0.974 013 318 541 907 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 907(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
0.974 013 318 541 907(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: 0.974 013 318 541 907.

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) 0.974 013 318 541 907 × 2 = 1 + 0.948 026 637 083 814;
  • 2) 0.948 026 637 083 814 × 2 = 1 + 0.896 053 274 167 628;
  • 3) 0.896 053 274 167 628 × 2 = 1 + 0.792 106 548 335 256;
  • 4) 0.792 106 548 335 256 × 2 = 1 + 0.584 213 096 670 512;
  • 5) 0.584 213 096 670 512 × 2 = 1 + 0.168 426 193 341 024;
  • 6) 0.168 426 193 341 024 × 2 = 0 + 0.336 852 386 682 048;
  • 7) 0.336 852 386 682 048 × 2 = 0 + 0.673 704 773 364 096;
  • 8) 0.673 704 773 364 096 × 2 = 1 + 0.347 409 546 728 192;
  • 9) 0.347 409 546 728 192 × 2 = 0 + 0.694 819 093 456 384;
  • 10) 0.694 819 093 456 384 × 2 = 1 + 0.389 638 186 912 768;
  • 11) 0.389 638 186 912 768 × 2 = 0 + 0.779 276 373 825 536;
  • 12) 0.779 276 373 825 536 × 2 = 1 + 0.558 552 747 651 072;
  • 13) 0.558 552 747 651 072 × 2 = 1 + 0.117 105 495 302 144;
  • 14) 0.117 105 495 302 144 × 2 = 0 + 0.234 210 990 604 288;
  • 15) 0.234 210 990 604 288 × 2 = 0 + 0.468 421 981 208 576;
  • 16) 0.468 421 981 208 576 × 2 = 0 + 0.936 843 962 417 152;
  • 17) 0.936 843 962 417 152 × 2 = 1 + 0.873 687 924 834 304;
  • 18) 0.873 687 924 834 304 × 2 = 1 + 0.747 375 849 668 608;
  • 19) 0.747 375 849 668 608 × 2 = 1 + 0.494 751 699 337 216;
  • 20) 0.494 751 699 337 216 × 2 = 0 + 0.989 503 398 674 432;
  • 21) 0.989 503 398 674 432 × 2 = 1 + 0.979 006 797 348 864;
  • 22) 0.979 006 797 348 864 × 2 = 1 + 0.958 013 594 697 728;
  • 23) 0.958 013 594 697 728 × 2 = 1 + 0.916 027 189 395 456;
  • 24) 0.916 027 189 395 456 × 2 = 1 + 0.832 054 378 790 912;
  • 25) 0.832 054 378 790 912 × 2 = 1 + 0.664 108 757 581 824;
  • 26) 0.664 108 757 581 824 × 2 = 1 + 0.328 217 515 163 648;
  • 27) 0.328 217 515 163 648 × 2 = 0 + 0.656 435 030 327 296;
  • 28) 0.656 435 030 327 296 × 2 = 1 + 0.312 870 060 654 592;
  • 29) 0.312 870 060 654 592 × 2 = 0 + 0.625 740 121 309 184;
  • 30) 0.625 740 121 309 184 × 2 = 1 + 0.251 480 242 618 368;
  • 31) 0.251 480 242 618 368 × 2 = 0 + 0.502 960 485 236 736;
  • 32) 0.502 960 485 236 736 × 2 = 1 + 0.005 920 970 473 472;
  • 33) 0.005 920 970 473 472 × 2 = 0 + 0.011 841 940 946 944;
  • 34) 0.011 841 940 946 944 × 2 = 0 + 0.023 683 881 893 888;
  • 35) 0.023 683 881 893 888 × 2 = 0 + 0.047 367 763 787 776;
  • 36) 0.047 367 763 787 776 × 2 = 0 + 0.094 735 527 575 552;
  • 37) 0.094 735 527 575 552 × 2 = 0 + 0.189 471 055 151 104;
  • 38) 0.189 471 055 151 104 × 2 = 0 + 0.378 942 110 302 208;
  • 39) 0.378 942 110 302 208 × 2 = 0 + 0.757 884 220 604 416;
  • 40) 0.757 884 220 604 416 × 2 = 1 + 0.515 768 441 208 832;
  • 41) 0.515 768 441 208 832 × 2 = 1 + 0.031 536 882 417 664;
  • 42) 0.031 536 882 417 664 × 2 = 0 + 0.063 073 764 835 328;
  • 43) 0.063 073 764 835 328 × 2 = 0 + 0.126 147 529 670 656;
  • 44) 0.126 147 529 670 656 × 2 = 0 + 0.252 295 059 341 312;
  • 45) 0.252 295 059 341 312 × 2 = 0 + 0.504 590 118 682 624;
  • 46) 0.504 590 118 682 624 × 2 = 1 + 0.009 180 237 365 248;
  • 47) 0.009 180 237 365 248 × 2 = 0 + 0.018 360 474 730 496;
  • 48) 0.018 360 474 730 496 × 2 = 0 + 0.036 720 949 460 992;
  • 49) 0.036 720 949 460 992 × 2 = 0 + 0.073 441 898 921 984;
  • 50) 0.073 441 898 921 984 × 2 = 0 + 0.146 883 797 843 968;
  • 51) 0.146 883 797 843 968 × 2 = 0 + 0.293 767 595 687 936;
  • 52) 0.293 767 595 687 936 × 2 = 0 + 0.587 535 191 375 872;
  • 53) 0.587 535 191 375 872 × 2 = 1 + 0.175 070 382 751 744;

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:


0.974 013 318 541 907(10) =

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

5. Positive number before normalization:

0.974 013 318 541 907(10) =

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


0.974 013 318 541 907(10) =

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

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

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0011 0000 1000 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 0011 0000 1000 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 0011 0000 1000 0001 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0011 0000 1000 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 0011 0000 1000 0001


Decimal number 0.974 013 318 541 907 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 0011 0000 1000 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