39.776 399 999 951 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 39.776 399 999 951 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
39.776 399 999 951 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: 39.
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;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

39(10) =

10 0111(2)


3. Convert to binary (base 2) the fractional part: 0.776 399 999 951 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) 0.776 399 999 951 1 × 2 = 1 + 0.552 799 999 902 2;
  • 2) 0.552 799 999 902 2 × 2 = 1 + 0.105 599 999 804 4;
  • 3) 0.105 599 999 804 4 × 2 = 0 + 0.211 199 999 608 8;
  • 4) 0.211 199 999 608 8 × 2 = 0 + 0.422 399 999 217 6;
  • 5) 0.422 399 999 217 6 × 2 = 0 + 0.844 799 998 435 2;
  • 6) 0.844 799 998 435 2 × 2 = 1 + 0.689 599 996 870 4;
  • 7) 0.689 599 996 870 4 × 2 = 1 + 0.379 199 993 740 8;
  • 8) 0.379 199 993 740 8 × 2 = 0 + 0.758 399 987 481 6;
  • 9) 0.758 399 987 481 6 × 2 = 1 + 0.516 799 974 963 2;
  • 10) 0.516 799 974 963 2 × 2 = 1 + 0.033 599 949 926 4;
  • 11) 0.033 599 949 926 4 × 2 = 0 + 0.067 199 899 852 8;
  • 12) 0.067 199 899 852 8 × 2 = 0 + 0.134 399 799 705 6;
  • 13) 0.134 399 799 705 6 × 2 = 0 + 0.268 799 599 411 2;
  • 14) 0.268 799 599 411 2 × 2 = 0 + 0.537 599 198 822 4;
  • 15) 0.537 599 198 822 4 × 2 = 1 + 0.075 198 397 644 8;
  • 16) 0.075 198 397 644 8 × 2 = 0 + 0.150 396 795 289 6;
  • 17) 0.150 396 795 289 6 × 2 = 0 + 0.300 793 590 579 2;
  • 18) 0.300 793 590 579 2 × 2 = 0 + 0.601 587 181 158 4;
  • 19) 0.601 587 181 158 4 × 2 = 1 + 0.203 174 362 316 8;
  • 20) 0.203 174 362 316 8 × 2 = 0 + 0.406 348 724 633 6;
  • 21) 0.406 348 724 633 6 × 2 = 0 + 0.812 697 449 267 2;
  • 22) 0.812 697 449 267 2 × 2 = 1 + 0.625 394 898 534 4;
  • 23) 0.625 394 898 534 4 × 2 = 1 + 0.250 789 797 068 8;
  • 24) 0.250 789 797 068 8 × 2 = 0 + 0.501 579 594 137 6;
  • 25) 0.501 579 594 137 6 × 2 = 1 + 0.003 159 188 275 2;
  • 26) 0.003 159 188 275 2 × 2 = 0 + 0.006 318 376 550 4;
  • 27) 0.006 318 376 550 4 × 2 = 0 + 0.012 636 753 100 8;
  • 28) 0.012 636 753 100 8 × 2 = 0 + 0.025 273 506 201 6;
  • 29) 0.025 273 506 201 6 × 2 = 0 + 0.050 547 012 403 2;
  • 30) 0.050 547 012 403 2 × 2 = 0 + 0.101 094 024 806 4;
  • 31) 0.101 094 024 806 4 × 2 = 0 + 0.202 188 049 612 8;
  • 32) 0.202 188 049 612 8 × 2 = 0 + 0.404 376 099 225 6;
  • 33) 0.404 376 099 225 6 × 2 = 0 + 0.808 752 198 451 2;
  • 34) 0.808 752 198 451 2 × 2 = 1 + 0.617 504 396 902 4;
  • 35) 0.617 504 396 902 4 × 2 = 1 + 0.235 008 793 804 8;
  • 36) 0.235 008 793 804 8 × 2 = 0 + 0.470 017 587 609 6;
  • 37) 0.470 017 587 609 6 × 2 = 0 + 0.940 035 175 219 2;
  • 38) 0.940 035 175 219 2 × 2 = 1 + 0.880 070 350 438 4;
  • 39) 0.880 070 350 438 4 × 2 = 1 + 0.760 140 700 876 8;
  • 40) 0.760 140 700 876 8 × 2 = 1 + 0.520 281 401 753 6;
  • 41) 0.520 281 401 753 6 × 2 = 1 + 0.040 562 803 507 2;
  • 42) 0.040 562 803 507 2 × 2 = 0 + 0.081 125 607 014 4;
  • 43) 0.081 125 607 014 4 × 2 = 0 + 0.162 251 214 028 8;
  • 44) 0.162 251 214 028 8 × 2 = 0 + 0.324 502 428 057 6;
  • 45) 0.324 502 428 057 6 × 2 = 0 + 0.649 004 856 115 2;
  • 46) 0.649 004 856 115 2 × 2 = 1 + 0.298 009 712 230 4;
  • 47) 0.298 009 712 230 4 × 2 = 0 + 0.596 019 424 460 8;
  • 48) 0.596 019 424 460 8 × 2 = 1 + 0.192 038 848 921 6;
  • 49) 0.192 038 848 921 6 × 2 = 0 + 0.384 077 697 843 2;
  • 50) 0.384 077 697 843 2 × 2 = 0 + 0.768 155 395 686 4;
  • 51) 0.768 155 395 686 4 × 2 = 1 + 0.536 310 791 372 8;
  • 52) 0.536 310 791 372 8 × 2 = 1 + 0.072 621 582 745 6;
  • 53) 0.072 621 582 745 6 × 2 = 0 + 0.145 243 165 491 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:


0.776 399 999 951 1(10) =

0.1100 0110 1100 0010 0010 0110 1000 0000 0110 0111 1000 0101 0011 0(2)

5. Positive number before normalization:

39.776 399 999 951 1(10) =

10 0111.1100 0110 1100 0010 0010 0110 1000 0000 0110 0111 1000 0101 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


39.776 399 999 951 1(10) =

10 0111.1100 0110 1100 0010 0010 0110 1000 0000 0110 0111 1000 0101 0011 0(2) =

10 0111.1100 0110 1100 0010 0010 0110 1000 0000 0110 0111 1000 0101 0011 0(2) × 20 =

1.0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010 1001 10(2) × 25


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): 5

Mantissa (not normalized):
1.0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010 1001 10


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1028(10) =

100 0000 0100(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. 0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010 10 0110 =

0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010


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) =
100 0000 0100

Mantissa (52 bits) =
0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010


Decimal number 39.776 399 999 951 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 0011 1110 0011 0110 0001 0001 0011 0100 0000 0011 0011 1100 0010

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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