3.141 592 653 689 793 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 793 38(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
3.141 592 653 689 793 38(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: 0.141 592 653 689 793 38.

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) 0.141 592 653 689 793 38 × 2 = 0 + 0.283 185 307 379 586 76;
  • 2) 0.283 185 307 379 586 76 × 2 = 0 + 0.566 370 614 759 173 52;
  • 3) 0.566 370 614 759 173 52 × 2 = 1 + 0.132 741 229 518 347 04;
  • 4) 0.132 741 229 518 347 04 × 2 = 0 + 0.265 482 459 036 694 08;
  • 5) 0.265 482 459 036 694 08 × 2 = 0 + 0.530 964 918 073 388 16;
  • 6) 0.530 964 918 073 388 16 × 2 = 1 + 0.061 929 836 146 776 32;
  • 7) 0.061 929 836 146 776 32 × 2 = 0 + 0.123 859 672 293 552 64;
  • 8) 0.123 859 672 293 552 64 × 2 = 0 + 0.247 719 344 587 105 28;
  • 9) 0.247 719 344 587 105 28 × 2 = 0 + 0.495 438 689 174 210 56;
  • 10) 0.495 438 689 174 210 56 × 2 = 0 + 0.990 877 378 348 421 12;
  • 11) 0.990 877 378 348 421 12 × 2 = 1 + 0.981 754 756 696 842 24;
  • 12) 0.981 754 756 696 842 24 × 2 = 1 + 0.963 509 513 393 684 48;
  • 13) 0.963 509 513 393 684 48 × 2 = 1 + 0.927 019 026 787 368 96;
  • 14) 0.927 019 026 787 368 96 × 2 = 1 + 0.854 038 053 574 737 92;
  • 15) 0.854 038 053 574 737 92 × 2 = 1 + 0.708 076 107 149 475 84;
  • 16) 0.708 076 107 149 475 84 × 2 = 1 + 0.416 152 214 298 951 68;
  • 17) 0.416 152 214 298 951 68 × 2 = 0 + 0.832 304 428 597 903 36;
  • 18) 0.832 304 428 597 903 36 × 2 = 1 + 0.664 608 857 195 806 72;
  • 19) 0.664 608 857 195 806 72 × 2 = 1 + 0.329 217 714 391 613 44;
  • 20) 0.329 217 714 391 613 44 × 2 = 0 + 0.658 435 428 783 226 88;
  • 21) 0.658 435 428 783 226 88 × 2 = 1 + 0.316 870 857 566 453 76;
  • 22) 0.316 870 857 566 453 76 × 2 = 0 + 0.633 741 715 132 907 52;
  • 23) 0.633 741 715 132 907 52 × 2 = 1 + 0.267 483 430 265 815 04;
  • 24) 0.267 483 430 265 815 04 × 2 = 0 + 0.534 966 860 531 630 08;
  • 25) 0.534 966 860 531 630 08 × 2 = 1 + 0.069 933 721 063 260 16;
  • 26) 0.069 933 721 063 260 16 × 2 = 0 + 0.139 867 442 126 520 32;
  • 27) 0.139 867 442 126 520 32 × 2 = 0 + 0.279 734 884 253 040 64;
  • 28) 0.279 734 884 253 040 64 × 2 = 0 + 0.559 469 768 506 081 28;
  • 29) 0.559 469 768 506 081 28 × 2 = 1 + 0.118 939 537 012 162 56;
  • 30) 0.118 939 537 012 162 56 × 2 = 0 + 0.237 879 074 024 325 12;
  • 31) 0.237 879 074 024 325 12 × 2 = 0 + 0.475 758 148 048 650 24;
  • 32) 0.475 758 148 048 650 24 × 2 = 0 + 0.951 516 296 097 300 48;
  • 33) 0.951 516 296 097 300 48 × 2 = 1 + 0.903 032 592 194 600 96;
  • 34) 0.903 032 592 194 600 96 × 2 = 1 + 0.806 065 184 389 201 92;
  • 35) 0.806 065 184 389 201 92 × 2 = 1 + 0.612 130 368 778 403 84;
  • 36) 0.612 130 368 778 403 84 × 2 = 1 + 0.224 260 737 556 807 68;
  • 37) 0.224 260 737 556 807 68 × 2 = 0 + 0.448 521 475 113 615 36;
  • 38) 0.448 521 475 113 615 36 × 2 = 0 + 0.897 042 950 227 230 72;
  • 39) 0.897 042 950 227 230 72 × 2 = 1 + 0.794 085 900 454 461 44;
  • 40) 0.794 085 900 454 461 44 × 2 = 1 + 0.588 171 800 908 922 88;
  • 41) 0.588 171 800 908 922 88 × 2 = 1 + 0.176 343 601 817 845 76;
  • 42) 0.176 343 601 817 845 76 × 2 = 0 + 0.352 687 203 635 691 52;
  • 43) 0.352 687 203 635 691 52 × 2 = 0 + 0.705 374 407 271 383 04;
  • 44) 0.705 374 407 271 383 04 × 2 = 1 + 0.410 748 814 542 766 08;
  • 45) 0.410 748 814 542 766 08 × 2 = 0 + 0.821 497 629 085 532 16;
  • 46) 0.821 497 629 085 532 16 × 2 = 1 + 0.642 995 258 171 064 32;
  • 47) 0.642 995 258 171 064 32 × 2 = 1 + 0.285 990 516 342 128 64;
  • 48) 0.285 990 516 342 128 64 × 2 = 0 + 0.571 981 032 684 257 28;
  • 49) 0.571 981 032 684 257 28 × 2 = 1 + 0.143 962 065 368 514 56;
  • 50) 0.143 962 065 368 514 56 × 2 = 0 + 0.287 924 130 737 029 12;
  • 51) 0.287 924 130 737 029 12 × 2 = 0 + 0.575 848 261 474 058 24;
  • 52) 0.575 848 261 474 058 24 × 2 = 1 + 0.151 696 522 948 116 48;
  • 53) 0.151 696 522 948 116 48 × 2 = 0 + 0.303 393 045 896 232 96;

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:


0.141 592 653 689 793 38(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 1001 0(2)

5. Positive number before normalization:

3.141 592 653 689 793 38(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 1001 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:


3.141 592 653 689 793 38(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 1001 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 1001 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100 10(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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100 10


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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100 10 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100


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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100


Decimal number 3.141 592 653 689 793 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0100

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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