3.141 592 653 689 793 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 793 09(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 09(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 09.

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 09 × 2 = 0 + 0.283 185 307 379 586 18;
  • 2) 0.283 185 307 379 586 18 × 2 = 0 + 0.566 370 614 759 172 36;
  • 3) 0.566 370 614 759 172 36 × 2 = 1 + 0.132 741 229 518 344 72;
  • 4) 0.132 741 229 518 344 72 × 2 = 0 + 0.265 482 459 036 689 44;
  • 5) 0.265 482 459 036 689 44 × 2 = 0 + 0.530 964 918 073 378 88;
  • 6) 0.530 964 918 073 378 88 × 2 = 1 + 0.061 929 836 146 757 76;
  • 7) 0.061 929 836 146 757 76 × 2 = 0 + 0.123 859 672 293 515 52;
  • 8) 0.123 859 672 293 515 52 × 2 = 0 + 0.247 719 344 587 031 04;
  • 9) 0.247 719 344 587 031 04 × 2 = 0 + 0.495 438 689 174 062 08;
  • 10) 0.495 438 689 174 062 08 × 2 = 0 + 0.990 877 378 348 124 16;
  • 11) 0.990 877 378 348 124 16 × 2 = 1 + 0.981 754 756 696 248 32;
  • 12) 0.981 754 756 696 248 32 × 2 = 1 + 0.963 509 513 392 496 64;
  • 13) 0.963 509 513 392 496 64 × 2 = 1 + 0.927 019 026 784 993 28;
  • 14) 0.927 019 026 784 993 28 × 2 = 1 + 0.854 038 053 569 986 56;
  • 15) 0.854 038 053 569 986 56 × 2 = 1 + 0.708 076 107 139 973 12;
  • 16) 0.708 076 107 139 973 12 × 2 = 1 + 0.416 152 214 279 946 24;
  • 17) 0.416 152 214 279 946 24 × 2 = 0 + 0.832 304 428 559 892 48;
  • 18) 0.832 304 428 559 892 48 × 2 = 1 + 0.664 608 857 119 784 96;
  • 19) 0.664 608 857 119 784 96 × 2 = 1 + 0.329 217 714 239 569 92;
  • 20) 0.329 217 714 239 569 92 × 2 = 0 + 0.658 435 428 479 139 84;
  • 21) 0.658 435 428 479 139 84 × 2 = 1 + 0.316 870 856 958 279 68;
  • 22) 0.316 870 856 958 279 68 × 2 = 0 + 0.633 741 713 916 559 36;
  • 23) 0.633 741 713 916 559 36 × 2 = 1 + 0.267 483 427 833 118 72;
  • 24) 0.267 483 427 833 118 72 × 2 = 0 + 0.534 966 855 666 237 44;
  • 25) 0.534 966 855 666 237 44 × 2 = 1 + 0.069 933 711 332 474 88;
  • 26) 0.069 933 711 332 474 88 × 2 = 0 + 0.139 867 422 664 949 76;
  • 27) 0.139 867 422 664 949 76 × 2 = 0 + 0.279 734 845 329 899 52;
  • 28) 0.279 734 845 329 899 52 × 2 = 0 + 0.559 469 690 659 799 04;
  • 29) 0.559 469 690 659 799 04 × 2 = 1 + 0.118 939 381 319 598 08;
  • 30) 0.118 939 381 319 598 08 × 2 = 0 + 0.237 878 762 639 196 16;
  • 31) 0.237 878 762 639 196 16 × 2 = 0 + 0.475 757 525 278 392 32;
  • 32) 0.475 757 525 278 392 32 × 2 = 0 + 0.951 515 050 556 784 64;
  • 33) 0.951 515 050 556 784 64 × 2 = 1 + 0.903 030 101 113 569 28;
  • 34) 0.903 030 101 113 569 28 × 2 = 1 + 0.806 060 202 227 138 56;
  • 35) 0.806 060 202 227 138 56 × 2 = 1 + 0.612 120 404 454 277 12;
  • 36) 0.612 120 404 454 277 12 × 2 = 1 + 0.224 240 808 908 554 24;
  • 37) 0.224 240 808 908 554 24 × 2 = 0 + 0.448 481 617 817 108 48;
  • 38) 0.448 481 617 817 108 48 × 2 = 0 + 0.896 963 235 634 216 96;
  • 39) 0.896 963 235 634 216 96 × 2 = 1 + 0.793 926 471 268 433 92;
  • 40) 0.793 926 471 268 433 92 × 2 = 1 + 0.587 852 942 536 867 84;
  • 41) 0.587 852 942 536 867 84 × 2 = 1 + 0.175 705 885 073 735 68;
  • 42) 0.175 705 885 073 735 68 × 2 = 0 + 0.351 411 770 147 471 36;
  • 43) 0.351 411 770 147 471 36 × 2 = 0 + 0.702 823 540 294 942 72;
  • 44) 0.702 823 540 294 942 72 × 2 = 1 + 0.405 647 080 589 885 44;
  • 45) 0.405 647 080 589 885 44 × 2 = 0 + 0.811 294 161 179 770 88;
  • 46) 0.811 294 161 179 770 88 × 2 = 1 + 0.622 588 322 359 541 76;
  • 47) 0.622 588 322 359 541 76 × 2 = 1 + 0.245 176 644 719 083 52;
  • 48) 0.245 176 644 719 083 52 × 2 = 0 + 0.490 353 289 438 167 04;
  • 49) 0.490 353 289 438 167 04 × 2 = 0 + 0.980 706 578 876 334 08;
  • 50) 0.980 706 578 876 334 08 × 2 = 1 + 0.961 413 157 752 668 16;
  • 51) 0.961 413 157 752 668 16 × 2 = 1 + 0.922 826 315 505 336 32;
  • 52) 0.922 826 315 505 336 32 × 2 = 1 + 0.845 652 631 010 672 64;
  • 53) 0.845 652 631 010 672 64 × 2 = 1 + 0.691 305 262 021 345 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 - 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 09(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 0111 1(2)

5. Positive number before normalization:

3.141 592 653 689 793 09(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 0111 1(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 09(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 0111 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0110 0111 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0011 11(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 0011 11


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 0011 11 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 0011


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 0011


Decimal number 3.141 592 653 689 793 09 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 0011

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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