0.974 013 318 541 726 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 726 8(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
0.974 013 318 541 726 8(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: 0.974 013 318 541 726 8.

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) 0.974 013 318 541 726 8 × 2 = 1 + 0.948 026 637 083 453 6;
  • 2) 0.948 026 637 083 453 6 × 2 = 1 + 0.896 053 274 166 907 2;
  • 3) 0.896 053 274 166 907 2 × 2 = 1 + 0.792 106 548 333 814 4;
  • 4) 0.792 106 548 333 814 4 × 2 = 1 + 0.584 213 096 667 628 8;
  • 5) 0.584 213 096 667 628 8 × 2 = 1 + 0.168 426 193 335 257 6;
  • 6) 0.168 426 193 335 257 6 × 2 = 0 + 0.336 852 386 670 515 2;
  • 7) 0.336 852 386 670 515 2 × 2 = 0 + 0.673 704 773 341 030 4;
  • 8) 0.673 704 773 341 030 4 × 2 = 1 + 0.347 409 546 682 060 8;
  • 9) 0.347 409 546 682 060 8 × 2 = 0 + 0.694 819 093 364 121 6;
  • 10) 0.694 819 093 364 121 6 × 2 = 1 + 0.389 638 186 728 243 2;
  • 11) 0.389 638 186 728 243 2 × 2 = 0 + 0.779 276 373 456 486 4;
  • 12) 0.779 276 373 456 486 4 × 2 = 1 + 0.558 552 746 912 972 8;
  • 13) 0.558 552 746 912 972 8 × 2 = 1 + 0.117 105 493 825 945 6;
  • 14) 0.117 105 493 825 945 6 × 2 = 0 + 0.234 210 987 651 891 2;
  • 15) 0.234 210 987 651 891 2 × 2 = 0 + 0.468 421 975 303 782 4;
  • 16) 0.468 421 975 303 782 4 × 2 = 0 + 0.936 843 950 607 564 8;
  • 17) 0.936 843 950 607 564 8 × 2 = 1 + 0.873 687 901 215 129 6;
  • 18) 0.873 687 901 215 129 6 × 2 = 1 + 0.747 375 802 430 259 2;
  • 19) 0.747 375 802 430 259 2 × 2 = 1 + 0.494 751 604 860 518 4;
  • 20) 0.494 751 604 860 518 4 × 2 = 0 + 0.989 503 209 721 036 8;
  • 21) 0.989 503 209 721 036 8 × 2 = 1 + 0.979 006 419 442 073 6;
  • 22) 0.979 006 419 442 073 6 × 2 = 1 + 0.958 012 838 884 147 2;
  • 23) 0.958 012 838 884 147 2 × 2 = 1 + 0.916 025 677 768 294 4;
  • 24) 0.916 025 677 768 294 4 × 2 = 1 + 0.832 051 355 536 588 8;
  • 25) 0.832 051 355 536 588 8 × 2 = 1 + 0.664 102 711 073 177 6;
  • 26) 0.664 102 711 073 177 6 × 2 = 1 + 0.328 205 422 146 355 2;
  • 27) 0.328 205 422 146 355 2 × 2 = 0 + 0.656 410 844 292 710 4;
  • 28) 0.656 410 844 292 710 4 × 2 = 1 + 0.312 821 688 585 420 8;
  • 29) 0.312 821 688 585 420 8 × 2 = 0 + 0.625 643 377 170 841 6;
  • 30) 0.625 643 377 170 841 6 × 2 = 1 + 0.251 286 754 341 683 2;
  • 31) 0.251 286 754 341 683 2 × 2 = 0 + 0.502 573 508 683 366 4;
  • 32) 0.502 573 508 683 366 4 × 2 = 1 + 0.005 147 017 366 732 8;
  • 33) 0.005 147 017 366 732 8 × 2 = 0 + 0.010 294 034 733 465 6;
  • 34) 0.010 294 034 733 465 6 × 2 = 0 + 0.020 588 069 466 931 2;
  • 35) 0.020 588 069 466 931 2 × 2 = 0 + 0.041 176 138 933 862 4;
  • 36) 0.041 176 138 933 862 4 × 2 = 0 + 0.082 352 277 867 724 8;
  • 37) 0.082 352 277 867 724 8 × 2 = 0 + 0.164 704 555 735 449 6;
  • 38) 0.164 704 555 735 449 6 × 2 = 0 + 0.329 409 111 470 899 2;
  • 39) 0.329 409 111 470 899 2 × 2 = 0 + 0.658 818 222 941 798 4;
  • 40) 0.658 818 222 941 798 4 × 2 = 1 + 0.317 636 445 883 596 8;
  • 41) 0.317 636 445 883 596 8 × 2 = 0 + 0.635 272 891 767 193 6;
  • 42) 0.635 272 891 767 193 6 × 2 = 1 + 0.270 545 783 534 387 2;
  • 43) 0.270 545 783 534 387 2 × 2 = 0 + 0.541 091 567 068 774 4;
  • 44) 0.541 091 567 068 774 4 × 2 = 1 + 0.082 183 134 137 548 8;
  • 45) 0.082 183 134 137 548 8 × 2 = 0 + 0.164 366 268 275 097 6;
  • 46) 0.164 366 268 275 097 6 × 2 = 0 + 0.328 732 536 550 195 2;
  • 47) 0.328 732 536 550 195 2 × 2 = 0 + 0.657 465 073 100 390 4;
  • 48) 0.657 465 073 100 390 4 × 2 = 1 + 0.314 930 146 200 780 8;
  • 49) 0.314 930 146 200 780 8 × 2 = 0 + 0.629 860 292 401 561 6;
  • 50) 0.629 860 292 401 561 6 × 2 = 1 + 0.259 720 584 803 123 2;
  • 51) 0.259 720 584 803 123 2 × 2 = 0 + 0.519 441 169 606 246 4;
  • 52) 0.519 441 169 606 246 4 × 2 = 1 + 0.038 882 339 212 492 8;
  • 53) 0.038 882 339 212 492 8 × 2 = 0 + 0.077 764 678 424 985 6;

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:


0.974 013 318 541 726 8(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0101 0(2)

5. Positive number before normalization:

0.974 013 318 541 726 8(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0101 0(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:


0.974 013 318 541 726 8(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0101 0(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 0101 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0010 1010(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 1010 1010 0000 0010 1010 0010 1010


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 1010 1010 0000 0010 1010 0010 1010 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0010 1010


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 1010 1010 0000 0010 1010 0010 1010


Decimal number 0.974 013 318 541 726 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0010 1010

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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