0.974 013 318 541 708 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 708 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
0.974 013 318 541 708 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: 0.974 013 318 541 708 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) 0.974 013 318 541 708 5 × 2 = 1 + 0.948 026 637 083 417;
  • 2) 0.948 026 637 083 417 × 2 = 1 + 0.896 053 274 166 834;
  • 3) 0.896 053 274 166 834 × 2 = 1 + 0.792 106 548 333 668;
  • 4) 0.792 106 548 333 668 × 2 = 1 + 0.584 213 096 667 336;
  • 5) 0.584 213 096 667 336 × 2 = 1 + 0.168 426 193 334 672;
  • 6) 0.168 426 193 334 672 × 2 = 0 + 0.336 852 386 669 344;
  • 7) 0.336 852 386 669 344 × 2 = 0 + 0.673 704 773 338 688;
  • 8) 0.673 704 773 338 688 × 2 = 1 + 0.347 409 546 677 376;
  • 9) 0.347 409 546 677 376 × 2 = 0 + 0.694 819 093 354 752;
  • 10) 0.694 819 093 354 752 × 2 = 1 + 0.389 638 186 709 504;
  • 11) 0.389 638 186 709 504 × 2 = 0 + 0.779 276 373 419 008;
  • 12) 0.779 276 373 419 008 × 2 = 1 + 0.558 552 746 838 016;
  • 13) 0.558 552 746 838 016 × 2 = 1 + 0.117 105 493 676 032;
  • 14) 0.117 105 493 676 032 × 2 = 0 + 0.234 210 987 352 064;
  • 15) 0.234 210 987 352 064 × 2 = 0 + 0.468 421 974 704 128;
  • 16) 0.468 421 974 704 128 × 2 = 0 + 0.936 843 949 408 256;
  • 17) 0.936 843 949 408 256 × 2 = 1 + 0.873 687 898 816 512;
  • 18) 0.873 687 898 816 512 × 2 = 1 + 0.747 375 797 633 024;
  • 19) 0.747 375 797 633 024 × 2 = 1 + 0.494 751 595 266 048;
  • 20) 0.494 751 595 266 048 × 2 = 0 + 0.989 503 190 532 096;
  • 21) 0.989 503 190 532 096 × 2 = 1 + 0.979 006 381 064 192;
  • 22) 0.979 006 381 064 192 × 2 = 1 + 0.958 012 762 128 384;
  • 23) 0.958 012 762 128 384 × 2 = 1 + 0.916 025 524 256 768;
  • 24) 0.916 025 524 256 768 × 2 = 1 + 0.832 051 048 513 536;
  • 25) 0.832 051 048 513 536 × 2 = 1 + 0.664 102 097 027 072;
  • 26) 0.664 102 097 027 072 × 2 = 1 + 0.328 204 194 054 144;
  • 27) 0.328 204 194 054 144 × 2 = 0 + 0.656 408 388 108 288;
  • 28) 0.656 408 388 108 288 × 2 = 1 + 0.312 816 776 216 576;
  • 29) 0.312 816 776 216 576 × 2 = 0 + 0.625 633 552 433 152;
  • 30) 0.625 633 552 433 152 × 2 = 1 + 0.251 267 104 866 304;
  • 31) 0.251 267 104 866 304 × 2 = 0 + 0.502 534 209 732 608;
  • 32) 0.502 534 209 732 608 × 2 = 1 + 0.005 068 419 465 216;
  • 33) 0.005 068 419 465 216 × 2 = 0 + 0.010 136 838 930 432;
  • 34) 0.010 136 838 930 432 × 2 = 0 + 0.020 273 677 860 864;
  • 35) 0.020 273 677 860 864 × 2 = 0 + 0.040 547 355 721 728;
  • 36) 0.040 547 355 721 728 × 2 = 0 + 0.081 094 711 443 456;
  • 37) 0.081 094 711 443 456 × 2 = 0 + 0.162 189 422 886 912;
  • 38) 0.162 189 422 886 912 × 2 = 0 + 0.324 378 845 773 824;
  • 39) 0.324 378 845 773 824 × 2 = 0 + 0.648 757 691 547 648;
  • 40) 0.648 757 691 547 648 × 2 = 1 + 0.297 515 383 095 296;
  • 41) 0.297 515 383 095 296 × 2 = 0 + 0.595 030 766 190 592;
  • 42) 0.595 030 766 190 592 × 2 = 1 + 0.190 061 532 381 184;
  • 43) 0.190 061 532 381 184 × 2 = 0 + 0.380 123 064 762 368;
  • 44) 0.380 123 064 762 368 × 2 = 0 + 0.760 246 129 524 736;
  • 45) 0.760 246 129 524 736 × 2 = 1 + 0.520 492 259 049 472;
  • 46) 0.520 492 259 049 472 × 2 = 1 + 0.040 984 518 098 944;
  • 47) 0.040 984 518 098 944 × 2 = 0 + 0.081 969 036 197 888;
  • 48) 0.081 969 036 197 888 × 2 = 0 + 0.163 938 072 395 776;
  • 49) 0.163 938 072 395 776 × 2 = 0 + 0.327 876 144 791 552;
  • 50) 0.327 876 144 791 552 × 2 = 0 + 0.655 752 289 583 104;
  • 51) 0.655 752 289 583 104 × 2 = 1 + 0.311 504 579 166 208;
  • 52) 0.311 504 579 166 208 × 2 = 0 + 0.623 009 158 332 416;
  • 53) 0.623 009 158 332 416 × 2 = 1 + 0.246 018 316 664 832;

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 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0010 1(2)

5. Positive number before normalization:

0.974 013 318 541 708 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0010 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:


0.974 013 318 541 708 5(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0010 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 1100 0010 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1001 1000 0101(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 0101


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 0101 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1001 1000 0101


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 0101


Decimal number 0.974 013 318 541 708 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 1001 1000 0101

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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