64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 976.562 627 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 976.562 627(10) converted and written in 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: 976.
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;
  • 976 ÷ 2 = 488 + 0;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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.


976(10) =


11 1101 0000(2)


3. Convert to binary (base 2) the fractional part: 0.562 627.

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) 0.562 627 × 2 = 1 + 0.125 254;
  • 2) 0.125 254 × 2 = 0 + 0.250 508;
  • 3) 0.250 508 × 2 = 0 + 0.501 016;
  • 4) 0.501 016 × 2 = 1 + 0.002 032;
  • 5) 0.002 032 × 2 = 0 + 0.004 064;
  • 6) 0.004 064 × 2 = 0 + 0.008 128;
  • 7) 0.008 128 × 2 = 0 + 0.016 256;
  • 8) 0.016 256 × 2 = 0 + 0.032 512;
  • 9) 0.032 512 × 2 = 0 + 0.065 024;
  • 10) 0.065 024 × 2 = 0 + 0.130 048;
  • 11) 0.130 048 × 2 = 0 + 0.260 096;
  • 12) 0.260 096 × 2 = 0 + 0.520 192;
  • 13) 0.520 192 × 2 = 1 + 0.040 384;
  • 14) 0.040 384 × 2 = 0 + 0.080 768;
  • 15) 0.080 768 × 2 = 0 + 0.161 536;
  • 16) 0.161 536 × 2 = 0 + 0.323 072;
  • 17) 0.323 072 × 2 = 0 + 0.646 144;
  • 18) 0.646 144 × 2 = 1 + 0.292 288;
  • 19) 0.292 288 × 2 = 0 + 0.584 576;
  • 20) 0.584 576 × 2 = 1 + 0.169 152;
  • 21) 0.169 152 × 2 = 0 + 0.338 304;
  • 22) 0.338 304 × 2 = 0 + 0.676 608;
  • 23) 0.676 608 × 2 = 1 + 0.353 216;
  • 24) 0.353 216 × 2 = 0 + 0.706 432;
  • 25) 0.706 432 × 2 = 1 + 0.412 864;
  • 26) 0.412 864 × 2 = 0 + 0.825 728;
  • 27) 0.825 728 × 2 = 1 + 0.651 456;
  • 28) 0.651 456 × 2 = 1 + 0.302 912;
  • 29) 0.302 912 × 2 = 0 + 0.605 824;
  • 30) 0.605 824 × 2 = 1 + 0.211 648;
  • 31) 0.211 648 × 2 = 0 + 0.423 296;
  • 32) 0.423 296 × 2 = 0 + 0.846 592;
  • 33) 0.846 592 × 2 = 1 + 0.693 184;
  • 34) 0.693 184 × 2 = 1 + 0.386 368;
  • 35) 0.386 368 × 2 = 0 + 0.772 736;
  • 36) 0.772 736 × 2 = 1 + 0.545 472;
  • 37) 0.545 472 × 2 = 1 + 0.090 944;
  • 38) 0.090 944 × 2 = 0 + 0.181 888;
  • 39) 0.181 888 × 2 = 0 + 0.363 776;
  • 40) 0.363 776 × 2 = 0 + 0.727 552;
  • 41) 0.727 552 × 2 = 1 + 0.455 104;
  • 42) 0.455 104 × 2 = 0 + 0.910 208;
  • 43) 0.910 208 × 2 = 1 + 0.820 416;
  • 44) 0.820 416 × 2 = 1 + 0.640 832;
  • 45) 0.640 832 × 2 = 1 + 0.281 664;
  • 46) 0.281 664 × 2 = 0 + 0.563 328;
  • 47) 0.563 328 × 2 = 1 + 0.126 656;
  • 48) 0.126 656 × 2 = 0 + 0.253 312;
  • 49) 0.253 312 × 2 = 0 + 0.506 624;
  • 50) 0.506 624 × 2 = 1 + 0.013 248;
  • 51) 0.013 248 × 2 = 0 + 0.026 496;
  • 52) 0.026 496 × 2 = 0 + 0.052 992;
  • 53) 0.052 992 × 2 = 0 + 0.105 984;

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...)


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:


0.562 627(10) =


0.1001 0000 0000 1000 0101 0010 1011 0100 1101 1000 1011 1010 0100 0(2)


5. Positive number before normalization:

976.562 627(10) =


11 1101 0000.1001 0000 0000 1000 0101 0010 1011 0100 1101 1000 1011 1010 0100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


976.562 627(10) =


11 1101 0000.1001 0000 0000 1000 0101 0010 1011 0100 1101 1000 1011 1010 0100 0(2) =


11 1101 0000.1001 0000 0000 1000 0101 0010 1011 0100 1101 1000 1011 1010 0100 0(2) × 20 =


1.1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101 1101 0010 00(2) × 29


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): 9


Mantissa (not normalized):
1.1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101 1101 0010 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


9 + 2(11-1) - 1 =


(9 + 1 023)(10) =


1 032(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;
  • 1 032 ÷ 2 = 516 + 0;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 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) =


1032(10) =


100 0000 1000(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. 1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101 11 0100 1000 =


1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101


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 1000


Mantissa (52 bits) =
1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101


The base ten decimal number 976.562 627 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1000 - 1110 1000 0100 1000 0000 0100 0010 1001 0101 1010 0110 1100 0101

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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