0.974 013 318 541 708 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 708(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
0.974 013 318 541 708(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: 0.974 013 318 541 708.

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) 0.974 013 318 541 708 × 2 = 1 + 0.948 026 637 083 416;
  • 2) 0.948 026 637 083 416 × 2 = 1 + 0.896 053 274 166 832;
  • 3) 0.896 053 274 166 832 × 2 = 1 + 0.792 106 548 333 664;
  • 4) 0.792 106 548 333 664 × 2 = 1 + 0.584 213 096 667 328;
  • 5) 0.584 213 096 667 328 × 2 = 1 + 0.168 426 193 334 656;
  • 6) 0.168 426 193 334 656 × 2 = 0 + 0.336 852 386 669 312;
  • 7) 0.336 852 386 669 312 × 2 = 0 + 0.673 704 773 338 624;
  • 8) 0.673 704 773 338 624 × 2 = 1 + 0.347 409 546 677 248;
  • 9) 0.347 409 546 677 248 × 2 = 0 + 0.694 819 093 354 496;
  • 10) 0.694 819 093 354 496 × 2 = 1 + 0.389 638 186 708 992;
  • 11) 0.389 638 186 708 992 × 2 = 0 + 0.779 276 373 417 984;
  • 12) 0.779 276 373 417 984 × 2 = 1 + 0.558 552 746 835 968;
  • 13) 0.558 552 746 835 968 × 2 = 1 + 0.117 105 493 671 936;
  • 14) 0.117 105 493 671 936 × 2 = 0 + 0.234 210 987 343 872;
  • 15) 0.234 210 987 343 872 × 2 = 0 + 0.468 421 974 687 744;
  • 16) 0.468 421 974 687 744 × 2 = 0 + 0.936 843 949 375 488;
  • 17) 0.936 843 949 375 488 × 2 = 1 + 0.873 687 898 750 976;
  • 18) 0.873 687 898 750 976 × 2 = 1 + 0.747 375 797 501 952;
  • 19) 0.747 375 797 501 952 × 2 = 1 + 0.494 751 595 003 904;
  • 20) 0.494 751 595 003 904 × 2 = 0 + 0.989 503 190 007 808;
  • 21) 0.989 503 190 007 808 × 2 = 1 + 0.979 006 380 015 616;
  • 22) 0.979 006 380 015 616 × 2 = 1 + 0.958 012 760 031 232;
  • 23) 0.958 012 760 031 232 × 2 = 1 + 0.916 025 520 062 464;
  • 24) 0.916 025 520 062 464 × 2 = 1 + 0.832 051 040 124 928;
  • 25) 0.832 051 040 124 928 × 2 = 1 + 0.664 102 080 249 856;
  • 26) 0.664 102 080 249 856 × 2 = 1 + 0.328 204 160 499 712;
  • 27) 0.328 204 160 499 712 × 2 = 0 + 0.656 408 320 999 424;
  • 28) 0.656 408 320 999 424 × 2 = 1 + 0.312 816 641 998 848;
  • 29) 0.312 816 641 998 848 × 2 = 0 + 0.625 633 283 997 696;
  • 30) 0.625 633 283 997 696 × 2 = 1 + 0.251 266 567 995 392;
  • 31) 0.251 266 567 995 392 × 2 = 0 + 0.502 533 135 990 784;
  • 32) 0.502 533 135 990 784 × 2 = 1 + 0.005 066 271 981 568;
  • 33) 0.005 066 271 981 568 × 2 = 0 + 0.010 132 543 963 136;
  • 34) 0.010 132 543 963 136 × 2 = 0 + 0.020 265 087 926 272;
  • 35) 0.020 265 087 926 272 × 2 = 0 + 0.040 530 175 852 544;
  • 36) 0.040 530 175 852 544 × 2 = 0 + 0.081 060 351 705 088;
  • 37) 0.081 060 351 705 088 × 2 = 0 + 0.162 120 703 410 176;
  • 38) 0.162 120 703 410 176 × 2 = 0 + 0.324 241 406 820 352;
  • 39) 0.324 241 406 820 352 × 2 = 0 + 0.648 482 813 640 704;
  • 40) 0.648 482 813 640 704 × 2 = 1 + 0.296 965 627 281 408;
  • 41) 0.296 965 627 281 408 × 2 = 0 + 0.593 931 254 562 816;
  • 42) 0.593 931 254 562 816 × 2 = 1 + 0.187 862 509 125 632;
  • 43) 0.187 862 509 125 632 × 2 = 0 + 0.375 725 018 251 264;
  • 44) 0.375 725 018 251 264 × 2 = 0 + 0.751 450 036 502 528;
  • 45) 0.751 450 036 502 528 × 2 = 1 + 0.502 900 073 005 056;
  • 46) 0.502 900 073 005 056 × 2 = 1 + 0.005 800 146 010 112;
  • 47) 0.005 800 146 010 112 × 2 = 0 + 0.011 600 292 020 224;
  • 48) 0.011 600 292 020 224 × 2 = 0 + 0.023 200 584 040 448;
  • 49) 0.023 200 584 040 448 × 2 = 0 + 0.046 401 168 080 896;
  • 50) 0.046 401 168 080 896 × 2 = 0 + 0.092 802 336 161 792;
  • 51) 0.092 802 336 161 792 × 2 = 0 + 0.185 604 672 323 584;
  • 52) 0.185 604 672 323 584 × 2 = 0 + 0.371 209 344 647 168;
  • 53) 0.371 209 344 647 168 × 2 = 0 + 0.742 418 689 294 336;

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:


0.974 013 318 541 708(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0000 0(2)

5. Positive number before normalization:

0.974 013 318 541 708(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 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:


0.974 013 318 541 708(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0000 0(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0000 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1001 1000 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 1001 1000 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 1001 1000 0000 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1001 1000 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 1001 1000 0000


Decimal number 0.974 013 318 541 708 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 1001 1000 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