3.141 592 653 589 793 238 462 094 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 094(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
3.141 592 653 589 793 238 462 094(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: 0.141 592 653 589 793 238 462 094.

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) 0.141 592 653 589 793 238 462 094 × 2 = 0 + 0.283 185 307 179 586 476 924 188;
  • 2) 0.283 185 307 179 586 476 924 188 × 2 = 0 + 0.566 370 614 359 172 953 848 376;
  • 3) 0.566 370 614 359 172 953 848 376 × 2 = 1 + 0.132 741 228 718 345 907 696 752;
  • 4) 0.132 741 228 718 345 907 696 752 × 2 = 0 + 0.265 482 457 436 691 815 393 504;
  • 5) 0.265 482 457 436 691 815 393 504 × 2 = 0 + 0.530 964 914 873 383 630 787 008;
  • 6) 0.530 964 914 873 383 630 787 008 × 2 = 1 + 0.061 929 829 746 767 261 574 016;
  • 7) 0.061 929 829 746 767 261 574 016 × 2 = 0 + 0.123 859 659 493 534 523 148 032;
  • 8) 0.123 859 659 493 534 523 148 032 × 2 = 0 + 0.247 719 318 987 069 046 296 064;
  • 9) 0.247 719 318 987 069 046 296 064 × 2 = 0 + 0.495 438 637 974 138 092 592 128;
  • 10) 0.495 438 637 974 138 092 592 128 × 2 = 0 + 0.990 877 275 948 276 185 184 256;
  • 11) 0.990 877 275 948 276 185 184 256 × 2 = 1 + 0.981 754 551 896 552 370 368 512;
  • 12) 0.981 754 551 896 552 370 368 512 × 2 = 1 + 0.963 509 103 793 104 740 737 024;
  • 13) 0.963 509 103 793 104 740 737 024 × 2 = 1 + 0.927 018 207 586 209 481 474 048;
  • 14) 0.927 018 207 586 209 481 474 048 × 2 = 1 + 0.854 036 415 172 418 962 948 096;
  • 15) 0.854 036 415 172 418 962 948 096 × 2 = 1 + 0.708 072 830 344 837 925 896 192;
  • 16) 0.708 072 830 344 837 925 896 192 × 2 = 1 + 0.416 145 660 689 675 851 792 384;
  • 17) 0.416 145 660 689 675 851 792 384 × 2 = 0 + 0.832 291 321 379 351 703 584 768;
  • 18) 0.832 291 321 379 351 703 584 768 × 2 = 1 + 0.664 582 642 758 703 407 169 536;
  • 19) 0.664 582 642 758 703 407 169 536 × 2 = 1 + 0.329 165 285 517 406 814 339 072;
  • 20) 0.329 165 285 517 406 814 339 072 × 2 = 0 + 0.658 330 571 034 813 628 678 144;
  • 21) 0.658 330 571 034 813 628 678 144 × 2 = 1 + 0.316 661 142 069 627 257 356 288;
  • 22) 0.316 661 142 069 627 257 356 288 × 2 = 0 + 0.633 322 284 139 254 514 712 576;
  • 23) 0.633 322 284 139 254 514 712 576 × 2 = 1 + 0.266 644 568 278 509 029 425 152;
  • 24) 0.266 644 568 278 509 029 425 152 × 2 = 0 + 0.533 289 136 557 018 058 850 304;
  • 25) 0.533 289 136 557 018 058 850 304 × 2 = 1 + 0.066 578 273 114 036 117 700 608;
  • 26) 0.066 578 273 114 036 117 700 608 × 2 = 0 + 0.133 156 546 228 072 235 401 216;
  • 27) 0.133 156 546 228 072 235 401 216 × 2 = 0 + 0.266 313 092 456 144 470 802 432;
  • 28) 0.266 313 092 456 144 470 802 432 × 2 = 0 + 0.532 626 184 912 288 941 604 864;
  • 29) 0.532 626 184 912 288 941 604 864 × 2 = 1 + 0.065 252 369 824 577 883 209 728;
  • 30) 0.065 252 369 824 577 883 209 728 × 2 = 0 + 0.130 504 739 649 155 766 419 456;
  • 31) 0.130 504 739 649 155 766 419 456 × 2 = 0 + 0.261 009 479 298 311 532 838 912;
  • 32) 0.261 009 479 298 311 532 838 912 × 2 = 0 + 0.522 018 958 596 623 065 677 824;
  • 33) 0.522 018 958 596 623 065 677 824 × 2 = 1 + 0.044 037 917 193 246 131 355 648;
  • 34) 0.044 037 917 193 246 131 355 648 × 2 = 0 + 0.088 075 834 386 492 262 711 296;
  • 35) 0.088 075 834 386 492 262 711 296 × 2 = 0 + 0.176 151 668 772 984 525 422 592;
  • 36) 0.176 151 668 772 984 525 422 592 × 2 = 0 + 0.352 303 337 545 969 050 845 184;
  • 37) 0.352 303 337 545 969 050 845 184 × 2 = 0 + 0.704 606 675 091 938 101 690 368;
  • 38) 0.704 606 675 091 938 101 690 368 × 2 = 1 + 0.409 213 350 183 876 203 380 736;
  • 39) 0.409 213 350 183 876 203 380 736 × 2 = 0 + 0.818 426 700 367 752 406 761 472;
  • 40) 0.818 426 700 367 752 406 761 472 × 2 = 1 + 0.636 853 400 735 504 813 522 944;
  • 41) 0.636 853 400 735 504 813 522 944 × 2 = 1 + 0.273 706 801 471 009 627 045 888;
  • 42) 0.273 706 801 471 009 627 045 888 × 2 = 0 + 0.547 413 602 942 019 254 091 776;
  • 43) 0.547 413 602 942 019 254 091 776 × 2 = 1 + 0.094 827 205 884 038 508 183 552;
  • 44) 0.094 827 205 884 038 508 183 552 × 2 = 0 + 0.189 654 411 768 077 016 367 104;
  • 45) 0.189 654 411 768 077 016 367 104 × 2 = 0 + 0.379 308 823 536 154 032 734 208;
  • 46) 0.379 308 823 536 154 032 734 208 × 2 = 0 + 0.758 617 647 072 308 065 468 416;
  • 47) 0.758 617 647 072 308 065 468 416 × 2 = 1 + 0.517 235 294 144 616 130 936 832;
  • 48) 0.517 235 294 144 616 130 936 832 × 2 = 1 + 0.034 470 588 289 232 261 873 664;
  • 49) 0.034 470 588 289 232 261 873 664 × 2 = 0 + 0.068 941 176 578 464 523 747 328;
  • 50) 0.068 941 176 578 464 523 747 328 × 2 = 0 + 0.137 882 353 156 929 047 494 656;
  • 51) 0.137 882 353 156 929 047 494 656 × 2 = 0 + 0.275 764 706 313 858 094 989 312;
  • 52) 0.275 764 706 313 858 094 989 312 × 2 = 0 + 0.551 529 412 627 716 189 978 624;
  • 53) 0.551 529 412 627 716 189 978 624 × 2 = 1 + 0.103 058 825 255 432 379 957 248;

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:


0.141 592 653 589 793 238 462 094(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 094(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:


3.141 592 653 589 793 238 462 094(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 3.141 592 653 589 793 238 462 094 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