64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.052 097 324 398 346 24 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 24(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 24.

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 24 × 2 = 0 + 0.104 194 648 796 692 48;
  • 2) 0.104 194 648 796 692 48 × 2 = 0 + 0.208 389 297 593 384 96;
  • 3) 0.208 389 297 593 384 96 × 2 = 0 + 0.416 778 595 186 769 92;
  • 4) 0.416 778 595 186 769 92 × 2 = 0 + 0.833 557 190 373 539 84;
  • 5) 0.833 557 190 373 539 84 × 2 = 1 + 0.667 114 380 747 079 68;
  • 6) 0.667 114 380 747 079 68 × 2 = 1 + 0.334 228 761 494 159 36;
  • 7) 0.334 228 761 494 159 36 × 2 = 0 + 0.668 457 522 988 318 72;
  • 8) 0.668 457 522 988 318 72 × 2 = 1 + 0.336 915 045 976 637 44;
  • 9) 0.336 915 045 976 637 44 × 2 = 0 + 0.673 830 091 953 274 88;
  • 10) 0.673 830 091 953 274 88 × 2 = 1 + 0.347 660 183 906 549 76;
  • 11) 0.347 660 183 906 549 76 × 2 = 0 + 0.695 320 367 813 099 52;
  • 12) 0.695 320 367 813 099 52 × 2 = 1 + 0.390 640 735 626 199 04;
  • 13) 0.390 640 735 626 199 04 × 2 = 0 + 0.781 281 471 252 398 08;
  • 14) 0.781 281 471 252 398 08 × 2 = 1 + 0.562 562 942 504 796 16;
  • 15) 0.562 562 942 504 796 16 × 2 = 1 + 0.125 125 885 009 592 32;
  • 16) 0.125 125 885 009 592 32 × 2 = 0 + 0.250 251 770 019 184 64;
  • 17) 0.250 251 770 019 184 64 × 2 = 0 + 0.500 503 540 038 369 28;
  • 18) 0.500 503 540 038 369 28 × 2 = 1 + 0.001 007 080 076 738 56;
  • 19) 0.001 007 080 076 738 56 × 2 = 0 + 0.002 014 160 153 477 12;
  • 20) 0.002 014 160 153 477 12 × 2 = 0 + 0.004 028 320 306 954 24;
  • 21) 0.004 028 320 306 954 24 × 2 = 0 + 0.008 056 640 613 908 48;
  • 22) 0.008 056 640 613 908 48 × 2 = 0 + 0.016 113 281 227 816 96;
  • 23) 0.016 113 281 227 816 96 × 2 = 0 + 0.032 226 562 455 633 92;
  • 24) 0.032 226 562 455 633 92 × 2 = 0 + 0.064 453 124 911 267 84;
  • 25) 0.064 453 124 911 267 84 × 2 = 0 + 0.128 906 249 822 535 68;
  • 26) 0.128 906 249 822 535 68 × 2 = 0 + 0.257 812 499 645 071 36;
  • 27) 0.257 812 499 645 071 36 × 2 = 0 + 0.515 624 999 290 142 72;
  • 28) 0.515 624 999 290 142 72 × 2 = 1 + 0.031 249 998 580 285 44;
  • 29) 0.031 249 998 580 285 44 × 2 = 0 + 0.062 499 997 160 570 88;
  • 30) 0.062 499 997 160 570 88 × 2 = 0 + 0.124 999 994 321 141 76;
  • 31) 0.124 999 994 321 141 76 × 2 = 0 + 0.249 999 988 642 283 52;
  • 32) 0.249 999 988 642 283 52 × 2 = 0 + 0.499 999 977 284 567 04;
  • 33) 0.499 999 977 284 567 04 × 2 = 0 + 0.999 999 954 569 134 08;
  • 34) 0.999 999 954 569 134 08 × 2 = 1 + 0.999 999 909 138 268 16;
  • 35) 0.999 999 909 138 268 16 × 2 = 1 + 0.999 999 818 276 536 32;
  • 36) 0.999 999 818 276 536 32 × 2 = 1 + 0.999 999 636 553 072 64;
  • 37) 0.999 999 636 553 072 64 × 2 = 1 + 0.999 999 273 106 145 28;
  • 38) 0.999 999 273 106 145 28 × 2 = 1 + 0.999 998 546 212 290 56;
  • 39) 0.999 998 546 212 290 56 × 2 = 1 + 0.999 997 092 424 581 12;
  • 40) 0.999 997 092 424 581 12 × 2 = 1 + 0.999 994 184 849 162 24;
  • 41) 0.999 994 184 849 162 24 × 2 = 1 + 0.999 988 369 698 324 48;
  • 42) 0.999 988 369 698 324 48 × 2 = 1 + 0.999 976 739 396 648 96;
  • 43) 0.999 976 739 396 648 96 × 2 = 1 + 0.999 953 478 793 297 92;
  • 44) 0.999 953 478 793 297 92 × 2 = 1 + 0.999 906 957 586 595 84;
  • 45) 0.999 906 957 586 595 84 × 2 = 1 + 0.999 813 915 173 191 68;
  • 46) 0.999 813 915 173 191 68 × 2 = 1 + 0.999 627 830 346 383 36;
  • 47) 0.999 627 830 346 383 36 × 2 = 1 + 0.999 255 660 692 766 72;
  • 48) 0.999 255 660 692 766 72 × 2 = 1 + 0.998 511 321 385 533 44;
  • 49) 0.998 511 321 385 533 44 × 2 = 1 + 0.997 022 642 771 066 88;
  • 50) 0.997 022 642 771 066 88 × 2 = 1 + 0.994 045 285 542 133 76;
  • 51) 0.994 045 285 542 133 76 × 2 = 1 + 0.988 090 571 084 267 52;
  • 52) 0.988 090 571 084 267 52 × 2 = 1 + 0.976 181 142 168 535 04;
  • 53) 0.976 181 142 168 535 04 × 2 = 1 + 0.952 362 284 337 070 08;
  • 54) 0.952 362 284 337 070 08 × 2 = 1 + 0.904 724 568 674 140 16;
  • 55) 0.904 724 568 674 140 16 × 2 = 1 + 0.809 449 137 348 280 32;
  • 56) 0.809 449 137 348 280 32 × 2 = 1 + 0.618 898 274 696 560 64;
  • 57) 0.618 898 274 696 560 64 × 2 = 1 + 0.237 796 549 393 121 28;

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 24(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 0111 1111 1111 1111 1111 1111 1(2)


5. Positive number before normalization:

0.052 097 324 398 346 24(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 0111 1111 1111 1111 1111 1111 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 24(10) =

0.0000 1101 0101 0110 0100 0000 0001 0000 0111 1111 1111 1111 1111 1111 1(2) =

0.0000 1101 0101 0110 0100 0000 0001 0000 0111 1111 1111 1111 1111 1111 1(2) × 20 =

1.1010 1010 1100 1000 0000 0010 0000 1111 1111 1111 1111 1111 1111(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 0000 1111 1111 1111 1111 1111 1111


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 0000 1111 1111 1111 1111 1111 1111 =

1010 1010 1100 1000 0000 0010 0000 1111 1111 1111 1111 1111 1111


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 0000 1111 1111 1111 1111 1111 1111


The base ten decimal number 0.052 097 324 398 346 24 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1010 - 1010 1010 1100 1000 0000 0010 0000 1111 1111 1111 1111 1111 1111

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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