1.785 714 285 715 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.785 714 285 715 6(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
1.785 714 285 715 6(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: 0.785 714 285 715 6.

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) 0.785 714 285 715 6 × 2 = 1 + 0.571 428 571 431 2;
  • 2) 0.571 428 571 431 2 × 2 = 1 + 0.142 857 142 862 4;
  • 3) 0.142 857 142 862 4 × 2 = 0 + 0.285 714 285 724 8;
  • 4) 0.285 714 285 724 8 × 2 = 0 + 0.571 428 571 449 6;
  • 5) 0.571 428 571 449 6 × 2 = 1 + 0.142 857 142 899 2;
  • 6) 0.142 857 142 899 2 × 2 = 0 + 0.285 714 285 798 4;
  • 7) 0.285 714 285 798 4 × 2 = 0 + 0.571 428 571 596 8;
  • 8) 0.571 428 571 596 8 × 2 = 1 + 0.142 857 143 193 6;
  • 9) 0.142 857 143 193 6 × 2 = 0 + 0.285 714 286 387 2;
  • 10) 0.285 714 286 387 2 × 2 = 0 + 0.571 428 572 774 4;
  • 11) 0.571 428 572 774 4 × 2 = 1 + 0.142 857 145 548 8;
  • 12) 0.142 857 145 548 8 × 2 = 0 + 0.285 714 291 097 6;
  • 13) 0.285 714 291 097 6 × 2 = 0 + 0.571 428 582 195 2;
  • 14) 0.571 428 582 195 2 × 2 = 1 + 0.142 857 164 390 4;
  • 15) 0.142 857 164 390 4 × 2 = 0 + 0.285 714 328 780 8;
  • 16) 0.285 714 328 780 8 × 2 = 0 + 0.571 428 657 561 6;
  • 17) 0.571 428 657 561 6 × 2 = 1 + 0.142 857 315 123 2;
  • 18) 0.142 857 315 123 2 × 2 = 0 + 0.285 714 630 246 4;
  • 19) 0.285 714 630 246 4 × 2 = 0 + 0.571 429 260 492 8;
  • 20) 0.571 429 260 492 8 × 2 = 1 + 0.142 858 520 985 6;
  • 21) 0.142 858 520 985 6 × 2 = 0 + 0.285 717 041 971 2;
  • 22) 0.285 717 041 971 2 × 2 = 0 + 0.571 434 083 942 4;
  • 23) 0.571 434 083 942 4 × 2 = 1 + 0.142 868 167 884 8;
  • 24) 0.142 868 167 884 8 × 2 = 0 + 0.285 736 335 769 6;
  • 25) 0.285 736 335 769 6 × 2 = 0 + 0.571 472 671 539 2;
  • 26) 0.571 472 671 539 2 × 2 = 1 + 0.142 945 343 078 4;
  • 27) 0.142 945 343 078 4 × 2 = 0 + 0.285 890 686 156 8;
  • 28) 0.285 890 686 156 8 × 2 = 0 + 0.571 781 372 313 6;
  • 29) 0.571 781 372 313 6 × 2 = 1 + 0.143 562 744 627 2;
  • 30) 0.143 562 744 627 2 × 2 = 0 + 0.287 125 489 254 4;
  • 31) 0.287 125 489 254 4 × 2 = 0 + 0.574 250 978 508 8;
  • 32) 0.574 250 978 508 8 × 2 = 1 + 0.148 501 957 017 6;
  • 33) 0.148 501 957 017 6 × 2 = 0 + 0.297 003 914 035 2;
  • 34) 0.297 003 914 035 2 × 2 = 0 + 0.594 007 828 070 4;
  • 35) 0.594 007 828 070 4 × 2 = 1 + 0.188 015 656 140 8;
  • 36) 0.188 015 656 140 8 × 2 = 0 + 0.376 031 312 281 6;
  • 37) 0.376 031 312 281 6 × 2 = 0 + 0.752 062 624 563 2;
  • 38) 0.752 062 624 563 2 × 2 = 1 + 0.504 125 249 126 4;
  • 39) 0.504 125 249 126 4 × 2 = 1 + 0.008 250 498 252 8;
  • 40) 0.008 250 498 252 8 × 2 = 0 + 0.016 500 996 505 6;
  • 41) 0.016 500 996 505 6 × 2 = 0 + 0.033 001 993 011 2;
  • 42) 0.033 001 993 011 2 × 2 = 0 + 0.066 003 986 022 4;
  • 43) 0.066 003 986 022 4 × 2 = 0 + 0.132 007 972 044 8;
  • 44) 0.132 007 972 044 8 × 2 = 0 + 0.264 015 944 089 6;
  • 45) 0.264 015 944 089 6 × 2 = 0 + 0.528 031 888 179 2;
  • 46) 0.528 031 888 179 2 × 2 = 1 + 0.056 063 776 358 4;
  • 47) 0.056 063 776 358 4 × 2 = 0 + 0.112 127 552 716 8;
  • 48) 0.112 127 552 716 8 × 2 = 0 + 0.224 255 105 433 6;
  • 49) 0.224 255 105 433 6 × 2 = 0 + 0.448 510 210 867 2;
  • 50) 0.448 510 210 867 2 × 2 = 0 + 0.897 020 421 734 4;
  • 51) 0.897 020 421 734 4 × 2 = 1 + 0.794 040 843 468 8;
  • 52) 0.794 040 843 468 8 × 2 = 1 + 0.588 081 686 937 6;
  • 53) 0.588 081 686 937 6 × 2 = 1 + 0.176 163 373 875 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:


0.785 714 285 715 6(10) =

0.1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 1(2)

5. Positive number before normalization:

1.785 714 285 715 6(10) =

1.1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 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:


1.785 714 285 715 6(10) =

1.1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 1(2) =

1.1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 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):
1.1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 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. 1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011 1 =

1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011


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) =
1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011


Decimal number 1.785 714 285 715 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0000 0100 0011

» Convert decimal 1.785 714 285 723 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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