3.141 592 653 589 793 238 462 019 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 019(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
3.141 592 653 589 793 238 462 019(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: 0.141 592 653 589 793 238 462 019.

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) 0.141 592 653 589 793 238 462 019 × 2 = 0 + 0.283 185 307 179 586 476 924 038;
  • 2) 0.283 185 307 179 586 476 924 038 × 2 = 0 + 0.566 370 614 359 172 953 848 076;
  • 3) 0.566 370 614 359 172 953 848 076 × 2 = 1 + 0.132 741 228 718 345 907 696 152;
  • 4) 0.132 741 228 718 345 907 696 152 × 2 = 0 + 0.265 482 457 436 691 815 392 304;
  • 5) 0.265 482 457 436 691 815 392 304 × 2 = 0 + 0.530 964 914 873 383 630 784 608;
  • 6) 0.530 964 914 873 383 630 784 608 × 2 = 1 + 0.061 929 829 746 767 261 569 216;
  • 7) 0.061 929 829 746 767 261 569 216 × 2 = 0 + 0.123 859 659 493 534 523 138 432;
  • 8) 0.123 859 659 493 534 523 138 432 × 2 = 0 + 0.247 719 318 987 069 046 276 864;
  • 9) 0.247 719 318 987 069 046 276 864 × 2 = 0 + 0.495 438 637 974 138 092 553 728;
  • 10) 0.495 438 637 974 138 092 553 728 × 2 = 0 + 0.990 877 275 948 276 185 107 456;
  • 11) 0.990 877 275 948 276 185 107 456 × 2 = 1 + 0.981 754 551 896 552 370 214 912;
  • 12) 0.981 754 551 896 552 370 214 912 × 2 = 1 + 0.963 509 103 793 104 740 429 824;
  • 13) 0.963 509 103 793 104 740 429 824 × 2 = 1 + 0.927 018 207 586 209 480 859 648;
  • 14) 0.927 018 207 586 209 480 859 648 × 2 = 1 + 0.854 036 415 172 418 961 719 296;
  • 15) 0.854 036 415 172 418 961 719 296 × 2 = 1 + 0.708 072 830 344 837 923 438 592;
  • 16) 0.708 072 830 344 837 923 438 592 × 2 = 1 + 0.416 145 660 689 675 846 877 184;
  • 17) 0.416 145 660 689 675 846 877 184 × 2 = 0 + 0.832 291 321 379 351 693 754 368;
  • 18) 0.832 291 321 379 351 693 754 368 × 2 = 1 + 0.664 582 642 758 703 387 508 736;
  • 19) 0.664 582 642 758 703 387 508 736 × 2 = 1 + 0.329 165 285 517 406 775 017 472;
  • 20) 0.329 165 285 517 406 775 017 472 × 2 = 0 + 0.658 330 571 034 813 550 034 944;
  • 21) 0.658 330 571 034 813 550 034 944 × 2 = 1 + 0.316 661 142 069 627 100 069 888;
  • 22) 0.316 661 142 069 627 100 069 888 × 2 = 0 + 0.633 322 284 139 254 200 139 776;
  • 23) 0.633 322 284 139 254 200 139 776 × 2 = 1 + 0.266 644 568 278 508 400 279 552;
  • 24) 0.266 644 568 278 508 400 279 552 × 2 = 0 + 0.533 289 136 557 016 800 559 104;
  • 25) 0.533 289 136 557 016 800 559 104 × 2 = 1 + 0.066 578 273 114 033 601 118 208;
  • 26) 0.066 578 273 114 033 601 118 208 × 2 = 0 + 0.133 156 546 228 067 202 236 416;
  • 27) 0.133 156 546 228 067 202 236 416 × 2 = 0 + 0.266 313 092 456 134 404 472 832;
  • 28) 0.266 313 092 456 134 404 472 832 × 2 = 0 + 0.532 626 184 912 268 808 945 664;
  • 29) 0.532 626 184 912 268 808 945 664 × 2 = 1 + 0.065 252 369 824 537 617 891 328;
  • 30) 0.065 252 369 824 537 617 891 328 × 2 = 0 + 0.130 504 739 649 075 235 782 656;
  • 31) 0.130 504 739 649 075 235 782 656 × 2 = 0 + 0.261 009 479 298 150 471 565 312;
  • 32) 0.261 009 479 298 150 471 565 312 × 2 = 0 + 0.522 018 958 596 300 943 130 624;
  • 33) 0.522 018 958 596 300 943 130 624 × 2 = 1 + 0.044 037 917 192 601 886 261 248;
  • 34) 0.044 037 917 192 601 886 261 248 × 2 = 0 + 0.088 075 834 385 203 772 522 496;
  • 35) 0.088 075 834 385 203 772 522 496 × 2 = 0 + 0.176 151 668 770 407 545 044 992;
  • 36) 0.176 151 668 770 407 545 044 992 × 2 = 0 + 0.352 303 337 540 815 090 089 984;
  • 37) 0.352 303 337 540 815 090 089 984 × 2 = 0 + 0.704 606 675 081 630 180 179 968;
  • 38) 0.704 606 675 081 630 180 179 968 × 2 = 1 + 0.409 213 350 163 260 360 359 936;
  • 39) 0.409 213 350 163 260 360 359 936 × 2 = 0 + 0.818 426 700 326 520 720 719 872;
  • 40) 0.818 426 700 326 520 720 719 872 × 2 = 1 + 0.636 853 400 653 041 441 439 744;
  • 41) 0.636 853 400 653 041 441 439 744 × 2 = 1 + 0.273 706 801 306 082 882 879 488;
  • 42) 0.273 706 801 306 082 882 879 488 × 2 = 0 + 0.547 413 602 612 165 765 758 976;
  • 43) 0.547 413 602 612 165 765 758 976 × 2 = 1 + 0.094 827 205 224 331 531 517 952;
  • 44) 0.094 827 205 224 331 531 517 952 × 2 = 0 + 0.189 654 410 448 663 063 035 904;
  • 45) 0.189 654 410 448 663 063 035 904 × 2 = 0 + 0.379 308 820 897 326 126 071 808;
  • 46) 0.379 308 820 897 326 126 071 808 × 2 = 0 + 0.758 617 641 794 652 252 143 616;
  • 47) 0.758 617 641 794 652 252 143 616 × 2 = 1 + 0.517 235 283 589 304 504 287 232;
  • 48) 0.517 235 283 589 304 504 287 232 × 2 = 1 + 0.034 470 567 178 609 008 574 464;
  • 49) 0.034 470 567 178 609 008 574 464 × 2 = 0 + 0.068 941 134 357 218 017 148 928;
  • 50) 0.068 941 134 357 218 017 148 928 × 2 = 0 + 0.137 882 268 714 436 034 297 856;
  • 51) 0.137 882 268 714 436 034 297 856 × 2 = 0 + 0.275 764 537 428 872 068 595 712;
  • 52) 0.275 764 537 428 872 068 595 712 × 2 = 0 + 0.551 529 074 857 744 137 191 424;
  • 53) 0.551 529 074 857 744 137 191 424 × 2 = 1 + 0.103 058 149 715 488 274 382 848;

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:


0.141 592 653 589 793 238 462 019(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 019(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:


3.141 592 653 589 793 238 462 019(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 3.141 592 653 589 793 238 462 019 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