3.141 592 653 589 793 237 819 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 237 819(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
3.141 592 653 589 793 237 819(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: 0.141 592 653 589 793 237 819.

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) 0.141 592 653 589 793 237 819 × 2 = 0 + 0.283 185 307 179 586 475 638;
  • 2) 0.283 185 307 179 586 475 638 × 2 = 0 + 0.566 370 614 359 172 951 276;
  • 3) 0.566 370 614 359 172 951 276 × 2 = 1 + 0.132 741 228 718 345 902 552;
  • 4) 0.132 741 228 718 345 902 552 × 2 = 0 + 0.265 482 457 436 691 805 104;
  • 5) 0.265 482 457 436 691 805 104 × 2 = 0 + 0.530 964 914 873 383 610 208;
  • 6) 0.530 964 914 873 383 610 208 × 2 = 1 + 0.061 929 829 746 767 220 416;
  • 7) 0.061 929 829 746 767 220 416 × 2 = 0 + 0.123 859 659 493 534 440 832;
  • 8) 0.123 859 659 493 534 440 832 × 2 = 0 + 0.247 719 318 987 068 881 664;
  • 9) 0.247 719 318 987 068 881 664 × 2 = 0 + 0.495 438 637 974 137 763 328;
  • 10) 0.495 438 637 974 137 763 328 × 2 = 0 + 0.990 877 275 948 275 526 656;
  • 11) 0.990 877 275 948 275 526 656 × 2 = 1 + 0.981 754 551 896 551 053 312;
  • 12) 0.981 754 551 896 551 053 312 × 2 = 1 + 0.963 509 103 793 102 106 624;
  • 13) 0.963 509 103 793 102 106 624 × 2 = 1 + 0.927 018 207 586 204 213 248;
  • 14) 0.927 018 207 586 204 213 248 × 2 = 1 + 0.854 036 415 172 408 426 496;
  • 15) 0.854 036 415 172 408 426 496 × 2 = 1 + 0.708 072 830 344 816 852 992;
  • 16) 0.708 072 830 344 816 852 992 × 2 = 1 + 0.416 145 660 689 633 705 984;
  • 17) 0.416 145 660 689 633 705 984 × 2 = 0 + 0.832 291 321 379 267 411 968;
  • 18) 0.832 291 321 379 267 411 968 × 2 = 1 + 0.664 582 642 758 534 823 936;
  • 19) 0.664 582 642 758 534 823 936 × 2 = 1 + 0.329 165 285 517 069 647 872;
  • 20) 0.329 165 285 517 069 647 872 × 2 = 0 + 0.658 330 571 034 139 295 744;
  • 21) 0.658 330 571 034 139 295 744 × 2 = 1 + 0.316 661 142 068 278 591 488;
  • 22) 0.316 661 142 068 278 591 488 × 2 = 0 + 0.633 322 284 136 557 182 976;
  • 23) 0.633 322 284 136 557 182 976 × 2 = 1 + 0.266 644 568 273 114 365 952;
  • 24) 0.266 644 568 273 114 365 952 × 2 = 0 + 0.533 289 136 546 228 731 904;
  • 25) 0.533 289 136 546 228 731 904 × 2 = 1 + 0.066 578 273 092 457 463 808;
  • 26) 0.066 578 273 092 457 463 808 × 2 = 0 + 0.133 156 546 184 914 927 616;
  • 27) 0.133 156 546 184 914 927 616 × 2 = 0 + 0.266 313 092 369 829 855 232;
  • 28) 0.266 313 092 369 829 855 232 × 2 = 0 + 0.532 626 184 739 659 710 464;
  • 29) 0.532 626 184 739 659 710 464 × 2 = 1 + 0.065 252 369 479 319 420 928;
  • 30) 0.065 252 369 479 319 420 928 × 2 = 0 + 0.130 504 738 958 638 841 856;
  • 31) 0.130 504 738 958 638 841 856 × 2 = 0 + 0.261 009 477 917 277 683 712;
  • 32) 0.261 009 477 917 277 683 712 × 2 = 0 + 0.522 018 955 834 555 367 424;
  • 33) 0.522 018 955 834 555 367 424 × 2 = 1 + 0.044 037 911 669 110 734 848;
  • 34) 0.044 037 911 669 110 734 848 × 2 = 0 + 0.088 075 823 338 221 469 696;
  • 35) 0.088 075 823 338 221 469 696 × 2 = 0 + 0.176 151 646 676 442 939 392;
  • 36) 0.176 151 646 676 442 939 392 × 2 = 0 + 0.352 303 293 352 885 878 784;
  • 37) 0.352 303 293 352 885 878 784 × 2 = 0 + 0.704 606 586 705 771 757 568;
  • 38) 0.704 606 586 705 771 757 568 × 2 = 1 + 0.409 213 173 411 543 515 136;
  • 39) 0.409 213 173 411 543 515 136 × 2 = 0 + 0.818 426 346 823 087 030 272;
  • 40) 0.818 426 346 823 087 030 272 × 2 = 1 + 0.636 852 693 646 174 060 544;
  • 41) 0.636 852 693 646 174 060 544 × 2 = 1 + 0.273 705 387 292 348 121 088;
  • 42) 0.273 705 387 292 348 121 088 × 2 = 0 + 0.547 410 774 584 696 242 176;
  • 43) 0.547 410 774 584 696 242 176 × 2 = 1 + 0.094 821 549 169 392 484 352;
  • 44) 0.094 821 549 169 392 484 352 × 2 = 0 + 0.189 643 098 338 784 968 704;
  • 45) 0.189 643 098 338 784 968 704 × 2 = 0 + 0.379 286 196 677 569 937 408;
  • 46) 0.379 286 196 677 569 937 408 × 2 = 0 + 0.758 572 393 355 139 874 816;
  • 47) 0.758 572 393 355 139 874 816 × 2 = 1 + 0.517 144 786 710 279 749 632;
  • 48) 0.517 144 786 710 279 749 632 × 2 = 1 + 0.034 289 573 420 559 499 264;
  • 49) 0.034 289 573 420 559 499 264 × 2 = 0 + 0.068 579 146 841 118 998 528;
  • 50) 0.068 579 146 841 118 998 528 × 2 = 0 + 0.137 158 293 682 237 997 056;
  • 51) 0.137 158 293 682 237 997 056 × 2 = 0 + 0.274 316 587 364 475 994 112;
  • 52) 0.274 316 587 364 475 994 112 × 2 = 0 + 0.548 633 174 728 951 988 224;
  • 53) 0.548 633 174 728 951 988 224 × 2 = 1 + 0.097 266 349 457 903 976 448;

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:


0.141 592 653 589 793 237 819(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 237 819(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:


3.141 592 653 589 793 237 819(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 3.141 592 653 589 793 237 819 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