1.400 000 000 370 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.400 000 000 370 77(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
1.400 000 000 370 77(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: 0.400 000 000 370 77.

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) 0.400 000 000 370 77 × 2 = 0 + 0.800 000 000 741 54;
  • 2) 0.800 000 000 741 54 × 2 = 1 + 0.600 000 001 483 08;
  • 3) 0.600 000 001 483 08 × 2 = 1 + 0.200 000 002 966 16;
  • 4) 0.200 000 002 966 16 × 2 = 0 + 0.400 000 005 932 32;
  • 5) 0.400 000 005 932 32 × 2 = 0 + 0.800 000 011 864 64;
  • 6) 0.800 000 011 864 64 × 2 = 1 + 0.600 000 023 729 28;
  • 7) 0.600 000 023 729 28 × 2 = 1 + 0.200 000 047 458 56;
  • 8) 0.200 000 047 458 56 × 2 = 0 + 0.400 000 094 917 12;
  • 9) 0.400 000 094 917 12 × 2 = 0 + 0.800 000 189 834 24;
  • 10) 0.800 000 189 834 24 × 2 = 1 + 0.600 000 379 668 48;
  • 11) 0.600 000 379 668 48 × 2 = 1 + 0.200 000 759 336 96;
  • 12) 0.200 000 759 336 96 × 2 = 0 + 0.400 001 518 673 92;
  • 13) 0.400 001 518 673 92 × 2 = 0 + 0.800 003 037 347 84;
  • 14) 0.800 003 037 347 84 × 2 = 1 + 0.600 006 074 695 68;
  • 15) 0.600 006 074 695 68 × 2 = 1 + 0.200 012 149 391 36;
  • 16) 0.200 012 149 391 36 × 2 = 0 + 0.400 024 298 782 72;
  • 17) 0.400 024 298 782 72 × 2 = 0 + 0.800 048 597 565 44;
  • 18) 0.800 048 597 565 44 × 2 = 1 + 0.600 097 195 130 88;
  • 19) 0.600 097 195 130 88 × 2 = 1 + 0.200 194 390 261 76;
  • 20) 0.200 194 390 261 76 × 2 = 0 + 0.400 388 780 523 52;
  • 21) 0.400 388 780 523 52 × 2 = 0 + 0.800 777 561 047 04;
  • 22) 0.800 777 561 047 04 × 2 = 1 + 0.601 555 122 094 08;
  • 23) 0.601 555 122 094 08 × 2 = 1 + 0.203 110 244 188 16;
  • 24) 0.203 110 244 188 16 × 2 = 0 + 0.406 220 488 376 32;
  • 25) 0.406 220 488 376 32 × 2 = 0 + 0.812 440 976 752 64;
  • 26) 0.812 440 976 752 64 × 2 = 1 + 0.624 881 953 505 28;
  • 27) 0.624 881 953 505 28 × 2 = 1 + 0.249 763 907 010 56;
  • 28) 0.249 763 907 010 56 × 2 = 0 + 0.499 527 814 021 12;
  • 29) 0.499 527 814 021 12 × 2 = 0 + 0.999 055 628 042 24;
  • 30) 0.999 055 628 042 24 × 2 = 1 + 0.998 111 256 084 48;
  • 31) 0.998 111 256 084 48 × 2 = 1 + 0.996 222 512 168 96;
  • 32) 0.996 222 512 168 96 × 2 = 1 + 0.992 445 024 337 92;
  • 33) 0.992 445 024 337 92 × 2 = 1 + 0.984 890 048 675 84;
  • 34) 0.984 890 048 675 84 × 2 = 1 + 0.969 780 097 351 68;
  • 35) 0.969 780 097 351 68 × 2 = 1 + 0.939 560 194 703 36;
  • 36) 0.939 560 194 703 36 × 2 = 1 + 0.879 120 389 406 72;
  • 37) 0.879 120 389 406 72 × 2 = 1 + 0.758 240 778 813 44;
  • 38) 0.758 240 778 813 44 × 2 = 1 + 0.516 481 557 626 88;
  • 39) 0.516 481 557 626 88 × 2 = 1 + 0.032 963 115 253 76;
  • 40) 0.032 963 115 253 76 × 2 = 0 + 0.065 926 230 507 52;
  • 41) 0.065 926 230 507 52 × 2 = 0 + 0.131 852 461 015 04;
  • 42) 0.131 852 461 015 04 × 2 = 0 + 0.263 704 922 030 08;
  • 43) 0.263 704 922 030 08 × 2 = 0 + 0.527 409 844 060 16;
  • 44) 0.527 409 844 060 16 × 2 = 1 + 0.054 819 688 120 32;
  • 45) 0.054 819 688 120 32 × 2 = 0 + 0.109 639 376 240 64;
  • 46) 0.109 639 376 240 64 × 2 = 0 + 0.219 278 752 481 28;
  • 47) 0.219 278 752 481 28 × 2 = 0 + 0.438 557 504 962 56;
  • 48) 0.438 557 504 962 56 × 2 = 0 + 0.877 115 009 925 12;
  • 49) 0.877 115 009 925 12 × 2 = 1 + 0.754 230 019 850 24;
  • 50) 0.754 230 019 850 24 × 2 = 1 + 0.508 460 039 700 48;
  • 51) 0.508 460 039 700 48 × 2 = 1 + 0.016 920 079 400 96;
  • 52) 0.016 920 079 400 96 × 2 = 0 + 0.033 840 158 801 92;
  • 53) 0.033 840 158 801 92 × 2 = 0 + 0.067 680 317 603 84;

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:


0.400 000 000 370 77(10) =

0.0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 0(2)

5. Positive number before normalization:

1.400 000 000 370 77(10) =

1.0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 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:


1.400 000 000 370 77(10) =

1.0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 0(2) =

1.0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 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):
1.0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 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. 0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110 0 =

0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110


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) =
0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110


Decimal number 1.400 000 000 370 77 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0110 0110 0110 0110 0110 0110 0110 0111 1111 1110 0001 0000 1110

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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