + 0.352 755 712;
  • 11) 0.352 755 712 × 2 = 0 + 0.705 511 424;
  • 12) 0.705 511 424 × 2 = 1 + 0.411 022 848;
  • 13) 0.411 022 848 × 2 = 0 + 0.822 045 696;
  • 14) 0.822 045 696 × 2 = 1 + 0.644 091 392;
  • 15) 0.644 091 392 × 2 = 1 + 0.288 182 784;
  • 16) 0.288 182 784 × 2 = 0 + 0.576 365 568;
  • 17) 0.576 365 568 × 2 = 1 + 0.152 731 136;
  • 18) 0.152 731 136 × 2 = 0 + 0.305 462 272;
  • 19) 0.305 462 272 × 2 = 0 + 0.610 924 544;
  • 20) 0.610 924 544 × 2 = 1 + 0.221 849 088;
  • 21) 0.221 849 088 × 2 = 0 + 0.443 698 176;
  • 22) 0.443 698 176 × 2 = 0 + 0.887 396 352;
  • 23) 0.887 396 352 × 2 = 1 + 0.774 792 704;
  • 24) 0.774 792 704 × 2 = 1 + 0.549 585 408;
  • 25) 0.549 585 408 × 2 = 1 + 0.099 170 816;
  • 26) 0.099 170 816 × 2 = 0 + 0.198 341 632;
  • 27) 0.198 341 632 × 2 = 0 + 0.396 683 264;
  • 28) 0.396 683 264 × 2 = 0 + 0.793 366 528;
  • 29) 0.793 366 528 × 2 = 1 + 0.586 733 056;
  • 30) 0.586 733 056 × 2 = 1 + 0.173 466 112;
  • 31) 0.173 466 112 × 2 = 0 + 0.346 932 224;
  • 32) 0.346 932 224 × 2 = 0 + 0.693 864 448;
  • 33) 0.693 864 448 × 2 = 1 + 0.387 728 896;
  • 34) 0.387 728 896 × 2 = 0 + 0.775 457 792;
  • 35) 0.775 457 792 × 2 = 1 + 0.550 915 584;
  • 36) 0.550 915 584 × 2 = 1 + 0.101 831 168;
  • 37) 0.101 831 168 × 2 = 0 + 0.203 662 336;
  • 38) 0.203 662 336 × 2 = 0 + 0.407 324 672;
  • 39) 0.407 324 672 × 2 = 0 + 0.814 649 344;
  • 40) 0.814 649 344 × 2 = 1 + 0.629 298 688;
  • 41) 0.629 298 688 × 2 = 1 + 0.258 597 376;
  • 42) 0.258 597 376 × 2 = 0 + 0.517 194 752;
  • 43) 0.517 194 752 × 2 = 1 + 0.034 389 504;
  • 44) 0.034 389 504 × 2 = 0 + 0.068 779 008;
  • 45) 0.068 779 008 × 2 = 0 + 0.137 558 016;
  • 46) 0.137 558 016 × 2 = 0 + 0.275 116 032;
  • 47) 0.275 116 032 × 2 = 0 + 0.550 232 064;
  • 48) 0.550 232 064 × 2 = 1 + 0.100 464 128;
  • 49) 0.100 464 128 × 2 = 0 + 0.200 928 256;
  • 50) 0.200 928 256 × 2 = 0 + 0.401 856 512;
  • 51) 0.401 856 512 × 2 = 0 + 0.803 713 024;
  • 52) 0.803 713 024 × 2 = 1 + 0.607 426 048;
  • 53) 0.607 426 048 × 2 = 1 + 0.214 852 096;

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...)


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:


0.010 110 113(10) =


0.0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 1(2)


5. Positive number before normalization:

1.010 110 113(10) =


1.0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 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:


1.010 110 113(10) =


1.0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 1(2) =


1.0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 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):
1.0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 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. 0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001 1 =


0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001


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) =
0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001


The base ten decimal number 1.010 110 113 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 0000 0010 1001 0110 1001 0011 1000 1100 1011 0001 1010 0001 0001

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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