3.133 333 333 333 333 328 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.133 333 333 333 333 328 51(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
3.133 333 333 333 333 328 51(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: 0.133 333 333 333 333 328 51.

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) 0.133 333 333 333 333 328 51 × 2 = 0 + 0.266 666 666 666 666 657 02;
  • 2) 0.266 666 666 666 666 657 02 × 2 = 0 + 0.533 333 333 333 333 314 04;
  • 3) 0.533 333 333 333 333 314 04 × 2 = 1 + 0.066 666 666 666 666 628 08;
  • 4) 0.066 666 666 666 666 628 08 × 2 = 0 + 0.133 333 333 333 333 256 16;
  • 5) 0.133 333 333 333 333 256 16 × 2 = 0 + 0.266 666 666 666 666 512 32;
  • 6) 0.266 666 666 666 666 512 32 × 2 = 0 + 0.533 333 333 333 333 024 64;
  • 7) 0.533 333 333 333 333 024 64 × 2 = 1 + 0.066 666 666 666 666 049 28;
  • 8) 0.066 666 666 666 666 049 28 × 2 = 0 + 0.133 333 333 333 332 098 56;
  • 9) 0.133 333 333 333 332 098 56 × 2 = 0 + 0.266 666 666 666 664 197 12;
  • 10) 0.266 666 666 666 664 197 12 × 2 = 0 + 0.533 333 333 333 328 394 24;
  • 11) 0.533 333 333 333 328 394 24 × 2 = 1 + 0.066 666 666 666 656 788 48;
  • 12) 0.066 666 666 666 656 788 48 × 2 = 0 + 0.133 333 333 333 313 576 96;
  • 13) 0.133 333 333 333 313 576 96 × 2 = 0 + 0.266 666 666 666 627 153 92;
  • 14) 0.266 666 666 666 627 153 92 × 2 = 0 + 0.533 333 333 333 254 307 84;
  • 15) 0.533 333 333 333 254 307 84 × 2 = 1 + 0.066 666 666 666 508 615 68;
  • 16) 0.066 666 666 666 508 615 68 × 2 = 0 + 0.133 333 333 333 017 231 36;
  • 17) 0.133 333 333 333 017 231 36 × 2 = 0 + 0.266 666 666 666 034 462 72;
  • 18) 0.266 666 666 666 034 462 72 × 2 = 0 + 0.533 333 333 332 068 925 44;
  • 19) 0.533 333 333 332 068 925 44 × 2 = 1 + 0.066 666 666 664 137 850 88;
  • 20) 0.066 666 666 664 137 850 88 × 2 = 0 + 0.133 333 333 328 275 701 76;
  • 21) 0.133 333 333 328 275 701 76 × 2 = 0 + 0.266 666 666 656 551 403 52;
  • 22) 0.266 666 666 656 551 403 52 × 2 = 0 + 0.533 333 333 313 102 807 04;
  • 23) 0.533 333 333 313 102 807 04 × 2 = 1 + 0.066 666 666 626 205 614 08;
  • 24) 0.066 666 666 626 205 614 08 × 2 = 0 + 0.133 333 333 252 411 228 16;
  • 25) 0.133 333 333 252 411 228 16 × 2 = 0 + 0.266 666 666 504 822 456 32;
  • 26) 0.266 666 666 504 822 456 32 × 2 = 0 + 0.533 333 333 009 644 912 64;
  • 27) 0.533 333 333 009 644 912 64 × 2 = 1 + 0.066 666 666 019 289 825 28;
  • 28) 0.066 666 666 019 289 825 28 × 2 = 0 + 0.133 333 332 038 579 650 56;
  • 29) 0.133 333 332 038 579 650 56 × 2 = 0 + 0.266 666 664 077 159 301 12;
  • 30) 0.266 666 664 077 159 301 12 × 2 = 0 + 0.533 333 328 154 318 602 24;
  • 31) 0.533 333 328 154 318 602 24 × 2 = 1 + 0.066 666 656 308 637 204 48;
  • 32) 0.066 666 656 308 637 204 48 × 2 = 0 + 0.133 333 312 617 274 408 96;
  • 33) 0.133 333 312 617 274 408 96 × 2 = 0 + 0.266 666 625 234 548 817 92;
  • 34) 0.266 666 625 234 548 817 92 × 2 = 0 + 0.533 333 250 469 097 635 84;
  • 35) 0.533 333 250 469 097 635 84 × 2 = 1 + 0.066 666 500 938 195 271 68;
  • 36) 0.066 666 500 938 195 271 68 × 2 = 0 + 0.133 333 001 876 390 543 36;
  • 37) 0.133 333 001 876 390 543 36 × 2 = 0 + 0.266 666 003 752 781 086 72;
  • 38) 0.266 666 003 752 781 086 72 × 2 = 0 + 0.533 332 007 505 562 173 44;
  • 39) 0.533 332 007 505 562 173 44 × 2 = 1 + 0.066 664 015 011 124 346 88;
  • 40) 0.066 664 015 011 124 346 88 × 2 = 0 + 0.133 328 030 022 248 693 76;
  • 41) 0.133 328 030 022 248 693 76 × 2 = 0 + 0.266 656 060 044 497 387 52;
  • 42) 0.266 656 060 044 497 387 52 × 2 = 0 + 0.533 312 120 088 994 775 04;
  • 43) 0.533 312 120 088 994 775 04 × 2 = 1 + 0.066 624 240 177 989 550 08;
  • 44) 0.066 624 240 177 989 550 08 × 2 = 0 + 0.133 248 480 355 979 100 16;
  • 45) 0.133 248 480 355 979 100 16 × 2 = 0 + 0.266 496 960 711 958 200 32;
  • 46) 0.266 496 960 711 958 200 32 × 2 = 0 + 0.532 993 921 423 916 400 64;
  • 47) 0.532 993 921 423 916 400 64 × 2 = 1 + 0.065 987 842 847 832 801 28;
  • 48) 0.065 987 842 847 832 801 28 × 2 = 0 + 0.131 975 685 695 665 602 56;
  • 49) 0.131 975 685 695 665 602 56 × 2 = 0 + 0.263 951 371 391 331 205 12;
  • 50) 0.263 951 371 391 331 205 12 × 2 = 0 + 0.527 902 742 782 662 410 24;
  • 51) 0.527 902 742 782 662 410 24 × 2 = 1 + 0.055 805 485 565 324 820 48;
  • 52) 0.055 805 485 565 324 820 48 × 2 = 0 + 0.111 610 971 130 649 640 96;
  • 53) 0.111 610 971 130 649 640 96 × 2 = 0 + 0.223 221 942 261 299 281 92;

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:


0.133 333 333 333 333 328 51(10) =

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

5. Positive number before normalization:

3.133 333 333 333 333 328 51(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:


3.133 333 333 333 333 328 51(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 3.133 333 333 333 333 328 51 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

Share

» Convert decimal 3.133 333 333 333 333 328 22 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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