1.307 100 402 906 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.307 100 402 906 1(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
1.307 100 402 906 1(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: 0.307 100 402 906 1.

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) 0.307 100 402 906 1 × 2 = 0 + 0.614 200 805 812 2;
  • 2) 0.614 200 805 812 2 × 2 = 1 + 0.228 401 611 624 4;
  • 3) 0.228 401 611 624 4 × 2 = 0 + 0.456 803 223 248 8;
  • 4) 0.456 803 223 248 8 × 2 = 0 + 0.913 606 446 497 6;
  • 5) 0.913 606 446 497 6 × 2 = 1 + 0.827 212 892 995 2;
  • 6) 0.827 212 892 995 2 × 2 = 1 + 0.654 425 785 990 4;
  • 7) 0.654 425 785 990 4 × 2 = 1 + 0.308 851 571 980 8;
  • 8) 0.308 851 571 980 8 × 2 = 0 + 0.617 703 143 961 6;
  • 9) 0.617 703 143 961 6 × 2 = 1 + 0.235 406 287 923 2;
  • 10) 0.235 406 287 923 2 × 2 = 0 + 0.470 812 575 846 4;
  • 11) 0.470 812 575 846 4 × 2 = 0 + 0.941 625 151 692 8;
  • 12) 0.941 625 151 692 8 × 2 = 1 + 0.883 250 303 385 6;
  • 13) 0.883 250 303 385 6 × 2 = 1 + 0.766 500 606 771 2;
  • 14) 0.766 500 606 771 2 × 2 = 1 + 0.533 001 213 542 4;
  • 15) 0.533 001 213 542 4 × 2 = 1 + 0.066 002 427 084 8;
  • 16) 0.066 002 427 084 8 × 2 = 0 + 0.132 004 854 169 6;
  • 17) 0.132 004 854 169 6 × 2 = 0 + 0.264 009 708 339 2;
  • 18) 0.264 009 708 339 2 × 2 = 0 + 0.528 019 416 678 4;
  • 19) 0.528 019 416 678 4 × 2 = 1 + 0.056 038 833 356 8;
  • 20) 0.056 038 833 356 8 × 2 = 0 + 0.112 077 666 713 6;
  • 21) 0.112 077 666 713 6 × 2 = 0 + 0.224 155 333 427 2;
  • 22) 0.224 155 333 427 2 × 2 = 0 + 0.448 310 666 854 4;
  • 23) 0.448 310 666 854 4 × 2 = 0 + 0.896 621 333 708 8;
  • 24) 0.896 621 333 708 8 × 2 = 1 + 0.793 242 667 417 6;
  • 25) 0.793 242 667 417 6 × 2 = 1 + 0.586 485 334 835 2;
  • 26) 0.586 485 334 835 2 × 2 = 1 + 0.172 970 669 670 4;
  • 27) 0.172 970 669 670 4 × 2 = 0 + 0.345 941 339 340 8;
  • 28) 0.345 941 339 340 8 × 2 = 0 + 0.691 882 678 681 6;
  • 29) 0.691 882 678 681 6 × 2 = 1 + 0.383 765 357 363 2;
  • 30) 0.383 765 357 363 2 × 2 = 0 + 0.767 530 714 726 4;
  • 31) 0.767 530 714 726 4 × 2 = 1 + 0.535 061 429 452 8;
  • 32) 0.535 061 429 452 8 × 2 = 1 + 0.070 122 858 905 6;
  • 33) 0.070 122 858 905 6 × 2 = 0 + 0.140 245 717 811 2;
  • 34) 0.140 245 717 811 2 × 2 = 0 + 0.280 491 435 622 4;
  • 35) 0.280 491 435 622 4 × 2 = 0 + 0.560 982 871 244 8;
  • 36) 0.560 982 871 244 8 × 2 = 1 + 0.121 965 742 489 6;
  • 37) 0.121 965 742 489 6 × 2 = 0 + 0.243 931 484 979 2;
  • 38) 0.243 931 484 979 2 × 2 = 0 + 0.487 862 969 958 4;
  • 39) 0.487 862 969 958 4 × 2 = 0 + 0.975 725 939 916 8;
  • 40) 0.975 725 939 916 8 × 2 = 1 + 0.951 451 879 833 6;
  • 41) 0.951 451 879 833 6 × 2 = 1 + 0.902 903 759 667 2;
  • 42) 0.902 903 759 667 2 × 2 = 1 + 0.805 807 519 334 4;
  • 43) 0.805 807 519 334 4 × 2 = 1 + 0.611 615 038 668 8;
  • 44) 0.611 615 038 668 8 × 2 = 1 + 0.223 230 077 337 6;
  • 45) 0.223 230 077 337 6 × 2 = 0 + 0.446 460 154 675 2;
  • 46) 0.446 460 154 675 2 × 2 = 0 + 0.892 920 309 350 4;
  • 47) 0.892 920 309 350 4 × 2 = 1 + 0.785 840 618 700 8;
  • 48) 0.785 840 618 700 8 × 2 = 1 + 0.571 681 237 401 6;
  • 49) 0.571 681 237 401 6 × 2 = 1 + 0.143 362 474 803 2;
  • 50) 0.143 362 474 803 2 × 2 = 0 + 0.286 724 949 606 4;
  • 51) 0.286 724 949 606 4 × 2 = 0 + 0.573 449 899 212 8;
  • 52) 0.573 449 899 212 8 × 2 = 1 + 0.146 899 798 425 6;
  • 53) 0.146 899 798 425 6 × 2 = 0 + 0.293 799 596 851 2;

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:


0.307 100 402 906 1(10) =

0.0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 0(2)

5. Positive number before normalization:

1.307 100 402 906 1(10) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 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:


1.307 100 402 906 1(10) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 0(2) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 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):
1.0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 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. 0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001 0 =

0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001


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) =
0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001


Decimal number 1.307 100 402 906 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0100 1110 1001 1110 0010 0001 1100 1011 0001 0001 1111 0011 1001

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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