0.974 013 318 541 739 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 739 9(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
0.974 013 318 541 739 9(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: 0.974 013 318 541 739 9.

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) 0.974 013 318 541 739 9 × 2 = 1 + 0.948 026 637 083 479 8;
  • 2) 0.948 026 637 083 479 8 × 2 = 1 + 0.896 053 274 166 959 6;
  • 3) 0.896 053 274 166 959 6 × 2 = 1 + 0.792 106 548 333 919 2;
  • 4) 0.792 106 548 333 919 2 × 2 = 1 + 0.584 213 096 667 838 4;
  • 5) 0.584 213 096 667 838 4 × 2 = 1 + 0.168 426 193 335 676 8;
  • 6) 0.168 426 193 335 676 8 × 2 = 0 + 0.336 852 386 671 353 6;
  • 7) 0.336 852 386 671 353 6 × 2 = 0 + 0.673 704 773 342 707 2;
  • 8) 0.673 704 773 342 707 2 × 2 = 1 + 0.347 409 546 685 414 4;
  • 9) 0.347 409 546 685 414 4 × 2 = 0 + 0.694 819 093 370 828 8;
  • 10) 0.694 819 093 370 828 8 × 2 = 1 + 0.389 638 186 741 657 6;
  • 11) 0.389 638 186 741 657 6 × 2 = 0 + 0.779 276 373 483 315 2;
  • 12) 0.779 276 373 483 315 2 × 2 = 1 + 0.558 552 746 966 630 4;
  • 13) 0.558 552 746 966 630 4 × 2 = 1 + 0.117 105 493 933 260 8;
  • 14) 0.117 105 493 933 260 8 × 2 = 0 + 0.234 210 987 866 521 6;
  • 15) 0.234 210 987 866 521 6 × 2 = 0 + 0.468 421 975 733 043 2;
  • 16) 0.468 421 975 733 043 2 × 2 = 0 + 0.936 843 951 466 086 4;
  • 17) 0.936 843 951 466 086 4 × 2 = 1 + 0.873 687 902 932 172 8;
  • 18) 0.873 687 902 932 172 8 × 2 = 1 + 0.747 375 805 864 345 6;
  • 19) 0.747 375 805 864 345 6 × 2 = 1 + 0.494 751 611 728 691 2;
  • 20) 0.494 751 611 728 691 2 × 2 = 0 + 0.989 503 223 457 382 4;
  • 21) 0.989 503 223 457 382 4 × 2 = 1 + 0.979 006 446 914 764 8;
  • 22) 0.979 006 446 914 764 8 × 2 = 1 + 0.958 012 893 829 529 6;
  • 23) 0.958 012 893 829 529 6 × 2 = 1 + 0.916 025 787 659 059 2;
  • 24) 0.916 025 787 659 059 2 × 2 = 1 + 0.832 051 575 318 118 4;
  • 25) 0.832 051 575 318 118 4 × 2 = 1 + 0.664 103 150 636 236 8;
  • 26) 0.664 103 150 636 236 8 × 2 = 1 + 0.328 206 301 272 473 6;
  • 27) 0.328 206 301 272 473 6 × 2 = 0 + 0.656 412 602 544 947 2;
  • 28) 0.656 412 602 544 947 2 × 2 = 1 + 0.312 825 205 089 894 4;
  • 29) 0.312 825 205 089 894 4 × 2 = 0 + 0.625 650 410 179 788 8;
  • 30) 0.625 650 410 179 788 8 × 2 = 1 + 0.251 300 820 359 577 6;
  • 31) 0.251 300 820 359 577 6 × 2 = 0 + 0.502 601 640 719 155 2;
  • 32) 0.502 601 640 719 155 2 × 2 = 1 + 0.005 203 281 438 310 4;
  • 33) 0.005 203 281 438 310 4 × 2 = 0 + 0.010 406 562 876 620 8;
  • 34) 0.010 406 562 876 620 8 × 2 = 0 + 0.020 813 125 753 241 6;
  • 35) 0.020 813 125 753 241 6 × 2 = 0 + 0.041 626 251 506 483 2;
  • 36) 0.041 626 251 506 483 2 × 2 = 0 + 0.083 252 503 012 966 4;
  • 37) 0.083 252 503 012 966 4 × 2 = 0 + 0.166 505 006 025 932 8;
  • 38) 0.166 505 006 025 932 8 × 2 = 0 + 0.333 010 012 051 865 6;
  • 39) 0.333 010 012 051 865 6 × 2 = 0 + 0.666 020 024 103 731 2;
  • 40) 0.666 020 024 103 731 2 × 2 = 1 + 0.332 040 048 207 462 4;
  • 41) 0.332 040 048 207 462 4 × 2 = 0 + 0.664 080 096 414 924 8;
  • 42) 0.664 080 096 414 924 8 × 2 = 1 + 0.328 160 192 829 849 6;
  • 43) 0.328 160 192 829 849 6 × 2 = 0 + 0.656 320 385 659 699 2;
  • 44) 0.656 320 385 659 699 2 × 2 = 1 + 0.312 640 771 319 398 4;
  • 45) 0.312 640 771 319 398 4 × 2 = 0 + 0.625 281 542 638 796 8;
  • 46) 0.625 281 542 638 796 8 × 2 = 1 + 0.250 563 085 277 593 6;
  • 47) 0.250 563 085 277 593 6 × 2 = 0 + 0.501 126 170 555 187 2;
  • 48) 0.501 126 170 555 187 2 × 2 = 1 + 0.002 252 341 110 374 4;
  • 49) 0.002 252 341 110 374 4 × 2 = 0 + 0.004 504 682 220 748 8;
  • 50) 0.004 504 682 220 748 8 × 2 = 0 + 0.009 009 364 441 497 6;
  • 51) 0.009 009 364 441 497 6 × 2 = 0 + 0.018 018 728 882 995 2;
  • 52) 0.018 018 728 882 995 2 × 2 = 0 + 0.036 037 457 765 990 4;
  • 53) 0.036 037 457 765 990 4 × 2 = 0 + 0.072 074 915 531 980 8;

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:


0.974 013 318 541 739 9(10) =

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

5. Positive number before normalization:

0.974 013 318 541 739 9(10) =

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


0.974 013 318 541 739 9(10) =

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

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

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


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 1010 0000 =

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


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 1010 0000


Decimal number 0.974 013 318 541 739 9 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 1010 0000

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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