0.974 013 318 541 745 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 745 4(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
0.974 013 318 541 745 4(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: 0.974 013 318 541 745 4.

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) 0.974 013 318 541 745 4 × 2 = 1 + 0.948 026 637 083 490 8;
  • 2) 0.948 026 637 083 490 8 × 2 = 1 + 0.896 053 274 166 981 6;
  • 3) 0.896 053 274 166 981 6 × 2 = 1 + 0.792 106 548 333 963 2;
  • 4) 0.792 106 548 333 963 2 × 2 = 1 + 0.584 213 096 667 926 4;
  • 5) 0.584 213 096 667 926 4 × 2 = 1 + 0.168 426 193 335 852 8;
  • 6) 0.168 426 193 335 852 8 × 2 = 0 + 0.336 852 386 671 705 6;
  • 7) 0.336 852 386 671 705 6 × 2 = 0 + 0.673 704 773 343 411 2;
  • 8) 0.673 704 773 343 411 2 × 2 = 1 + 0.347 409 546 686 822 4;
  • 9) 0.347 409 546 686 822 4 × 2 = 0 + 0.694 819 093 373 644 8;
  • 10) 0.694 819 093 373 644 8 × 2 = 1 + 0.389 638 186 747 289 6;
  • 11) 0.389 638 186 747 289 6 × 2 = 0 + 0.779 276 373 494 579 2;
  • 12) 0.779 276 373 494 579 2 × 2 = 1 + 0.558 552 746 989 158 4;
  • 13) 0.558 552 746 989 158 4 × 2 = 1 + 0.117 105 493 978 316 8;
  • 14) 0.117 105 493 978 316 8 × 2 = 0 + 0.234 210 987 956 633 6;
  • 15) 0.234 210 987 956 633 6 × 2 = 0 + 0.468 421 975 913 267 2;
  • 16) 0.468 421 975 913 267 2 × 2 = 0 + 0.936 843 951 826 534 4;
  • 17) 0.936 843 951 826 534 4 × 2 = 1 + 0.873 687 903 653 068 8;
  • 18) 0.873 687 903 653 068 8 × 2 = 1 + 0.747 375 807 306 137 6;
  • 19) 0.747 375 807 306 137 6 × 2 = 1 + 0.494 751 614 612 275 2;
  • 20) 0.494 751 614 612 275 2 × 2 = 0 + 0.989 503 229 224 550 4;
  • 21) 0.989 503 229 224 550 4 × 2 = 1 + 0.979 006 458 449 100 8;
  • 22) 0.979 006 458 449 100 8 × 2 = 1 + 0.958 012 916 898 201 6;
  • 23) 0.958 012 916 898 201 6 × 2 = 1 + 0.916 025 833 796 403 2;
  • 24) 0.916 025 833 796 403 2 × 2 = 1 + 0.832 051 667 592 806 4;
  • 25) 0.832 051 667 592 806 4 × 2 = 1 + 0.664 103 335 185 612 8;
  • 26) 0.664 103 335 185 612 8 × 2 = 1 + 0.328 206 670 371 225 6;
  • 27) 0.328 206 670 371 225 6 × 2 = 0 + 0.656 413 340 742 451 2;
  • 28) 0.656 413 340 742 451 2 × 2 = 1 + 0.312 826 681 484 902 4;
  • 29) 0.312 826 681 484 902 4 × 2 = 0 + 0.625 653 362 969 804 8;
  • 30) 0.625 653 362 969 804 8 × 2 = 1 + 0.251 306 725 939 609 6;
  • 31) 0.251 306 725 939 609 6 × 2 = 0 + 0.502 613 451 879 219 2;
  • 32) 0.502 613 451 879 219 2 × 2 = 1 + 0.005 226 903 758 438 4;
  • 33) 0.005 226 903 758 438 4 × 2 = 0 + 0.010 453 807 516 876 8;
  • 34) 0.010 453 807 516 876 8 × 2 = 0 + 0.020 907 615 033 753 6;
  • 35) 0.020 907 615 033 753 6 × 2 = 0 + 0.041 815 230 067 507 2;
  • 36) 0.041 815 230 067 507 2 × 2 = 0 + 0.083 630 460 135 014 4;
  • 37) 0.083 630 460 135 014 4 × 2 = 0 + 0.167 260 920 270 028 8;
  • 38) 0.167 260 920 270 028 8 × 2 = 0 + 0.334 521 840 540 057 6;
  • 39) 0.334 521 840 540 057 6 × 2 = 0 + 0.669 043 681 080 115 2;
  • 40) 0.669 043 681 080 115 2 × 2 = 1 + 0.338 087 362 160 230 4;
  • 41) 0.338 087 362 160 230 4 × 2 = 0 + 0.676 174 724 320 460 8;
  • 42) 0.676 174 724 320 460 8 × 2 = 1 + 0.352 349 448 640 921 6;
  • 43) 0.352 349 448 640 921 6 × 2 = 0 + 0.704 698 897 281 843 2;
  • 44) 0.704 698 897 281 843 2 × 2 = 1 + 0.409 397 794 563 686 4;
  • 45) 0.409 397 794 563 686 4 × 2 = 0 + 0.818 795 589 127 372 8;
  • 46) 0.818 795 589 127 372 8 × 2 = 1 + 0.637 591 178 254 745 6;
  • 47) 0.637 591 178 254 745 6 × 2 = 1 + 0.275 182 356 509 491 2;
  • 48) 0.275 182 356 509 491 2 × 2 = 0 + 0.550 364 713 018 982 4;
  • 49) 0.550 364 713 018 982 4 × 2 = 1 + 0.100 729 426 037 964 8;
  • 50) 0.100 729 426 037 964 8 × 2 = 0 + 0.201 458 852 075 929 6;
  • 51) 0.201 458 852 075 929 6 × 2 = 0 + 0.402 917 704 151 859 2;
  • 52) 0.402 917 704 151 859 2 × 2 = 0 + 0.805 835 408 303 718 4;
  • 53) 0.805 835 408 303 718 4 × 2 = 1 + 0.611 670 816 607 436 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:


0.974 013 318 541 745 4(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0110 1000 1(2)

5. Positive number before normalization:

0.974 013 318 541 745 4(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0110 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:


0.974 013 318 541 745 4(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0110 1000 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0110 1000 1(2) × 20 =

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

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


Decimal number 0.974 013 318 541 745 4 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 1101 0001

Share

» Convert decimal 0.974 013 318 541 744 1 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 01, 2026 [January]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: January - by our visitors

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