0.974 013 318 541 669 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 669(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
0.974 013 318 541 669(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: 0.974 013 318 541 669.

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) 0.974 013 318 541 669 × 2 = 1 + 0.948 026 637 083 338;
  • 2) 0.948 026 637 083 338 × 2 = 1 + 0.896 053 274 166 676;
  • 3) 0.896 053 274 166 676 × 2 = 1 + 0.792 106 548 333 352;
  • 4) 0.792 106 548 333 352 × 2 = 1 + 0.584 213 096 666 704;
  • 5) 0.584 213 096 666 704 × 2 = 1 + 0.168 426 193 333 408;
  • 6) 0.168 426 193 333 408 × 2 = 0 + 0.336 852 386 666 816;
  • 7) 0.336 852 386 666 816 × 2 = 0 + 0.673 704 773 333 632;
  • 8) 0.673 704 773 333 632 × 2 = 1 + 0.347 409 546 667 264;
  • 9) 0.347 409 546 667 264 × 2 = 0 + 0.694 819 093 334 528;
  • 10) 0.694 819 093 334 528 × 2 = 1 + 0.389 638 186 669 056;
  • 11) 0.389 638 186 669 056 × 2 = 0 + 0.779 276 373 338 112;
  • 12) 0.779 276 373 338 112 × 2 = 1 + 0.558 552 746 676 224;
  • 13) 0.558 552 746 676 224 × 2 = 1 + 0.117 105 493 352 448;
  • 14) 0.117 105 493 352 448 × 2 = 0 + 0.234 210 986 704 896;
  • 15) 0.234 210 986 704 896 × 2 = 0 + 0.468 421 973 409 792;
  • 16) 0.468 421 973 409 792 × 2 = 0 + 0.936 843 946 819 584;
  • 17) 0.936 843 946 819 584 × 2 = 1 + 0.873 687 893 639 168;
  • 18) 0.873 687 893 639 168 × 2 = 1 + 0.747 375 787 278 336;
  • 19) 0.747 375 787 278 336 × 2 = 1 + 0.494 751 574 556 672;
  • 20) 0.494 751 574 556 672 × 2 = 0 + 0.989 503 149 113 344;
  • 21) 0.989 503 149 113 344 × 2 = 1 + 0.979 006 298 226 688;
  • 22) 0.979 006 298 226 688 × 2 = 1 + 0.958 012 596 453 376;
  • 23) 0.958 012 596 453 376 × 2 = 1 + 0.916 025 192 906 752;
  • 24) 0.916 025 192 906 752 × 2 = 1 + 0.832 050 385 813 504;
  • 25) 0.832 050 385 813 504 × 2 = 1 + 0.664 100 771 627 008;
  • 26) 0.664 100 771 627 008 × 2 = 1 + 0.328 201 543 254 016;
  • 27) 0.328 201 543 254 016 × 2 = 0 + 0.656 403 086 508 032;
  • 28) 0.656 403 086 508 032 × 2 = 1 + 0.312 806 173 016 064;
  • 29) 0.312 806 173 016 064 × 2 = 0 + 0.625 612 346 032 128;
  • 30) 0.625 612 346 032 128 × 2 = 1 + 0.251 224 692 064 256;
  • 31) 0.251 224 692 064 256 × 2 = 0 + 0.502 449 384 128 512;
  • 32) 0.502 449 384 128 512 × 2 = 1 + 0.004 898 768 257 024;
  • 33) 0.004 898 768 257 024 × 2 = 0 + 0.009 797 536 514 048;
  • 34) 0.009 797 536 514 048 × 2 = 0 + 0.019 595 073 028 096;
  • 35) 0.019 595 073 028 096 × 2 = 0 + 0.039 190 146 056 192;
  • 36) 0.039 190 146 056 192 × 2 = 0 + 0.078 380 292 112 384;
  • 37) 0.078 380 292 112 384 × 2 = 0 + 0.156 760 584 224 768;
  • 38) 0.156 760 584 224 768 × 2 = 0 + 0.313 521 168 449 536;
  • 39) 0.313 521 168 449 536 × 2 = 0 + 0.627 042 336 899 072;
  • 40) 0.627 042 336 899 072 × 2 = 1 + 0.254 084 673 798 144;
  • 41) 0.254 084 673 798 144 × 2 = 0 + 0.508 169 347 596 288;
  • 42) 0.508 169 347 596 288 × 2 = 1 + 0.016 338 695 192 576;
  • 43) 0.016 338 695 192 576 × 2 = 0 + 0.032 677 390 385 152;
  • 44) 0.032 677 390 385 152 × 2 = 0 + 0.065 354 780 770 304;
  • 45) 0.065 354 780 770 304 × 2 = 0 + 0.130 709 561 540 608;
  • 46) 0.130 709 561 540 608 × 2 = 0 + 0.261 419 123 081 216;
  • 47) 0.261 419 123 081 216 × 2 = 0 + 0.522 838 246 162 432;
  • 48) 0.522 838 246 162 432 × 2 = 1 + 0.045 676 492 324 864;
  • 49) 0.045 676 492 324 864 × 2 = 0 + 0.091 352 984 649 728;
  • 50) 0.091 352 984 649 728 × 2 = 0 + 0.182 705 969 299 456;
  • 51) 0.182 705 969 299 456 × 2 = 0 + 0.365 411 938 598 912;
  • 52) 0.365 411 938 598 912 × 2 = 0 + 0.730 823 877 197 824;
  • 53) 0.730 823 877 197 824 × 2 = 1 + 0.461 647 754 395 648;

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:


0.974 013 318 541 669(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0001 0000 1(2)

5. Positive number before normalization:

0.974 013 318 541 669(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0001 0000 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:


0.974 013 318 541 669(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0001 0000 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0001 0000 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0010 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 1000 0010 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 1000 0010 0001 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0010 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 1000 0010 0001


Decimal number 0.974 013 318 541 669 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 1000 0010 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