3.141 592 653 689 793 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 793 4(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 4(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 4.

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 4 × 2 = 0 + 0.283 185 307 379 586 8;
  • 2) 0.283 185 307 379 586 8 × 2 = 0 + 0.566 370 614 759 173 6;
  • 3) 0.566 370 614 759 173 6 × 2 = 1 + 0.132 741 229 518 347 2;
  • 4) 0.132 741 229 518 347 2 × 2 = 0 + 0.265 482 459 036 694 4;
  • 5) 0.265 482 459 036 694 4 × 2 = 0 + 0.530 964 918 073 388 8;
  • 6) 0.530 964 918 073 388 8 × 2 = 1 + 0.061 929 836 146 777 6;
  • 7) 0.061 929 836 146 777 6 × 2 = 0 + 0.123 859 672 293 555 2;
  • 8) 0.123 859 672 293 555 2 × 2 = 0 + 0.247 719 344 587 110 4;
  • 9) 0.247 719 344 587 110 4 × 2 = 0 + 0.495 438 689 174 220 8;
  • 10) 0.495 438 689 174 220 8 × 2 = 0 + 0.990 877 378 348 441 6;
  • 11) 0.990 877 378 348 441 6 × 2 = 1 + 0.981 754 756 696 883 2;
  • 12) 0.981 754 756 696 883 2 × 2 = 1 + 0.963 509 513 393 766 4;
  • 13) 0.963 509 513 393 766 4 × 2 = 1 + 0.927 019 026 787 532 8;
  • 14) 0.927 019 026 787 532 8 × 2 = 1 + 0.854 038 053 575 065 6;
  • 15) 0.854 038 053 575 065 6 × 2 = 1 + 0.708 076 107 150 131 2;
  • 16) 0.708 076 107 150 131 2 × 2 = 1 + 0.416 152 214 300 262 4;
  • 17) 0.416 152 214 300 262 4 × 2 = 0 + 0.832 304 428 600 524 8;
  • 18) 0.832 304 428 600 524 8 × 2 = 1 + 0.664 608 857 201 049 6;
  • 19) 0.664 608 857 201 049 6 × 2 = 1 + 0.329 217 714 402 099 2;
  • 20) 0.329 217 714 402 099 2 × 2 = 0 + 0.658 435 428 804 198 4;
  • 21) 0.658 435 428 804 198 4 × 2 = 1 + 0.316 870 857 608 396 8;
  • 22) 0.316 870 857 608 396 8 × 2 = 0 + 0.633 741 715 216 793 6;
  • 23) 0.633 741 715 216 793 6 × 2 = 1 + 0.267 483 430 433 587 2;
  • 24) 0.267 483 430 433 587 2 × 2 = 0 + 0.534 966 860 867 174 4;
  • 25) 0.534 966 860 867 174 4 × 2 = 1 + 0.069 933 721 734 348 8;
  • 26) 0.069 933 721 734 348 8 × 2 = 0 + 0.139 867 443 468 697 6;
  • 27) 0.139 867 443 468 697 6 × 2 = 0 + 0.279 734 886 937 395 2;
  • 28) 0.279 734 886 937 395 2 × 2 = 0 + 0.559 469 773 874 790 4;
  • 29) 0.559 469 773 874 790 4 × 2 = 1 + 0.118 939 547 749 580 8;
  • 30) 0.118 939 547 749 580 8 × 2 = 0 + 0.237 879 095 499 161 6;
  • 31) 0.237 879 095 499 161 6 × 2 = 0 + 0.475 758 190 998 323 2;
  • 32) 0.475 758 190 998 323 2 × 2 = 0 + 0.951 516 381 996 646 4;
  • 33) 0.951 516 381 996 646 4 × 2 = 1 + 0.903 032 763 993 292 8;
  • 34) 0.903 032 763 993 292 8 × 2 = 1 + 0.806 065 527 986 585 6;
  • 35) 0.806 065 527 986 585 6 × 2 = 1 + 0.612 131 055 973 171 2;
  • 36) 0.612 131 055 973 171 2 × 2 = 1 + 0.224 262 111 946 342 4;
  • 37) 0.224 262 111 946 342 4 × 2 = 0 + 0.448 524 223 892 684 8;
  • 38) 0.448 524 223 892 684 8 × 2 = 0 + 0.897 048 447 785 369 6;
  • 39) 0.897 048 447 785 369 6 × 2 = 1 + 0.794 096 895 570 739 2;
  • 40) 0.794 096 895 570 739 2 × 2 = 1 + 0.588 193 791 141 478 4;
  • 41) 0.588 193 791 141 478 4 × 2 = 1 + 0.176 387 582 282 956 8;
  • 42) 0.176 387 582 282 956 8 × 2 = 0 + 0.352 775 164 565 913 6;
  • 43) 0.352 775 164 565 913 6 × 2 = 0 + 0.705 550 329 131 827 2;
  • 44) 0.705 550 329 131 827 2 × 2 = 1 + 0.411 100 658 263 654 4;
  • 45) 0.411 100 658 263 654 4 × 2 = 0 + 0.822 201 316 527 308 8;
  • 46) 0.822 201 316 527 308 8 × 2 = 1 + 0.644 402 633 054 617 6;
  • 47) 0.644 402 633 054 617 6 × 2 = 1 + 0.288 805 266 109 235 2;
  • 48) 0.288 805 266 109 235 2 × 2 = 0 + 0.577 610 532 218 470 4;
  • 49) 0.577 610 532 218 470 4 × 2 = 1 + 0.155 221 064 436 940 8;
  • 50) 0.155 221 064 436 940 8 × 2 = 0 + 0.310 442 128 873 881 6;
  • 51) 0.310 442 128 873 881 6 × 2 = 0 + 0.620 884 257 747 763 2;
  • 52) 0.620 884 257 747 763 2 × 2 = 1 + 0.241 768 515 495 526 4;
  • 53) 0.241 768 515 495 526 4 × 2 = 0 + 0.483 537 030 991 052 8;

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 4(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 4(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 4(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 4 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

» Convert decimal 3.141 592 653 689 802 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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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