1.307 100 402 893 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.307 100 402 893 9(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
1.307 100 402 893 9(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: 0.307 100 402 893 9.

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) 0.307 100 402 893 9 × 2 = 0 + 0.614 200 805 787 8;
  • 2) 0.614 200 805 787 8 × 2 = 1 + 0.228 401 611 575 6;
  • 3) 0.228 401 611 575 6 × 2 = 0 + 0.456 803 223 151 2;
  • 4) 0.456 803 223 151 2 × 2 = 0 + 0.913 606 446 302 4;
  • 5) 0.913 606 446 302 4 × 2 = 1 + 0.827 212 892 604 8;
  • 6) 0.827 212 892 604 8 × 2 = 1 + 0.654 425 785 209 6;
  • 7) 0.654 425 785 209 6 × 2 = 1 + 0.308 851 570 419 2;
  • 8) 0.308 851 570 419 2 × 2 = 0 + 0.617 703 140 838 4;
  • 9) 0.617 703 140 838 4 × 2 = 1 + 0.235 406 281 676 8;
  • 10) 0.235 406 281 676 8 × 2 = 0 + 0.470 812 563 353 6;
  • 11) 0.470 812 563 353 6 × 2 = 0 + 0.941 625 126 707 2;
  • 12) 0.941 625 126 707 2 × 2 = 1 + 0.883 250 253 414 4;
  • 13) 0.883 250 253 414 4 × 2 = 1 + 0.766 500 506 828 8;
  • 14) 0.766 500 506 828 8 × 2 = 1 + 0.533 001 013 657 6;
  • 15) 0.533 001 013 657 6 × 2 = 1 + 0.066 002 027 315 2;
  • 16) 0.066 002 027 315 2 × 2 = 0 + 0.132 004 054 630 4;
  • 17) 0.132 004 054 630 4 × 2 = 0 + 0.264 008 109 260 8;
  • 18) 0.264 008 109 260 8 × 2 = 0 + 0.528 016 218 521 6;
  • 19) 0.528 016 218 521 6 × 2 = 1 + 0.056 032 437 043 2;
  • 20) 0.056 032 437 043 2 × 2 = 0 + 0.112 064 874 086 4;
  • 21) 0.112 064 874 086 4 × 2 = 0 + 0.224 129 748 172 8;
  • 22) 0.224 129 748 172 8 × 2 = 0 + 0.448 259 496 345 6;
  • 23) 0.448 259 496 345 6 × 2 = 0 + 0.896 518 992 691 2;
  • 24) 0.896 518 992 691 2 × 2 = 1 + 0.793 037 985 382 4;
  • 25) 0.793 037 985 382 4 × 2 = 1 + 0.586 075 970 764 8;
  • 26) 0.586 075 970 764 8 × 2 = 1 + 0.172 151 941 529 6;
  • 27) 0.172 151 941 529 6 × 2 = 0 + 0.344 303 883 059 2;
  • 28) 0.344 303 883 059 2 × 2 = 0 + 0.688 607 766 118 4;
  • 29) 0.688 607 766 118 4 × 2 = 1 + 0.377 215 532 236 8;
  • 30) 0.377 215 532 236 8 × 2 = 0 + 0.754 431 064 473 6;
  • 31) 0.754 431 064 473 6 × 2 = 1 + 0.508 862 128 947 2;
  • 32) 0.508 862 128 947 2 × 2 = 1 + 0.017 724 257 894 4;
  • 33) 0.017 724 257 894 4 × 2 = 0 + 0.035 448 515 788 8;
  • 34) 0.035 448 515 788 8 × 2 = 0 + 0.070 897 031 577 6;
  • 35) 0.070 897 031 577 6 × 2 = 0 + 0.141 794 063 155 2;
  • 36) 0.141 794 063 155 2 × 2 = 0 + 0.283 588 126 310 4;
  • 37) 0.283 588 126 310 4 × 2 = 0 + 0.567 176 252 620 8;
  • 38) 0.567 176 252 620 8 × 2 = 1 + 0.134 352 505 241 6;
  • 39) 0.134 352 505 241 6 × 2 = 0 + 0.268 705 010 483 2;
  • 40) 0.268 705 010 483 2 × 2 = 0 + 0.537 410 020 966 4;
  • 41) 0.537 410 020 966 4 × 2 = 1 + 0.074 820 041 932 8;
  • 42) 0.074 820 041 932 8 × 2 = 0 + 0.149 640 083 865 6;
  • 43) 0.149 640 083 865 6 × 2 = 0 + 0.299 280 167 731 2;
  • 44) 0.299 280 167 731 2 × 2 = 0 + 0.598 560 335 462 4;
  • 45) 0.598 560 335 462 4 × 2 = 1 + 0.197 120 670 924 8;
  • 46) 0.197 120 670 924 8 × 2 = 0 + 0.394 241 341 849 6;
  • 47) 0.394 241 341 849 6 × 2 = 0 + 0.788 482 683 699 2;
  • 48) 0.788 482 683 699 2 × 2 = 1 + 0.576 965 367 398 4;
  • 49) 0.576 965 367 398 4 × 2 = 1 + 0.153 930 734 796 8;
  • 50) 0.153 930 734 796 8 × 2 = 0 + 0.307 861 469 593 6;
  • 51) 0.307 861 469 593 6 × 2 = 0 + 0.615 722 939 187 2;
  • 52) 0.615 722 939 187 2 × 2 = 1 + 0.231 445 878 374 4;
  • 53) 0.231 445 878 374 4 × 2 = 0 + 0.462 891 756 748 8;

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:


0.307 100 402 893 9(10) =

0.0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 1001 0(2)

5. Positive number before normalization:

1.307 100 402 893 9(10) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 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:


1.307 100 402 893 9(10) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 1001 0(2) =

1.0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 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 0000 0100 1000 1001 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 0000 0100 1000 1001 1001 0 =

0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 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 0000 0100 1000 1001 1001


Decimal number 1.307 100 402 893 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0100 1110 1001 1110 0010 0001 1100 1011 0000 0100 1000 1001 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