3.141 592 653 689 788 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 788 5(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
3.141 592 653 689 788 5(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: 0.141 592 653 689 788 5.

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) 0.141 592 653 689 788 5 × 2 = 0 + 0.283 185 307 379 577;
  • 2) 0.283 185 307 379 577 × 2 = 0 + 0.566 370 614 759 154;
  • 3) 0.566 370 614 759 154 × 2 = 1 + 0.132 741 229 518 308;
  • 4) 0.132 741 229 518 308 × 2 = 0 + 0.265 482 459 036 616;
  • 5) 0.265 482 459 036 616 × 2 = 0 + 0.530 964 918 073 232;
  • 6) 0.530 964 918 073 232 × 2 = 1 + 0.061 929 836 146 464;
  • 7) 0.061 929 836 146 464 × 2 = 0 + 0.123 859 672 292 928;
  • 8) 0.123 859 672 292 928 × 2 = 0 + 0.247 719 344 585 856;
  • 9) 0.247 719 344 585 856 × 2 = 0 + 0.495 438 689 171 712;
  • 10) 0.495 438 689 171 712 × 2 = 0 + 0.990 877 378 343 424;
  • 11) 0.990 877 378 343 424 × 2 = 1 + 0.981 754 756 686 848;
  • 12) 0.981 754 756 686 848 × 2 = 1 + 0.963 509 513 373 696;
  • 13) 0.963 509 513 373 696 × 2 = 1 + 0.927 019 026 747 392;
  • 14) 0.927 019 026 747 392 × 2 = 1 + 0.854 038 053 494 784;
  • 15) 0.854 038 053 494 784 × 2 = 1 + 0.708 076 106 989 568;
  • 16) 0.708 076 106 989 568 × 2 = 1 + 0.416 152 213 979 136;
  • 17) 0.416 152 213 979 136 × 2 = 0 + 0.832 304 427 958 272;
  • 18) 0.832 304 427 958 272 × 2 = 1 + 0.664 608 855 916 544;
  • 19) 0.664 608 855 916 544 × 2 = 1 + 0.329 217 711 833 088;
  • 20) 0.329 217 711 833 088 × 2 = 0 + 0.658 435 423 666 176;
  • 21) 0.658 435 423 666 176 × 2 = 1 + 0.316 870 847 332 352;
  • 22) 0.316 870 847 332 352 × 2 = 0 + 0.633 741 694 664 704;
  • 23) 0.633 741 694 664 704 × 2 = 1 + 0.267 483 389 329 408;
  • 24) 0.267 483 389 329 408 × 2 = 0 + 0.534 966 778 658 816;
  • 25) 0.534 966 778 658 816 × 2 = 1 + 0.069 933 557 317 632;
  • 26) 0.069 933 557 317 632 × 2 = 0 + 0.139 867 114 635 264;
  • 27) 0.139 867 114 635 264 × 2 = 0 + 0.279 734 229 270 528;
  • 28) 0.279 734 229 270 528 × 2 = 0 + 0.559 468 458 541 056;
  • 29) 0.559 468 458 541 056 × 2 = 1 + 0.118 936 917 082 112;
  • 30) 0.118 936 917 082 112 × 2 = 0 + 0.237 873 834 164 224;
  • 31) 0.237 873 834 164 224 × 2 = 0 + 0.475 747 668 328 448;
  • 32) 0.475 747 668 328 448 × 2 = 0 + 0.951 495 336 656 896;
  • 33) 0.951 495 336 656 896 × 2 = 1 + 0.902 990 673 313 792;
  • 34) 0.902 990 673 313 792 × 2 = 1 + 0.805 981 346 627 584;
  • 35) 0.805 981 346 627 584 × 2 = 1 + 0.611 962 693 255 168;
  • 36) 0.611 962 693 255 168 × 2 = 1 + 0.223 925 386 510 336;
  • 37) 0.223 925 386 510 336 × 2 = 0 + 0.447 850 773 020 672;
  • 38) 0.447 850 773 020 672 × 2 = 0 + 0.895 701 546 041 344;
  • 39) 0.895 701 546 041 344 × 2 = 1 + 0.791 403 092 082 688;
  • 40) 0.791 403 092 082 688 × 2 = 1 + 0.582 806 184 165 376;
  • 41) 0.582 806 184 165 376 × 2 = 1 + 0.165 612 368 330 752;
  • 42) 0.165 612 368 330 752 × 2 = 0 + 0.331 224 736 661 504;
  • 43) 0.331 224 736 661 504 × 2 = 0 + 0.662 449 473 323 008;
  • 44) 0.662 449 473 323 008 × 2 = 1 + 0.324 898 946 646 016;
  • 45) 0.324 898 946 646 016 × 2 = 0 + 0.649 797 893 292 032;
  • 46) 0.649 797 893 292 032 × 2 = 1 + 0.299 595 786 584 064;
  • 47) 0.299 595 786 584 064 × 2 = 0 + 0.599 191 573 168 128;
  • 48) 0.599 191 573 168 128 × 2 = 1 + 0.198 383 146 336 256;
  • 49) 0.198 383 146 336 256 × 2 = 0 + 0.396 766 292 672 512;
  • 50) 0.396 766 292 672 512 × 2 = 0 + 0.793 532 585 345 024;
  • 51) 0.793 532 585 345 024 × 2 = 1 + 0.587 065 170 690 048;
  • 52) 0.587 065 170 690 048 × 2 = 1 + 0.174 130 341 380 096;
  • 53) 0.174 130 341 380 096 × 2 = 0 + 0.348 260 682 760 192;

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:


0.141 592 653 689 788 5(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0101 0011 0(2)

5. Positive number before normalization:

3.141 592 653 689 788 5(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0101 0011 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:


3.141 592 653 689 788 5(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0101 0011 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0101 0011 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1010 1001 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 1010 1001 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 1010 1001 10 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1010 1001


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 1010 1001


Decimal number 3.141 592 653 689 788 5 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 1010 1001

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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