3.133 333 333 333 333 329 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.133 333 333 333 333 329 31(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
3.133 333 333 333 333 329 31(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.133 333 333 333 333 329 31.

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) 0.133 333 333 333 333 329 31 × 2 = 0 + 0.266 666 666 666 666 658 62;
  • 2) 0.266 666 666 666 666 658 62 × 2 = 0 + 0.533 333 333 333 333 317 24;
  • 3) 0.533 333 333 333 333 317 24 × 2 = 1 + 0.066 666 666 666 666 634 48;
  • 4) 0.066 666 666 666 666 634 48 × 2 = 0 + 0.133 333 333 333 333 268 96;
  • 5) 0.133 333 333 333 333 268 96 × 2 = 0 + 0.266 666 666 666 666 537 92;
  • 6) 0.266 666 666 666 666 537 92 × 2 = 0 + 0.533 333 333 333 333 075 84;
  • 7) 0.533 333 333 333 333 075 84 × 2 = 1 + 0.066 666 666 666 666 151 68;
  • 8) 0.066 666 666 666 666 151 68 × 2 = 0 + 0.133 333 333 333 332 303 36;
  • 9) 0.133 333 333 333 332 303 36 × 2 = 0 + 0.266 666 666 666 664 606 72;
  • 10) 0.266 666 666 666 664 606 72 × 2 = 0 + 0.533 333 333 333 329 213 44;
  • 11) 0.533 333 333 333 329 213 44 × 2 = 1 + 0.066 666 666 666 658 426 88;
  • 12) 0.066 666 666 666 658 426 88 × 2 = 0 + 0.133 333 333 333 316 853 76;
  • 13) 0.133 333 333 333 316 853 76 × 2 = 0 + 0.266 666 666 666 633 707 52;
  • 14) 0.266 666 666 666 633 707 52 × 2 = 0 + 0.533 333 333 333 267 415 04;
  • 15) 0.533 333 333 333 267 415 04 × 2 = 1 + 0.066 666 666 666 534 830 08;
  • 16) 0.066 666 666 666 534 830 08 × 2 = 0 + 0.133 333 333 333 069 660 16;
  • 17) 0.133 333 333 333 069 660 16 × 2 = 0 + 0.266 666 666 666 139 320 32;
  • 18) 0.266 666 666 666 139 320 32 × 2 = 0 + 0.533 333 333 332 278 640 64;
  • 19) 0.533 333 333 332 278 640 64 × 2 = 1 + 0.066 666 666 664 557 281 28;
  • 20) 0.066 666 666 664 557 281 28 × 2 = 0 + 0.133 333 333 329 114 562 56;
  • 21) 0.133 333 333 329 114 562 56 × 2 = 0 + 0.266 666 666 658 229 125 12;
  • 22) 0.266 666 666 658 229 125 12 × 2 = 0 + 0.533 333 333 316 458 250 24;
  • 23) 0.533 333 333 316 458 250 24 × 2 = 1 + 0.066 666 666 632 916 500 48;
  • 24) 0.066 666 666 632 916 500 48 × 2 = 0 + 0.133 333 333 265 833 000 96;
  • 25) 0.133 333 333 265 833 000 96 × 2 = 0 + 0.266 666 666 531 666 001 92;
  • 26) 0.266 666 666 531 666 001 92 × 2 = 0 + 0.533 333 333 063 332 003 84;
  • 27) 0.533 333 333 063 332 003 84 × 2 = 1 + 0.066 666 666 126 664 007 68;
  • 28) 0.066 666 666 126 664 007 68 × 2 = 0 + 0.133 333 332 253 328 015 36;
  • 29) 0.133 333 332 253 328 015 36 × 2 = 0 + 0.266 666 664 506 656 030 72;
  • 30) 0.266 666 664 506 656 030 72 × 2 = 0 + 0.533 333 329 013 312 061 44;
  • 31) 0.533 333 329 013 312 061 44 × 2 = 1 + 0.066 666 658 026 624 122 88;
  • 32) 0.066 666 658 026 624 122 88 × 2 = 0 + 0.133 333 316 053 248 245 76;
  • 33) 0.133 333 316 053 248 245 76 × 2 = 0 + 0.266 666 632 106 496 491 52;
  • 34) 0.266 666 632 106 496 491 52 × 2 = 0 + 0.533 333 264 212 992 983 04;
  • 35) 0.533 333 264 212 992 983 04 × 2 = 1 + 0.066 666 528 425 985 966 08;
  • 36) 0.066 666 528 425 985 966 08 × 2 = 0 + 0.133 333 056 851 971 932 16;
  • 37) 0.133 333 056 851 971 932 16 × 2 = 0 + 0.266 666 113 703 943 864 32;
  • 38) 0.266 666 113 703 943 864 32 × 2 = 0 + 0.533 332 227 407 887 728 64;
  • 39) 0.533 332 227 407 887 728 64 × 2 = 1 + 0.066 664 454 815 775 457 28;
  • 40) 0.066 664 454 815 775 457 28 × 2 = 0 + 0.133 328 909 631 550 914 56;
  • 41) 0.133 328 909 631 550 914 56 × 2 = 0 + 0.266 657 819 263 101 829 12;
  • 42) 0.266 657 819 263 101 829 12 × 2 = 0 + 0.533 315 638 526 203 658 24;
  • 43) 0.533 315 638 526 203 658 24 × 2 = 1 + 0.066 631 277 052 407 316 48;
  • 44) 0.066 631 277 052 407 316 48 × 2 = 0 + 0.133 262 554 104 814 632 96;
  • 45) 0.133 262 554 104 814 632 96 × 2 = 0 + 0.266 525 108 209 629 265 92;
  • 46) 0.266 525 108 209 629 265 92 × 2 = 0 + 0.533 050 216 419 258 531 84;
  • 47) 0.533 050 216 419 258 531 84 × 2 = 1 + 0.066 100 432 838 517 063 68;
  • 48) 0.066 100 432 838 517 063 68 × 2 = 0 + 0.132 200 865 677 034 127 36;
  • 49) 0.132 200 865 677 034 127 36 × 2 = 0 + 0.264 401 731 354 068 254 72;
  • 50) 0.264 401 731 354 068 254 72 × 2 = 0 + 0.528 803 462 708 136 509 44;
  • 51) 0.528 803 462 708 136 509 44 × 2 = 1 + 0.057 606 925 416 273 018 88;
  • 52) 0.057 606 925 416 273 018 88 × 2 = 0 + 0.115 213 850 832 546 037 76;
  • 53) 0.115 213 850 832 546 037 76 × 2 = 0 + 0.230 427 701 665 092 075 52;

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:


0.133 333 333 333 333 329 31(10) =

0.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2)

5. Positive number before normalization:

3.133 333 333 333 333 329 31(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.133 333 333 333 333 329 31(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) × 20 =

1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 00(2) × 21


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):
1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 00


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 00 =

1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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) =
100 0000 0000

Mantissa (52 bits) =
1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


Decimal number 3.133 333 333 333 333 329 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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