3.141 592 653 589 793 238 461 94 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 461 94(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
3.141 592 653 589 793 238 461 94(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: 0.141 592 653 589 793 238 461 94.

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) 0.141 592 653 589 793 238 461 94 × 2 = 0 + 0.283 185 307 179 586 476 923 88;
  • 2) 0.283 185 307 179 586 476 923 88 × 2 = 0 + 0.566 370 614 359 172 953 847 76;
  • 3) 0.566 370 614 359 172 953 847 76 × 2 = 1 + 0.132 741 228 718 345 907 695 52;
  • 4) 0.132 741 228 718 345 907 695 52 × 2 = 0 + 0.265 482 457 436 691 815 391 04;
  • 5) 0.265 482 457 436 691 815 391 04 × 2 = 0 + 0.530 964 914 873 383 630 782 08;
  • 6) 0.530 964 914 873 383 630 782 08 × 2 = 1 + 0.061 929 829 746 767 261 564 16;
  • 7) 0.061 929 829 746 767 261 564 16 × 2 = 0 + 0.123 859 659 493 534 523 128 32;
  • 8) 0.123 859 659 493 534 523 128 32 × 2 = 0 + 0.247 719 318 987 069 046 256 64;
  • 9) 0.247 719 318 987 069 046 256 64 × 2 = 0 + 0.495 438 637 974 138 092 513 28;
  • 10) 0.495 438 637 974 138 092 513 28 × 2 = 0 + 0.990 877 275 948 276 185 026 56;
  • 11) 0.990 877 275 948 276 185 026 56 × 2 = 1 + 0.981 754 551 896 552 370 053 12;
  • 12) 0.981 754 551 896 552 370 053 12 × 2 = 1 + 0.963 509 103 793 104 740 106 24;
  • 13) 0.963 509 103 793 104 740 106 24 × 2 = 1 + 0.927 018 207 586 209 480 212 48;
  • 14) 0.927 018 207 586 209 480 212 48 × 2 = 1 + 0.854 036 415 172 418 960 424 96;
  • 15) 0.854 036 415 172 418 960 424 96 × 2 = 1 + 0.708 072 830 344 837 920 849 92;
  • 16) 0.708 072 830 344 837 920 849 92 × 2 = 1 + 0.416 145 660 689 675 841 699 84;
  • 17) 0.416 145 660 689 675 841 699 84 × 2 = 0 + 0.832 291 321 379 351 683 399 68;
  • 18) 0.832 291 321 379 351 683 399 68 × 2 = 1 + 0.664 582 642 758 703 366 799 36;
  • 19) 0.664 582 642 758 703 366 799 36 × 2 = 1 + 0.329 165 285 517 406 733 598 72;
  • 20) 0.329 165 285 517 406 733 598 72 × 2 = 0 + 0.658 330 571 034 813 467 197 44;
  • 21) 0.658 330 571 034 813 467 197 44 × 2 = 1 + 0.316 661 142 069 626 934 394 88;
  • 22) 0.316 661 142 069 626 934 394 88 × 2 = 0 + 0.633 322 284 139 253 868 789 76;
  • 23) 0.633 322 284 139 253 868 789 76 × 2 = 1 + 0.266 644 568 278 507 737 579 52;
  • 24) 0.266 644 568 278 507 737 579 52 × 2 = 0 + 0.533 289 136 557 015 475 159 04;
  • 25) 0.533 289 136 557 015 475 159 04 × 2 = 1 + 0.066 578 273 114 030 950 318 08;
  • 26) 0.066 578 273 114 030 950 318 08 × 2 = 0 + 0.133 156 546 228 061 900 636 16;
  • 27) 0.133 156 546 228 061 900 636 16 × 2 = 0 + 0.266 313 092 456 123 801 272 32;
  • 28) 0.266 313 092 456 123 801 272 32 × 2 = 0 + 0.532 626 184 912 247 602 544 64;
  • 29) 0.532 626 184 912 247 602 544 64 × 2 = 1 + 0.065 252 369 824 495 205 089 28;
  • 30) 0.065 252 369 824 495 205 089 28 × 2 = 0 + 0.130 504 739 648 990 410 178 56;
  • 31) 0.130 504 739 648 990 410 178 56 × 2 = 0 + 0.261 009 479 297 980 820 357 12;
  • 32) 0.261 009 479 297 980 820 357 12 × 2 = 0 + 0.522 018 958 595 961 640 714 24;
  • 33) 0.522 018 958 595 961 640 714 24 × 2 = 1 + 0.044 037 917 191 923 281 428 48;
  • 34) 0.044 037 917 191 923 281 428 48 × 2 = 0 + 0.088 075 834 383 846 562 856 96;
  • 35) 0.088 075 834 383 846 562 856 96 × 2 = 0 + 0.176 151 668 767 693 125 713 92;
  • 36) 0.176 151 668 767 693 125 713 92 × 2 = 0 + 0.352 303 337 535 386 251 427 84;
  • 37) 0.352 303 337 535 386 251 427 84 × 2 = 0 + 0.704 606 675 070 772 502 855 68;
  • 38) 0.704 606 675 070 772 502 855 68 × 2 = 1 + 0.409 213 350 141 545 005 711 36;
  • 39) 0.409 213 350 141 545 005 711 36 × 2 = 0 + 0.818 426 700 283 090 011 422 72;
  • 40) 0.818 426 700 283 090 011 422 72 × 2 = 1 + 0.636 853 400 566 180 022 845 44;
  • 41) 0.636 853 400 566 180 022 845 44 × 2 = 1 + 0.273 706 801 132 360 045 690 88;
  • 42) 0.273 706 801 132 360 045 690 88 × 2 = 0 + 0.547 413 602 264 720 091 381 76;
  • 43) 0.547 413 602 264 720 091 381 76 × 2 = 1 + 0.094 827 204 529 440 182 763 52;
  • 44) 0.094 827 204 529 440 182 763 52 × 2 = 0 + 0.189 654 409 058 880 365 527 04;
  • 45) 0.189 654 409 058 880 365 527 04 × 2 = 0 + 0.379 308 818 117 760 731 054 08;
  • 46) 0.379 308 818 117 760 731 054 08 × 2 = 0 + 0.758 617 636 235 521 462 108 16;
  • 47) 0.758 617 636 235 521 462 108 16 × 2 = 1 + 0.517 235 272 471 042 924 216 32;
  • 48) 0.517 235 272 471 042 924 216 32 × 2 = 1 + 0.034 470 544 942 085 848 432 64;
  • 49) 0.034 470 544 942 085 848 432 64 × 2 = 0 + 0.068 941 089 884 171 696 865 28;
  • 50) 0.068 941 089 884 171 696 865 28 × 2 = 0 + 0.137 882 179 768 343 393 730 56;
  • 51) 0.137 882 179 768 343 393 730 56 × 2 = 0 + 0.275 764 359 536 686 787 461 12;
  • 52) 0.275 764 359 536 686 787 461 12 × 2 = 0 + 0.551 528 719 073 373 574 922 24;
  • 53) 0.551 528 719 073 373 574 922 24 × 2 = 1 + 0.103 057 438 146 747 149 844 48;

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:


0.141 592 653 589 793 238 461 94(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 461 94(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 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:


3.141 592 653 589 793 238 461 94(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 01(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 0100 0010 1101 0001 1000 01


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 0100 0010 1101 0001 1000 01 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


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 0100 0010 1101 0001 1000


Decimal number 3.141 592 653 589 793 238 461 94 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

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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