64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1 721 212 122 115 241 421 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1 721 212 122 115 241 421(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. 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;
  • 1 721 212 122 115 241 421 ÷ 2 = 860 606 061 057 620 710 + 1;
  • 860 606 061 057 620 710 ÷ 2 = 430 303 030 528 810 355 + 0;
  • 430 303 030 528 810 355 ÷ 2 = 215 151 515 264 405 177 + 1;
  • 215 151 515 264 405 177 ÷ 2 = 107 575 757 632 202 588 + 1;
  • 107 575 757 632 202 588 ÷ 2 = 53 787 878 816 101 294 + 0;
  • 53 787 878 816 101 294 ÷ 2 = 26 893 939 408 050 647 + 0;
  • 26 893 939 408 050 647 ÷ 2 = 13 446 969 704 025 323 + 1;
  • 13 446 969 704 025 323 ÷ 2 = 6 723 484 852 012 661 + 1;
  • 6 723 484 852 012 661 ÷ 2 = 3 361 742 426 006 330 + 1;
  • 3 361 742 426 006 330 ÷ 2 = 1 680 871 213 003 165 + 0;
  • 1 680 871 213 003 165 ÷ 2 = 840 435 606 501 582 + 1;
  • 840 435 606 501 582 ÷ 2 = 420 217 803 250 791 + 0;
  • 420 217 803 250 791 ÷ 2 = 210 108 901 625 395 + 1;
  • 210 108 901 625 395 ÷ 2 = 105 054 450 812 697 + 1;
  • 105 054 450 812 697 ÷ 2 = 52 527 225 406 348 + 1;
  • 52 527 225 406 348 ÷ 2 = 26 263 612 703 174 + 0;
  • 26 263 612 703 174 ÷ 2 = 13 131 806 351 587 + 0;
  • 13 131 806 351 587 ÷ 2 = 6 565 903 175 793 + 1;
  • 6 565 903 175 793 ÷ 2 = 3 282 951 587 896 + 1;
  • 3 282 951 587 896 ÷ 2 = 1 641 475 793 948 + 0;
  • 1 641 475 793 948 ÷ 2 = 820 737 896 974 + 0;
  • 820 737 896 974 ÷ 2 = 410 368 948 487 + 0;
  • 410 368 948 487 ÷ 2 = 205 184 474 243 + 1;
  • 205 184 474 243 ÷ 2 = 102 592 237 121 + 1;
  • 102 592 237 121 ÷ 2 = 51 296 118 560 + 1;
  • 51 296 118 560 ÷ 2 = 25 648 059 280 + 0;
  • 25 648 059 280 ÷ 2 = 12 824 029 640 + 0;
  • 12 824 029 640 ÷ 2 = 6 412 014 820 + 0;
  • 6 412 014 820 ÷ 2 = 3 206 007 410 + 0;
  • 3 206 007 410 ÷ 2 = 1 603 003 705 + 0;
  • 1 603 003 705 ÷ 2 = 801 501 852 + 1;
  • 801 501 852 ÷ 2 = 400 750 926 + 0;
  • 400 750 926 ÷ 2 = 200 375 463 + 0;
  • 200 375 463 ÷ 2 = 100 187 731 + 1;
  • 100 187 731 ÷ 2 = 50 093 865 + 1;
  • 50 093 865 ÷ 2 = 25 046 932 + 1;
  • 25 046 932 ÷ 2 = 12 523 466 + 0;
  • 12 523 466 ÷ 2 = 6 261 733 + 0;
  • 6 261 733 ÷ 2 = 3 130 866 + 1;
  • 3 130 866 ÷ 2 = 1 565 433 + 0;
  • 1 565 433 ÷ 2 = 782 716 + 1;
  • 782 716 ÷ 2 = 391 358 + 0;
  • 391 358 ÷ 2 = 195 679 + 0;
  • 195 679 ÷ 2 = 97 839 + 1;
  • 97 839 ÷ 2 = 48 919 + 1;
  • 48 919 ÷ 2 = 24 459 + 1;
  • 24 459 ÷ 2 = 12 229 + 1;
  • 12 229 ÷ 2 = 6 114 + 1;
  • 6 114 ÷ 2 = 3 057 + 0;
  • 3 057 ÷ 2 = 1 528 + 1;
  • 1 528 ÷ 2 = 764 + 0;
  • 764 ÷ 2 = 382 + 0;
  • 382 ÷ 2 = 191 + 0;
  • 191 ÷ 2 = 95 + 1;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.


1 721 212 122 115 241 421(10) =


1 0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 60 positions to the left, so that only one non zero digit remains to the left of it:


1 721 212 122 115 241 421(10) =


1 0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101(2) =


1 0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101(2) × 20 =


1.0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101(2) × 260


4. 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): 60


Mantissa (not normalized):
1.0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


60 + 2(11-1) - 1 =


(60 + 1 023)(10) =


1 083(10)


6. 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;
  • 1 083 ÷ 2 = 541 + 1;
  • 541 ÷ 2 = 270 + 1;
  • 270 ÷ 2 = 135 + 0;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1083(10) =


100 0011 1011(2)


8. 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. 0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101 1100 1101 =


0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101


9. 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 0011 1011


Mantissa (52 bits) =
0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101


The base ten decimal number 1 721 212 122 115 241 421 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0011 1011 - 0111 1110 0010 1111 1001 0100 1110 0100 0001 1100 0110 0111 0101

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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