0.974 013 318 501 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 501(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
0.974 013 318 501(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: 0.974 013 318 501.

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) 0.974 013 318 501 × 2 = 1 + 0.948 026 637 002;
  • 2) 0.948 026 637 002 × 2 = 1 + 0.896 053 274 004;
  • 3) 0.896 053 274 004 × 2 = 1 + 0.792 106 548 008;
  • 4) 0.792 106 548 008 × 2 = 1 + 0.584 213 096 016;
  • 5) 0.584 213 096 016 × 2 = 1 + 0.168 426 192 032;
  • 6) 0.168 426 192 032 × 2 = 0 + 0.336 852 384 064;
  • 7) 0.336 852 384 064 × 2 = 0 + 0.673 704 768 128;
  • 8) 0.673 704 768 128 × 2 = 1 + 0.347 409 536 256;
  • 9) 0.347 409 536 256 × 2 = 0 + 0.694 819 072 512;
  • 10) 0.694 819 072 512 × 2 = 1 + 0.389 638 145 024;
  • 11) 0.389 638 145 024 × 2 = 0 + 0.779 276 290 048;
  • 12) 0.779 276 290 048 × 2 = 1 + 0.558 552 580 096;
  • 13) 0.558 552 580 096 × 2 = 1 + 0.117 105 160 192;
  • 14) 0.117 105 160 192 × 2 = 0 + 0.234 210 320 384;
  • 15) 0.234 210 320 384 × 2 = 0 + 0.468 420 640 768;
  • 16) 0.468 420 640 768 × 2 = 0 + 0.936 841 281 536;
  • 17) 0.936 841 281 536 × 2 = 1 + 0.873 682 563 072;
  • 18) 0.873 682 563 072 × 2 = 1 + 0.747 365 126 144;
  • 19) 0.747 365 126 144 × 2 = 1 + 0.494 730 252 288;
  • 20) 0.494 730 252 288 × 2 = 0 + 0.989 460 504 576;
  • 21) 0.989 460 504 576 × 2 = 1 + 0.978 921 009 152;
  • 22) 0.978 921 009 152 × 2 = 1 + 0.957 842 018 304;
  • 23) 0.957 842 018 304 × 2 = 1 + 0.915 684 036 608;
  • 24) 0.915 684 036 608 × 2 = 1 + 0.831 368 073 216;
  • 25) 0.831 368 073 216 × 2 = 1 + 0.662 736 146 432;
  • 26) 0.662 736 146 432 × 2 = 1 + 0.325 472 292 864;
  • 27) 0.325 472 292 864 × 2 = 0 + 0.650 944 585 728;
  • 28) 0.650 944 585 728 × 2 = 1 + 0.301 889 171 456;
  • 29) 0.301 889 171 456 × 2 = 0 + 0.603 778 342 912;
  • 30) 0.603 778 342 912 × 2 = 1 + 0.207 556 685 824;
  • 31) 0.207 556 685 824 × 2 = 0 + 0.415 113 371 648;
  • 32) 0.415 113 371 648 × 2 = 0 + 0.830 226 743 296;
  • 33) 0.830 226 743 296 × 2 = 1 + 0.660 453 486 592;
  • 34) 0.660 453 486 592 × 2 = 1 + 0.320 906 973 184;
  • 35) 0.320 906 973 184 × 2 = 0 + 0.641 813 946 368;
  • 36) 0.641 813 946 368 × 2 = 1 + 0.283 627 892 736;
  • 37) 0.283 627 892 736 × 2 = 0 + 0.567 255 785 472;
  • 38) 0.567 255 785 472 × 2 = 1 + 0.134 511 570 944;
  • 39) 0.134 511 570 944 × 2 = 0 + 0.269 023 141 888;
  • 40) 0.269 023 141 888 × 2 = 0 + 0.538 046 283 776;
  • 41) 0.538 046 283 776 × 2 = 1 + 0.076 092 567 552;
  • 42) 0.076 092 567 552 × 2 = 0 + 0.152 185 135 104;
  • 43) 0.152 185 135 104 × 2 = 0 + 0.304 370 270 208;
  • 44) 0.304 370 270 208 × 2 = 0 + 0.608 740 540 416;
  • 45) 0.608 740 540 416 × 2 = 1 + 0.217 481 080 832;
  • 46) 0.217 481 080 832 × 2 = 0 + 0.434 962 161 664;
  • 47) 0.434 962 161 664 × 2 = 0 + 0.869 924 323 328;
  • 48) 0.869 924 323 328 × 2 = 1 + 0.739 848 646 656;
  • 49) 0.739 848 646 656 × 2 = 1 + 0.479 697 293 312;
  • 50) 0.479 697 293 312 × 2 = 0 + 0.959 394 586 624;
  • 51) 0.959 394 586 624 × 2 = 1 + 0.918 789 173 248;
  • 52) 0.918 789 173 248 × 2 = 1 + 0.837 578 346 496;
  • 53) 0.837 578 346 496 × 2 = 1 + 0.675 156 692 992;

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:


0.974 013 318 501(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1101 0100 1000 1001 1011 1(2)

5. Positive number before normalization:

0.974 013 318 501(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1101 0100 1000 1001 1011 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:


0.974 013 318 501(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1101 0100 1000 1001 1011 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1101 0100 1000 1001 1011 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1001 1010 1001 0001 0011 0111(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 1001 1010 1001 0001 0011 0111


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 1001 1010 1001 0001 0011 0111 =

1111 0010 1011 0001 1101 1111 1010 1001 1010 1001 0001 0011 0111


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 1001 1010 1001 0001 0011 0111


Decimal number 0.974 013 318 501 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1001 1010 1001 0001 0011 0111

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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