1.324 234 234 354 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.324 234 234 354(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
1.324 234 234 354(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: 0.324 234 234 354.

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) 0.324 234 234 354 × 2 = 0 + 0.648 468 468 708;
  • 2) 0.648 468 468 708 × 2 = 1 + 0.296 936 937 416;
  • 3) 0.296 936 937 416 × 2 = 0 + 0.593 873 874 832;
  • 4) 0.593 873 874 832 × 2 = 1 + 0.187 747 749 664;
  • 5) 0.187 747 749 664 × 2 = 0 + 0.375 495 499 328;
  • 6) 0.375 495 499 328 × 2 = 0 + 0.750 990 998 656;
  • 7) 0.750 990 998 656 × 2 = 1 + 0.501 981 997 312;
  • 8) 0.501 981 997 312 × 2 = 1 + 0.003 963 994 624;
  • 9) 0.003 963 994 624 × 2 = 0 + 0.007 927 989 248;
  • 10) 0.007 927 989 248 × 2 = 0 + 0.015 855 978 496;
  • 11) 0.015 855 978 496 × 2 = 0 + 0.031 711 956 992;
  • 12) 0.031 711 956 992 × 2 = 0 + 0.063 423 913 984;
  • 13) 0.063 423 913 984 × 2 = 0 + 0.126 847 827 968;
  • 14) 0.126 847 827 968 × 2 = 0 + 0.253 695 655 936;
  • 15) 0.253 695 655 936 × 2 = 0 + 0.507 391 311 872;
  • 16) 0.507 391 311 872 × 2 = 1 + 0.014 782 623 744;
  • 17) 0.014 782 623 744 × 2 = 0 + 0.029 565 247 488;
  • 18) 0.029 565 247 488 × 2 = 0 + 0.059 130 494 976;
  • 19) 0.059 130 494 976 × 2 = 0 + 0.118 260 989 952;
  • 20) 0.118 260 989 952 × 2 = 0 + 0.236 521 979 904;
  • 21) 0.236 521 979 904 × 2 = 0 + 0.473 043 959 808;
  • 22) 0.473 043 959 808 × 2 = 0 + 0.946 087 919 616;
  • 23) 0.946 087 919 616 × 2 = 1 + 0.892 175 839 232;
  • 24) 0.892 175 839 232 × 2 = 1 + 0.784 351 678 464;
  • 25) 0.784 351 678 464 × 2 = 1 + 0.568 703 356 928;
  • 26) 0.568 703 356 928 × 2 = 1 + 0.137 406 713 856;
  • 27) 0.137 406 713 856 × 2 = 0 + 0.274 813 427 712;
  • 28) 0.274 813 427 712 × 2 = 0 + 0.549 626 855 424;
  • 29) 0.549 626 855 424 × 2 = 1 + 0.099 253 710 848;
  • 30) 0.099 253 710 848 × 2 = 0 + 0.198 507 421 696;
  • 31) 0.198 507 421 696 × 2 = 0 + 0.397 014 843 392;
  • 32) 0.397 014 843 392 × 2 = 0 + 0.794 029 686 784;
  • 33) 0.794 029 686 784 × 2 = 1 + 0.588 059 373 568;
  • 34) 0.588 059 373 568 × 2 = 1 + 0.176 118 747 136;
  • 35) 0.176 118 747 136 × 2 = 0 + 0.352 237 494 272;
  • 36) 0.352 237 494 272 × 2 = 0 + 0.704 474 988 544;
  • 37) 0.704 474 988 544 × 2 = 1 + 0.408 949 977 088;
  • 38) 0.408 949 977 088 × 2 = 0 + 0.817 899 954 176;
  • 39) 0.817 899 954 176 × 2 = 1 + 0.635 799 908 352;
  • 40) 0.635 799 908 352 × 2 = 1 + 0.271 599 816 704;
  • 41) 0.271 599 816 704 × 2 = 0 + 0.543 199 633 408;
  • 42) 0.543 199 633 408 × 2 = 1 + 0.086 399 266 816;
  • 43) 0.086 399 266 816 × 2 = 0 + 0.172 798 533 632;
  • 44) 0.172 798 533 632 × 2 = 0 + 0.345 597 067 264;
  • 45) 0.345 597 067 264 × 2 = 0 + 0.691 194 134 528;
  • 46) 0.691 194 134 528 × 2 = 1 + 0.382 388 269 056;
  • 47) 0.382 388 269 056 × 2 = 0 + 0.764 776 538 112;
  • 48) 0.764 776 538 112 × 2 = 1 + 0.529 553 076 224;
  • 49) 0.529 553 076 224 × 2 = 1 + 0.059 106 152 448;
  • 50) 0.059 106 152 448 × 2 = 0 + 0.118 212 304 896;
  • 51) 0.118 212 304 896 × 2 = 0 + 0.236 424 609 792;
  • 52) 0.236 424 609 792 × 2 = 0 + 0.472 849 219 584;
  • 53) 0.472 849 219 584 × 2 = 0 + 0.945 698 439 168;

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:


0.324 234 234 354(10) =

0.0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0(2)

5. Positive number before normalization:

1.324 234 234 354(10) =

1.0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0(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:


1.324 234 234 354(10) =

1.0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0(2) =

1.0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0(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):
1.0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0


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. 0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000 0 =

0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000


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) =
0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000


Decimal number 1.324 234 234 354 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0101 0011 0000 0001 0000 0011 1100 1000 1100 1011 0100 0101 1000

» Convert decimal 1.324 234 234 335 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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