0.974 013 316 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 316 93(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
0.974 013 316 93(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.974 013 316 93.

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) 0.974 013 316 93 × 2 = 1 + 0.948 026 633 86;
  • 2) 0.948 026 633 86 × 2 = 1 + 0.896 053 267 72;
  • 3) 0.896 053 267 72 × 2 = 1 + 0.792 106 535 44;
  • 4) 0.792 106 535 44 × 2 = 1 + 0.584 213 070 88;
  • 5) 0.584 213 070 88 × 2 = 1 + 0.168 426 141 76;
  • 6) 0.168 426 141 76 × 2 = 0 + 0.336 852 283 52;
  • 7) 0.336 852 283 52 × 2 = 0 + 0.673 704 567 04;
  • 8) 0.673 704 567 04 × 2 = 1 + 0.347 409 134 08;
  • 9) 0.347 409 134 08 × 2 = 0 + 0.694 818 268 16;
  • 10) 0.694 818 268 16 × 2 = 1 + 0.389 636 536 32;
  • 11) 0.389 636 536 32 × 2 = 0 + 0.779 273 072 64;
  • 12) 0.779 273 072 64 × 2 = 1 + 0.558 546 145 28;
  • 13) 0.558 546 145 28 × 2 = 1 + 0.117 092 290 56;
  • 14) 0.117 092 290 56 × 2 = 0 + 0.234 184 581 12;
  • 15) 0.234 184 581 12 × 2 = 0 + 0.468 369 162 24;
  • 16) 0.468 369 162 24 × 2 = 0 + 0.936 738 324 48;
  • 17) 0.936 738 324 48 × 2 = 1 + 0.873 476 648 96;
  • 18) 0.873 476 648 96 × 2 = 1 + 0.746 953 297 92;
  • 19) 0.746 953 297 92 × 2 = 1 + 0.493 906 595 84;
  • 20) 0.493 906 595 84 × 2 = 0 + 0.987 813 191 68;
  • 21) 0.987 813 191 68 × 2 = 1 + 0.975 626 383 36;
  • 22) 0.975 626 383 36 × 2 = 1 + 0.951 252 766 72;
  • 23) 0.951 252 766 72 × 2 = 1 + 0.902 505 533 44;
  • 24) 0.902 505 533 44 × 2 = 1 + 0.805 011 066 88;
  • 25) 0.805 011 066 88 × 2 = 1 + 0.610 022 133 76;
  • 26) 0.610 022 133 76 × 2 = 1 + 0.220 044 267 52;
  • 27) 0.220 044 267 52 × 2 = 0 + 0.440 088 535 04;
  • 28) 0.440 088 535 04 × 2 = 0 + 0.880 177 070 08;
  • 29) 0.880 177 070 08 × 2 = 1 + 0.760 354 140 16;
  • 30) 0.760 354 140 16 × 2 = 1 + 0.520 708 280 32;
  • 31) 0.520 708 280 32 × 2 = 1 + 0.041 416 560 64;
  • 32) 0.041 416 560 64 × 2 = 0 + 0.082 833 121 28;
  • 33) 0.082 833 121 28 × 2 = 0 + 0.165 666 242 56;
  • 34) 0.165 666 242 56 × 2 = 0 + 0.331 332 485 12;
  • 35) 0.331 332 485 12 × 2 = 0 + 0.662 664 970 24;
  • 36) 0.662 664 970 24 × 2 = 1 + 0.325 329 940 48;
  • 37) 0.325 329 940 48 × 2 = 0 + 0.650 659 880 96;
  • 38) 0.650 659 880 96 × 2 = 1 + 0.301 319 761 92;
  • 39) 0.301 319 761 92 × 2 = 0 + 0.602 639 523 84;
  • 40) 0.602 639 523 84 × 2 = 1 + 0.205 279 047 68;
  • 41) 0.205 279 047 68 × 2 = 0 + 0.410 558 095 36;
  • 42) 0.410 558 095 36 × 2 = 0 + 0.821 116 190 72;
  • 43) 0.821 116 190 72 × 2 = 1 + 0.642 232 381 44;
  • 44) 0.642 232 381 44 × 2 = 1 + 0.284 464 762 88;
  • 45) 0.284 464 762 88 × 2 = 0 + 0.568 929 525 76;
  • 46) 0.568 929 525 76 × 2 = 1 + 0.137 859 051 52;
  • 47) 0.137 859 051 52 × 2 = 0 + 0.275 718 103 04;
  • 48) 0.275 718 103 04 × 2 = 0 + 0.551 436 206 08;
  • 49) 0.551 436 206 08 × 2 = 1 + 0.102 872 412 16;
  • 50) 0.102 872 412 16 × 2 = 0 + 0.205 744 824 32;
  • 51) 0.205 744 824 32 × 2 = 0 + 0.411 489 648 64;
  • 52) 0.411 489 648 64 × 2 = 0 + 0.822 979 297 28;
  • 53) 0.822 979 297 28 × 2 = 1 + 0.645 958 594 56;

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:


0.974 013 316 93(10) =

0.1111 1001 0101 1000 1110 1111 1100 1110 0001 0101 0011 0100 1000 1(2)

5. Positive number before normalization:

0.974 013 316 93(10) =

0.1111 1001 0101 1000 1110 1111 1100 1110 0001 0101 0011 0100 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.974 013 316 93(10) =

0.1111 1001 0101 1000 1110 1111 1100 1110 0001 0101 0011 0100 1000 1(2) =

0.1111 1001 0101 1000 1110 1111 1100 1110 0001 0101 0011 0100 1000 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001(2) × 2-1


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): -1

Mantissa (not normalized):
1.1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001 =

1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001


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 1110

Mantissa (52 bits) =
1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001


Decimal number 0.974 013 316 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1001 1100 0010 1010 0110 1001 0001

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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