3.141 592 653 589 793 126 462 127 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 126 462 127(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
3.141 592 653 589 793 126 462 127(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: 0.141 592 653 589 793 126 462 127.

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) 0.141 592 653 589 793 126 462 127 × 2 = 0 + 0.283 185 307 179 586 252 924 254;
  • 2) 0.283 185 307 179 586 252 924 254 × 2 = 0 + 0.566 370 614 359 172 505 848 508;
  • 3) 0.566 370 614 359 172 505 848 508 × 2 = 1 + 0.132 741 228 718 345 011 697 016;
  • 4) 0.132 741 228 718 345 011 697 016 × 2 = 0 + 0.265 482 457 436 690 023 394 032;
  • 5) 0.265 482 457 436 690 023 394 032 × 2 = 0 + 0.530 964 914 873 380 046 788 064;
  • 6) 0.530 964 914 873 380 046 788 064 × 2 = 1 + 0.061 929 829 746 760 093 576 128;
  • 7) 0.061 929 829 746 760 093 576 128 × 2 = 0 + 0.123 859 659 493 520 187 152 256;
  • 8) 0.123 859 659 493 520 187 152 256 × 2 = 0 + 0.247 719 318 987 040 374 304 512;
  • 9) 0.247 719 318 987 040 374 304 512 × 2 = 0 + 0.495 438 637 974 080 748 609 024;
  • 10) 0.495 438 637 974 080 748 609 024 × 2 = 0 + 0.990 877 275 948 161 497 218 048;
  • 11) 0.990 877 275 948 161 497 218 048 × 2 = 1 + 0.981 754 551 896 322 994 436 096;
  • 12) 0.981 754 551 896 322 994 436 096 × 2 = 1 + 0.963 509 103 792 645 988 872 192;
  • 13) 0.963 509 103 792 645 988 872 192 × 2 = 1 + 0.927 018 207 585 291 977 744 384;
  • 14) 0.927 018 207 585 291 977 744 384 × 2 = 1 + 0.854 036 415 170 583 955 488 768;
  • 15) 0.854 036 415 170 583 955 488 768 × 2 = 1 + 0.708 072 830 341 167 910 977 536;
  • 16) 0.708 072 830 341 167 910 977 536 × 2 = 1 + 0.416 145 660 682 335 821 955 072;
  • 17) 0.416 145 660 682 335 821 955 072 × 2 = 0 + 0.832 291 321 364 671 643 910 144;
  • 18) 0.832 291 321 364 671 643 910 144 × 2 = 1 + 0.664 582 642 729 343 287 820 288;
  • 19) 0.664 582 642 729 343 287 820 288 × 2 = 1 + 0.329 165 285 458 686 575 640 576;
  • 20) 0.329 165 285 458 686 575 640 576 × 2 = 0 + 0.658 330 570 917 373 151 281 152;
  • 21) 0.658 330 570 917 373 151 281 152 × 2 = 1 + 0.316 661 141 834 746 302 562 304;
  • 22) 0.316 661 141 834 746 302 562 304 × 2 = 0 + 0.633 322 283 669 492 605 124 608;
  • 23) 0.633 322 283 669 492 605 124 608 × 2 = 1 + 0.266 644 567 338 985 210 249 216;
  • 24) 0.266 644 567 338 985 210 249 216 × 2 = 0 + 0.533 289 134 677 970 420 498 432;
  • 25) 0.533 289 134 677 970 420 498 432 × 2 = 1 + 0.066 578 269 355 940 840 996 864;
  • 26) 0.066 578 269 355 940 840 996 864 × 2 = 0 + 0.133 156 538 711 881 681 993 728;
  • 27) 0.133 156 538 711 881 681 993 728 × 2 = 0 + 0.266 313 077 423 763 363 987 456;
  • 28) 0.266 313 077 423 763 363 987 456 × 2 = 0 + 0.532 626 154 847 526 727 974 912;
  • 29) 0.532 626 154 847 526 727 974 912 × 2 = 1 + 0.065 252 309 695 053 455 949 824;
  • 30) 0.065 252 309 695 053 455 949 824 × 2 = 0 + 0.130 504 619 390 106 911 899 648;
  • 31) 0.130 504 619 390 106 911 899 648 × 2 = 0 + 0.261 009 238 780 213 823 799 296;
  • 32) 0.261 009 238 780 213 823 799 296 × 2 = 0 + 0.522 018 477 560 427 647 598 592;
  • 33) 0.522 018 477 560 427 647 598 592 × 2 = 1 + 0.044 036 955 120 855 295 197 184;
  • 34) 0.044 036 955 120 855 295 197 184 × 2 = 0 + 0.088 073 910 241 710 590 394 368;
  • 35) 0.088 073 910 241 710 590 394 368 × 2 = 0 + 0.176 147 820 483 421 180 788 736;
  • 36) 0.176 147 820 483 421 180 788 736 × 2 = 0 + 0.352 295 640 966 842 361 577 472;
  • 37) 0.352 295 640 966 842 361 577 472 × 2 = 0 + 0.704 591 281 933 684 723 154 944;
  • 38) 0.704 591 281 933 684 723 154 944 × 2 = 1 + 0.409 182 563 867 369 446 309 888;
  • 39) 0.409 182 563 867 369 446 309 888 × 2 = 0 + 0.818 365 127 734 738 892 619 776;
  • 40) 0.818 365 127 734 738 892 619 776 × 2 = 1 + 0.636 730 255 469 477 785 239 552;
  • 41) 0.636 730 255 469 477 785 239 552 × 2 = 1 + 0.273 460 510 938 955 570 479 104;
  • 42) 0.273 460 510 938 955 570 479 104 × 2 = 0 + 0.546 921 021 877 911 140 958 208;
  • 43) 0.546 921 021 877 911 140 958 208 × 2 = 1 + 0.093 842 043 755 822 281 916 416;
  • 44) 0.093 842 043 755 822 281 916 416 × 2 = 0 + 0.187 684 087 511 644 563 832 832;
  • 45) 0.187 684 087 511 644 563 832 832 × 2 = 0 + 0.375 368 175 023 289 127 665 664;
  • 46) 0.375 368 175 023 289 127 665 664 × 2 = 0 + 0.750 736 350 046 578 255 331 328;
  • 47) 0.750 736 350 046 578 255 331 328 × 2 = 1 + 0.501 472 700 093 156 510 662 656;
  • 48) 0.501 472 700 093 156 510 662 656 × 2 = 1 + 0.002 945 400 186 313 021 325 312;
  • 49) 0.002 945 400 186 313 021 325 312 × 2 = 0 + 0.005 890 800 372 626 042 650 624;
  • 50) 0.005 890 800 372 626 042 650 624 × 2 = 0 + 0.011 781 600 745 252 085 301 248;
  • 51) 0.011 781 600 745 252 085 301 248 × 2 = 0 + 0.023 563 201 490 504 170 602 496;
  • 52) 0.023 563 201 490 504 170 602 496 × 2 = 0 + 0.047 126 402 981 008 341 204 992;
  • 53) 0.047 126 402 981 008 341 204 992 × 2 = 0 + 0.094 252 805 962 016 682 409 984;

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:


0.141 592 653 589 793 126 462 127(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2)

5. Positive number before normalization:

3.141 592 653 589 793 126 462 127(10) =

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


3.141 592 653 589 793 126 462 127(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 00(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 00


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 00 =

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 3.141 592 653 589 793 126 462 127 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