64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.052 097 324 398 346 6 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.052 097 324 398 346 6(10) converted and written in 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: 0.052 097 324 398 346 6.

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) 0.052 097 324 398 346 6 × 2 = 0 + 0.104 194 648 796 693 2;
  • 2) 0.104 194 648 796 693 2 × 2 = 0 + 0.208 389 297 593 386 4;
  • 3) 0.208 389 297 593 386 4 × 2 = 0 + 0.416 778 595 186 772 8;
  • 4) 0.416 778 595 186 772 8 × 2 = 0 + 0.833 557 190 373 545 6;
  • 5) 0.833 557 190 373 545 6 × 2 = 1 + 0.667 114 380 747 091 2;
  • 6) 0.667 114 380 747 091 2 × 2 = 1 + 0.334 228 761 494 182 4;
  • 7) 0.334 228 761 494 182 4 × 2 = 0 + 0.668 457 522 988 364 8;
  • 8) 0.668 457 522 988 364 8 × 2 = 1 + 0.336 915 045 976 729 6;
  • 9) 0.336 915 045 976 729 6 × 2 = 0 + 0.673 830 091 953 459 2;
  • 10) 0.673 830 091 953 459 2 × 2 = 1 + 0.347 660 183 906 918 4;
  • 11) 0.347 660 183 906 918 4 × 2 = 0 + 0.695 320 367 813 836 8;
  • 12) 0.695 320 367 813 836 8 × 2 = 1 + 0.390 640 735 627 673 6;
  • 13) 0.390 640 735 627 673 6 × 2 = 0 + 0.781 281 471 255 347 2;
  • 14) 0.781 281 471 255 347 2 × 2 = 1 + 0.562 562 942 510 694 4;
  • 15) 0.562 562 942 510 694 4 × 2 = 1 + 0.125 125 885 021 388 8;
  • 16) 0.125 125 885 021 388 8 × 2 = 0 + 0.250 251 770 042 777 6;
  • 17) 0.250 251 770 042 777 6 × 2 = 0 + 0.500 503 540 085 555 2;
  • 18) 0.500 503 540 085 555 2 × 2 = 1 + 0.001 007 080 171 110 4;
  • 19) 0.001 007 080 171 110 4 × 2 = 0 + 0.002 014 160 342 220 8;
  • 20) 0.002 014 160 342 220 8 × 2 = 0 + 0.004 028 320 684 441 6;
  • 21) 0.004 028 320 684 441 6 × 2 = 0 + 0.008 056 641 368 883 2;
  • 22) 0.008 056 641 368 883 2 × 2 = 0 + 0.016 113 282 737 766 4;
  • 23) 0.016 113 282 737 766 4 × 2 = 0 + 0.032 226 565 475 532 8;
  • 24) 0.032 226 565 475 532 8 × 2 = 0 + 0.064 453 130 951 065 6;
  • 25) 0.064 453 130 951 065 6 × 2 = 0 + 0.128 906 261 902 131 2;
  • 26) 0.128 906 261 902 131 2 × 2 = 0 + 0.257 812 523 804 262 4;
  • 27) 0.257 812 523 804 262 4 × 2 = 0 + 0.515 625 047 608 524 8;
  • 28) 0.515 625 047 608 524 8 × 2 = 1 + 0.031 250 095 217 049 6;
  • 29) 0.031 250 095 217 049 6 × 2 = 0 + 0.062 500 190 434 099 2;
  • 30) 0.062 500 190 434 099 2 × 2 = 0 + 0.125 000 380 868 198 4;
  • 31) 0.125 000 380 868 198 4 × 2 = 0 + 0.250 000 761 736 396 8;
  • 32) 0.250 000 761 736 396 8 × 2 = 0 + 0.500 001 523 472 793 6;
  • 33) 0.500 001 523 472 793 6 × 2 = 1 + 0.000 003 046 945 587 2;
  • 34) 0.000 003 046 945 587 2 × 2 = 0 + 0.000 006 093 891 174 4;
  • 35) 0.000 006 093 891 174 4 × 2 = 0 + 0.000 012 187 782 348 8;
  • 36) 0.000 012 187 782 348 8 × 2 = 0 + 0.000 024 375 564 697 6;
  • 37) 0.000 024 375 564 697 6 × 2 = 0 + 0.000 048 751 129 395 2;
  • 38) 0.000 048 751 129 395 2 × 2 = 0 + 0.000 097 502 258 790 4;
  • 39) 0.000 097 502 258 790 4 × 2 = 0 + 0.000 195 004 517 580 8;
  • 40) 0.000 195 004 517 580 8 × 2 = 0 + 0.000 390 009 035 161 6;
  • 41) 0.000 390 009 035 161 6 × 2 = 0 + 0.000 780 018 070 323 2;
  • 42) 0.000 780 018 070 323 2 × 2 = 0 + 0.001 560 036 140 646 4;
  • 43) 0.001 560 036 140 646 4 × 2 = 0 + 0.003 120 072 281 292 8;
  • 44) 0.003 120 072 281 292 8 × 2 = 0 + 0.006 240 144 562 585 6;
  • 45) 0.006 240 144 562 585 6 × 2 = 0 + 0.012 480 289 125 171 2;
  • 46) 0.012 480 289 125 171 2 × 2 = 0 + 0.024 960 578 250 342 4;
  • 47) 0.024 960 578 250 342 4 × 2 = 0 + 0.049 921 156 500 684 8;
  • 48) 0.049 921 156 500 684 8 × 2 = 0 + 0.099 842 313 001 369 6;
  • 49) 0.099 842 313 001 369 6 × 2 = 0 + 0.199 684 626 002 739 2;
  • 50) 0.199 684 626 002 739 2 × 2 = 0 + 0.399 369 252 005 478 4;
  • 51) 0.399 369 252 005 478 4 × 2 = 0 + 0.798 738 504 010 956 8;
  • 52) 0.798 738 504 010 956 8 × 2 = 1 + 0.597 477 008 021 913 6;
  • 53) 0.597 477 008 021 913 6 × 2 = 1 + 0.194 954 016 043 827 2;
  • 54) 0.194 954 016 043 827 2 × 2 = 0 + 0.389 908 032 087 654 4;
  • 55) 0.389 908 032 087 654 4 × 2 = 0 + 0.779 816 064 175 308 8;
  • 56) 0.779 816 064 175 308 8 × 2 = 1 + 0.559 632 128 350 617 6;
  • 57) 0.559 632 128 350 617 6 × 2 = 1 + 0.119 264 256 701 235 2;

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...)


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:


0.052 097 324 398 346 6(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 1000 0000 0000 0000 0001 1001 1(2)


5. Positive number before normalization:

0.052 097 324 398 346 6(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 1000 0000 0000 0000 0001 1001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.052 097 324 398 346 6(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 1000 0000 0000 0000 0001 1001 1(2) =

0.0000 1101 0101 0110 0100 0000 0001 0000 1000 0000 0000 0000 0001 1001 1(2) × 20 =

1.1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011(2) × 2-5


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): -5

Mantissa (not normalized):
1.1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1018(10) =

011 1111 1010(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. 1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011 =

1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011


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 1010

Mantissa (52 bits) =
1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011


The base ten decimal number 0.052 097 324 398 346 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1010 - 1010 1010 1100 1000 0000 0010 0001 0000 0000 0000 0000 0011 0011

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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