3.141 592 653 589 793 238 462 062 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 062(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
3.141 592 653 589 793 238 462 062(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: 0.141 592 653 589 793 238 462 062.

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) 0.141 592 653 589 793 238 462 062 × 2 = 0 + 0.283 185 307 179 586 476 924 124;
  • 2) 0.283 185 307 179 586 476 924 124 × 2 = 0 + 0.566 370 614 359 172 953 848 248;
  • 3) 0.566 370 614 359 172 953 848 248 × 2 = 1 + 0.132 741 228 718 345 907 696 496;
  • 4) 0.132 741 228 718 345 907 696 496 × 2 = 0 + 0.265 482 457 436 691 815 392 992;
  • 5) 0.265 482 457 436 691 815 392 992 × 2 = 0 + 0.530 964 914 873 383 630 785 984;
  • 6) 0.530 964 914 873 383 630 785 984 × 2 = 1 + 0.061 929 829 746 767 261 571 968;
  • 7) 0.061 929 829 746 767 261 571 968 × 2 = 0 + 0.123 859 659 493 534 523 143 936;
  • 8) 0.123 859 659 493 534 523 143 936 × 2 = 0 + 0.247 719 318 987 069 046 287 872;
  • 9) 0.247 719 318 987 069 046 287 872 × 2 = 0 + 0.495 438 637 974 138 092 575 744;
  • 10) 0.495 438 637 974 138 092 575 744 × 2 = 0 + 0.990 877 275 948 276 185 151 488;
  • 11) 0.990 877 275 948 276 185 151 488 × 2 = 1 + 0.981 754 551 896 552 370 302 976;
  • 12) 0.981 754 551 896 552 370 302 976 × 2 = 1 + 0.963 509 103 793 104 740 605 952;
  • 13) 0.963 509 103 793 104 740 605 952 × 2 = 1 + 0.927 018 207 586 209 481 211 904;
  • 14) 0.927 018 207 586 209 481 211 904 × 2 = 1 + 0.854 036 415 172 418 962 423 808;
  • 15) 0.854 036 415 172 418 962 423 808 × 2 = 1 + 0.708 072 830 344 837 924 847 616;
  • 16) 0.708 072 830 344 837 924 847 616 × 2 = 1 + 0.416 145 660 689 675 849 695 232;
  • 17) 0.416 145 660 689 675 849 695 232 × 2 = 0 + 0.832 291 321 379 351 699 390 464;
  • 18) 0.832 291 321 379 351 699 390 464 × 2 = 1 + 0.664 582 642 758 703 398 780 928;
  • 19) 0.664 582 642 758 703 398 780 928 × 2 = 1 + 0.329 165 285 517 406 797 561 856;
  • 20) 0.329 165 285 517 406 797 561 856 × 2 = 0 + 0.658 330 571 034 813 595 123 712;
  • 21) 0.658 330 571 034 813 595 123 712 × 2 = 1 + 0.316 661 142 069 627 190 247 424;
  • 22) 0.316 661 142 069 627 190 247 424 × 2 = 0 + 0.633 322 284 139 254 380 494 848;
  • 23) 0.633 322 284 139 254 380 494 848 × 2 = 1 + 0.266 644 568 278 508 760 989 696;
  • 24) 0.266 644 568 278 508 760 989 696 × 2 = 0 + 0.533 289 136 557 017 521 979 392;
  • 25) 0.533 289 136 557 017 521 979 392 × 2 = 1 + 0.066 578 273 114 035 043 958 784;
  • 26) 0.066 578 273 114 035 043 958 784 × 2 = 0 + 0.133 156 546 228 070 087 917 568;
  • 27) 0.133 156 546 228 070 087 917 568 × 2 = 0 + 0.266 313 092 456 140 175 835 136;
  • 28) 0.266 313 092 456 140 175 835 136 × 2 = 0 + 0.532 626 184 912 280 351 670 272;
  • 29) 0.532 626 184 912 280 351 670 272 × 2 = 1 + 0.065 252 369 824 560 703 340 544;
  • 30) 0.065 252 369 824 560 703 340 544 × 2 = 0 + 0.130 504 739 649 121 406 681 088;
  • 31) 0.130 504 739 649 121 406 681 088 × 2 = 0 + 0.261 009 479 298 242 813 362 176;
  • 32) 0.261 009 479 298 242 813 362 176 × 2 = 0 + 0.522 018 958 596 485 626 724 352;
  • 33) 0.522 018 958 596 485 626 724 352 × 2 = 1 + 0.044 037 917 192 971 253 448 704;
  • 34) 0.044 037 917 192 971 253 448 704 × 2 = 0 + 0.088 075 834 385 942 506 897 408;
  • 35) 0.088 075 834 385 942 506 897 408 × 2 = 0 + 0.176 151 668 771 885 013 794 816;
  • 36) 0.176 151 668 771 885 013 794 816 × 2 = 0 + 0.352 303 337 543 770 027 589 632;
  • 37) 0.352 303 337 543 770 027 589 632 × 2 = 0 + 0.704 606 675 087 540 055 179 264;
  • 38) 0.704 606 675 087 540 055 179 264 × 2 = 1 + 0.409 213 350 175 080 110 358 528;
  • 39) 0.409 213 350 175 080 110 358 528 × 2 = 0 + 0.818 426 700 350 160 220 717 056;
  • 40) 0.818 426 700 350 160 220 717 056 × 2 = 1 + 0.636 853 400 700 320 441 434 112;
  • 41) 0.636 853 400 700 320 441 434 112 × 2 = 1 + 0.273 706 801 400 640 882 868 224;
  • 42) 0.273 706 801 400 640 882 868 224 × 2 = 0 + 0.547 413 602 801 281 765 736 448;
  • 43) 0.547 413 602 801 281 765 736 448 × 2 = 1 + 0.094 827 205 602 563 531 472 896;
  • 44) 0.094 827 205 602 563 531 472 896 × 2 = 0 + 0.189 654 411 205 127 062 945 792;
  • 45) 0.189 654 411 205 127 062 945 792 × 2 = 0 + 0.379 308 822 410 254 125 891 584;
  • 46) 0.379 308 822 410 254 125 891 584 × 2 = 0 + 0.758 617 644 820 508 251 783 168;
  • 47) 0.758 617 644 820 508 251 783 168 × 2 = 1 + 0.517 235 289 641 016 503 566 336;
  • 48) 0.517 235 289 641 016 503 566 336 × 2 = 1 + 0.034 470 579 282 033 007 132 672;
  • 49) 0.034 470 579 282 033 007 132 672 × 2 = 0 + 0.068 941 158 564 066 014 265 344;
  • 50) 0.068 941 158 564 066 014 265 344 × 2 = 0 + 0.137 882 317 128 132 028 530 688;
  • 51) 0.137 882 317 128 132 028 530 688 × 2 = 0 + 0.275 764 634 256 264 057 061 376;
  • 52) 0.275 764 634 256 264 057 061 376 × 2 = 0 + 0.551 529 268 512 528 114 122 752;
  • 53) 0.551 529 268 512 528 114 122 752 × 2 = 1 + 0.103 058 537 025 056 228 245 504;

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:


0.141 592 653 589 793 238 462 062(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 062(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:


3.141 592 653 589 793 238 462 062(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 3.141 592 653 589 793 238 462 062 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