3.141 592 653 589 793 238 462 262 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 262(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
3.141 592 653 589 793 238 462 262(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: 0.141 592 653 589 793 238 462 262.

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) 0.141 592 653 589 793 238 462 262 × 2 = 0 + 0.283 185 307 179 586 476 924 524;
  • 2) 0.283 185 307 179 586 476 924 524 × 2 = 0 + 0.566 370 614 359 172 953 849 048;
  • 3) 0.566 370 614 359 172 953 849 048 × 2 = 1 + 0.132 741 228 718 345 907 698 096;
  • 4) 0.132 741 228 718 345 907 698 096 × 2 = 0 + 0.265 482 457 436 691 815 396 192;
  • 5) 0.265 482 457 436 691 815 396 192 × 2 = 0 + 0.530 964 914 873 383 630 792 384;
  • 6) 0.530 964 914 873 383 630 792 384 × 2 = 1 + 0.061 929 829 746 767 261 584 768;
  • 7) 0.061 929 829 746 767 261 584 768 × 2 = 0 + 0.123 859 659 493 534 523 169 536;
  • 8) 0.123 859 659 493 534 523 169 536 × 2 = 0 + 0.247 719 318 987 069 046 339 072;
  • 9) 0.247 719 318 987 069 046 339 072 × 2 = 0 + 0.495 438 637 974 138 092 678 144;
  • 10) 0.495 438 637 974 138 092 678 144 × 2 = 0 + 0.990 877 275 948 276 185 356 288;
  • 11) 0.990 877 275 948 276 185 356 288 × 2 = 1 + 0.981 754 551 896 552 370 712 576;
  • 12) 0.981 754 551 896 552 370 712 576 × 2 = 1 + 0.963 509 103 793 104 741 425 152;
  • 13) 0.963 509 103 793 104 741 425 152 × 2 = 1 + 0.927 018 207 586 209 482 850 304;
  • 14) 0.927 018 207 586 209 482 850 304 × 2 = 1 + 0.854 036 415 172 418 965 700 608;
  • 15) 0.854 036 415 172 418 965 700 608 × 2 = 1 + 0.708 072 830 344 837 931 401 216;
  • 16) 0.708 072 830 344 837 931 401 216 × 2 = 1 + 0.416 145 660 689 675 862 802 432;
  • 17) 0.416 145 660 689 675 862 802 432 × 2 = 0 + 0.832 291 321 379 351 725 604 864;
  • 18) 0.832 291 321 379 351 725 604 864 × 2 = 1 + 0.664 582 642 758 703 451 209 728;
  • 19) 0.664 582 642 758 703 451 209 728 × 2 = 1 + 0.329 165 285 517 406 902 419 456;
  • 20) 0.329 165 285 517 406 902 419 456 × 2 = 0 + 0.658 330 571 034 813 804 838 912;
  • 21) 0.658 330 571 034 813 804 838 912 × 2 = 1 + 0.316 661 142 069 627 609 677 824;
  • 22) 0.316 661 142 069 627 609 677 824 × 2 = 0 + 0.633 322 284 139 255 219 355 648;
  • 23) 0.633 322 284 139 255 219 355 648 × 2 = 1 + 0.266 644 568 278 510 438 711 296;
  • 24) 0.266 644 568 278 510 438 711 296 × 2 = 0 + 0.533 289 136 557 020 877 422 592;
  • 25) 0.533 289 136 557 020 877 422 592 × 2 = 1 + 0.066 578 273 114 041 754 845 184;
  • 26) 0.066 578 273 114 041 754 845 184 × 2 = 0 + 0.133 156 546 228 083 509 690 368;
  • 27) 0.133 156 546 228 083 509 690 368 × 2 = 0 + 0.266 313 092 456 167 019 380 736;
  • 28) 0.266 313 092 456 167 019 380 736 × 2 = 0 + 0.532 626 184 912 334 038 761 472;
  • 29) 0.532 626 184 912 334 038 761 472 × 2 = 1 + 0.065 252 369 824 668 077 522 944;
  • 30) 0.065 252 369 824 668 077 522 944 × 2 = 0 + 0.130 504 739 649 336 155 045 888;
  • 31) 0.130 504 739 649 336 155 045 888 × 2 = 0 + 0.261 009 479 298 672 310 091 776;
  • 32) 0.261 009 479 298 672 310 091 776 × 2 = 0 + 0.522 018 958 597 344 620 183 552;
  • 33) 0.522 018 958 597 344 620 183 552 × 2 = 1 + 0.044 037 917 194 689 240 367 104;
  • 34) 0.044 037 917 194 689 240 367 104 × 2 = 0 + 0.088 075 834 389 378 480 734 208;
  • 35) 0.088 075 834 389 378 480 734 208 × 2 = 0 + 0.176 151 668 778 756 961 468 416;
  • 36) 0.176 151 668 778 756 961 468 416 × 2 = 0 + 0.352 303 337 557 513 922 936 832;
  • 37) 0.352 303 337 557 513 922 936 832 × 2 = 0 + 0.704 606 675 115 027 845 873 664;
  • 38) 0.704 606 675 115 027 845 873 664 × 2 = 1 + 0.409 213 350 230 055 691 747 328;
  • 39) 0.409 213 350 230 055 691 747 328 × 2 = 0 + 0.818 426 700 460 111 383 494 656;
  • 40) 0.818 426 700 460 111 383 494 656 × 2 = 1 + 0.636 853 400 920 222 766 989 312;
  • 41) 0.636 853 400 920 222 766 989 312 × 2 = 1 + 0.273 706 801 840 445 533 978 624;
  • 42) 0.273 706 801 840 445 533 978 624 × 2 = 0 + 0.547 413 603 680 891 067 957 248;
  • 43) 0.547 413 603 680 891 067 957 248 × 2 = 1 + 0.094 827 207 361 782 135 914 496;
  • 44) 0.094 827 207 361 782 135 914 496 × 2 = 0 + 0.189 654 414 723 564 271 828 992;
  • 45) 0.189 654 414 723 564 271 828 992 × 2 = 0 + 0.379 308 829 447 128 543 657 984;
  • 46) 0.379 308 829 447 128 543 657 984 × 2 = 0 + 0.758 617 658 894 257 087 315 968;
  • 47) 0.758 617 658 894 257 087 315 968 × 2 = 1 + 0.517 235 317 788 514 174 631 936;
  • 48) 0.517 235 317 788 514 174 631 936 × 2 = 1 + 0.034 470 635 577 028 349 263 872;
  • 49) 0.034 470 635 577 028 349 263 872 × 2 = 0 + 0.068 941 271 154 056 698 527 744;
  • 50) 0.068 941 271 154 056 698 527 744 × 2 = 0 + 0.137 882 542 308 113 397 055 488;
  • 51) 0.137 882 542 308 113 397 055 488 × 2 = 0 + 0.275 765 084 616 226 794 110 976;
  • 52) 0.275 765 084 616 226 794 110 976 × 2 = 0 + 0.551 530 169 232 453 588 221 952;
  • 53) 0.551 530 169 232 453 588 221 952 × 2 = 1 + 0.103 060 338 464 907 176 443 904;

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:


0.141 592 653 589 793 238 462 262(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 262(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:


3.141 592 653 589 793 238 462 262(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 3.141 592 653 589 793 238 462 262 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