0.974 013 318 541 728 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 728 5(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
0.974 013 318 541 728 5(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: 0.974 013 318 541 728 5.

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) 0.974 013 318 541 728 5 × 2 = 1 + 0.948 026 637 083 457;
  • 2) 0.948 026 637 083 457 × 2 = 1 + 0.896 053 274 166 914;
  • 3) 0.896 053 274 166 914 × 2 = 1 + 0.792 106 548 333 828;
  • 4) 0.792 106 548 333 828 × 2 = 1 + 0.584 213 096 667 656;
  • 5) 0.584 213 096 667 656 × 2 = 1 + 0.168 426 193 335 312;
  • 6) 0.168 426 193 335 312 × 2 = 0 + 0.336 852 386 670 624;
  • 7) 0.336 852 386 670 624 × 2 = 0 + 0.673 704 773 341 248;
  • 8) 0.673 704 773 341 248 × 2 = 1 + 0.347 409 546 682 496;
  • 9) 0.347 409 546 682 496 × 2 = 0 + 0.694 819 093 364 992;
  • 10) 0.694 819 093 364 992 × 2 = 1 + 0.389 638 186 729 984;
  • 11) 0.389 638 186 729 984 × 2 = 0 + 0.779 276 373 459 968;
  • 12) 0.779 276 373 459 968 × 2 = 1 + 0.558 552 746 919 936;
  • 13) 0.558 552 746 919 936 × 2 = 1 + 0.117 105 493 839 872;
  • 14) 0.117 105 493 839 872 × 2 = 0 + 0.234 210 987 679 744;
  • 15) 0.234 210 987 679 744 × 2 = 0 + 0.468 421 975 359 488;
  • 16) 0.468 421 975 359 488 × 2 = 0 + 0.936 843 950 718 976;
  • 17) 0.936 843 950 718 976 × 2 = 1 + 0.873 687 901 437 952;
  • 18) 0.873 687 901 437 952 × 2 = 1 + 0.747 375 802 875 904;
  • 19) 0.747 375 802 875 904 × 2 = 1 + 0.494 751 605 751 808;
  • 20) 0.494 751 605 751 808 × 2 = 0 + 0.989 503 211 503 616;
  • 21) 0.989 503 211 503 616 × 2 = 1 + 0.979 006 423 007 232;
  • 22) 0.979 006 423 007 232 × 2 = 1 + 0.958 012 846 014 464;
  • 23) 0.958 012 846 014 464 × 2 = 1 + 0.916 025 692 028 928;
  • 24) 0.916 025 692 028 928 × 2 = 1 + 0.832 051 384 057 856;
  • 25) 0.832 051 384 057 856 × 2 = 1 + 0.664 102 768 115 712;
  • 26) 0.664 102 768 115 712 × 2 = 1 + 0.328 205 536 231 424;
  • 27) 0.328 205 536 231 424 × 2 = 0 + 0.656 411 072 462 848;
  • 28) 0.656 411 072 462 848 × 2 = 1 + 0.312 822 144 925 696;
  • 29) 0.312 822 144 925 696 × 2 = 0 + 0.625 644 289 851 392;
  • 30) 0.625 644 289 851 392 × 2 = 1 + 0.251 288 579 702 784;
  • 31) 0.251 288 579 702 784 × 2 = 0 + 0.502 577 159 405 568;
  • 32) 0.502 577 159 405 568 × 2 = 1 + 0.005 154 318 811 136;
  • 33) 0.005 154 318 811 136 × 2 = 0 + 0.010 308 637 622 272;
  • 34) 0.010 308 637 622 272 × 2 = 0 + 0.020 617 275 244 544;
  • 35) 0.020 617 275 244 544 × 2 = 0 + 0.041 234 550 489 088;
  • 36) 0.041 234 550 489 088 × 2 = 0 + 0.082 469 100 978 176;
  • 37) 0.082 469 100 978 176 × 2 = 0 + 0.164 938 201 956 352;
  • 38) 0.164 938 201 956 352 × 2 = 0 + 0.329 876 403 912 704;
  • 39) 0.329 876 403 912 704 × 2 = 0 + 0.659 752 807 825 408;
  • 40) 0.659 752 807 825 408 × 2 = 1 + 0.319 505 615 650 816;
  • 41) 0.319 505 615 650 816 × 2 = 0 + 0.639 011 231 301 632;
  • 42) 0.639 011 231 301 632 × 2 = 1 + 0.278 022 462 603 264;
  • 43) 0.278 022 462 603 264 × 2 = 0 + 0.556 044 925 206 528;
  • 44) 0.556 044 925 206 528 × 2 = 1 + 0.112 089 850 413 056;
  • 45) 0.112 089 850 413 056 × 2 = 0 + 0.224 179 700 826 112;
  • 46) 0.224 179 700 826 112 × 2 = 0 + 0.448 359 401 652 224;
  • 47) 0.448 359 401 652 224 × 2 = 0 + 0.896 718 803 304 448;
  • 48) 0.896 718 803 304 448 × 2 = 1 + 0.793 437 606 608 896;
  • 49) 0.793 437 606 608 896 × 2 = 1 + 0.586 875 213 217 792;
  • 50) 0.586 875 213 217 792 × 2 = 1 + 0.173 750 426 435 584;
  • 51) 0.173 750 426 435 584 × 2 = 0 + 0.347 500 852 871 168;
  • 52) 0.347 500 852 871 168 × 2 = 0 + 0.695 001 705 742 336;
  • 53) 0.695 001 705 742 336 × 2 = 1 + 0.390 003 411 484 672;

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:


0.974 013 318 541 728 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1100 1(2)

5. Positive number before normalization:

0.974 013 318 541 728 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1100 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:


0.974 013 318 541 728 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1100 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0101 0001 1100 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 1001(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 0011 1001


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 0011 1001 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1010 0011 1001


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 0011 1001


Decimal number 0.974 013 318 541 728 5 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 0011 1001

Share

» Convert decimal 0.974 013 318 541 721 7 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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