3.141 592 653 589 793 238 461 959 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 461 959(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
3.141 592 653 589 793 238 461 959(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: 0.141 592 653 589 793 238 461 959.

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) 0.141 592 653 589 793 238 461 959 × 2 = 0 + 0.283 185 307 179 586 476 923 918;
  • 2) 0.283 185 307 179 586 476 923 918 × 2 = 0 + 0.566 370 614 359 172 953 847 836;
  • 3) 0.566 370 614 359 172 953 847 836 × 2 = 1 + 0.132 741 228 718 345 907 695 672;
  • 4) 0.132 741 228 718 345 907 695 672 × 2 = 0 + 0.265 482 457 436 691 815 391 344;
  • 5) 0.265 482 457 436 691 815 391 344 × 2 = 0 + 0.530 964 914 873 383 630 782 688;
  • 6) 0.530 964 914 873 383 630 782 688 × 2 = 1 + 0.061 929 829 746 767 261 565 376;
  • 7) 0.061 929 829 746 767 261 565 376 × 2 = 0 + 0.123 859 659 493 534 523 130 752;
  • 8) 0.123 859 659 493 534 523 130 752 × 2 = 0 + 0.247 719 318 987 069 046 261 504;
  • 9) 0.247 719 318 987 069 046 261 504 × 2 = 0 + 0.495 438 637 974 138 092 523 008;
  • 10) 0.495 438 637 974 138 092 523 008 × 2 = 0 + 0.990 877 275 948 276 185 046 016;
  • 11) 0.990 877 275 948 276 185 046 016 × 2 = 1 + 0.981 754 551 896 552 370 092 032;
  • 12) 0.981 754 551 896 552 370 092 032 × 2 = 1 + 0.963 509 103 793 104 740 184 064;
  • 13) 0.963 509 103 793 104 740 184 064 × 2 = 1 + 0.927 018 207 586 209 480 368 128;
  • 14) 0.927 018 207 586 209 480 368 128 × 2 = 1 + 0.854 036 415 172 418 960 736 256;
  • 15) 0.854 036 415 172 418 960 736 256 × 2 = 1 + 0.708 072 830 344 837 921 472 512;
  • 16) 0.708 072 830 344 837 921 472 512 × 2 = 1 + 0.416 145 660 689 675 842 945 024;
  • 17) 0.416 145 660 689 675 842 945 024 × 2 = 0 + 0.832 291 321 379 351 685 890 048;
  • 18) 0.832 291 321 379 351 685 890 048 × 2 = 1 + 0.664 582 642 758 703 371 780 096;
  • 19) 0.664 582 642 758 703 371 780 096 × 2 = 1 + 0.329 165 285 517 406 743 560 192;
  • 20) 0.329 165 285 517 406 743 560 192 × 2 = 0 + 0.658 330 571 034 813 487 120 384;
  • 21) 0.658 330 571 034 813 487 120 384 × 2 = 1 + 0.316 661 142 069 626 974 240 768;
  • 22) 0.316 661 142 069 626 974 240 768 × 2 = 0 + 0.633 322 284 139 253 948 481 536;
  • 23) 0.633 322 284 139 253 948 481 536 × 2 = 1 + 0.266 644 568 278 507 896 963 072;
  • 24) 0.266 644 568 278 507 896 963 072 × 2 = 0 + 0.533 289 136 557 015 793 926 144;
  • 25) 0.533 289 136 557 015 793 926 144 × 2 = 1 + 0.066 578 273 114 031 587 852 288;
  • 26) 0.066 578 273 114 031 587 852 288 × 2 = 0 + 0.133 156 546 228 063 175 704 576;
  • 27) 0.133 156 546 228 063 175 704 576 × 2 = 0 + 0.266 313 092 456 126 351 409 152;
  • 28) 0.266 313 092 456 126 351 409 152 × 2 = 0 + 0.532 626 184 912 252 702 818 304;
  • 29) 0.532 626 184 912 252 702 818 304 × 2 = 1 + 0.065 252 369 824 505 405 636 608;
  • 30) 0.065 252 369 824 505 405 636 608 × 2 = 0 + 0.130 504 739 649 010 811 273 216;
  • 31) 0.130 504 739 649 010 811 273 216 × 2 = 0 + 0.261 009 479 298 021 622 546 432;
  • 32) 0.261 009 479 298 021 622 546 432 × 2 = 0 + 0.522 018 958 596 043 245 092 864;
  • 33) 0.522 018 958 596 043 245 092 864 × 2 = 1 + 0.044 037 917 192 086 490 185 728;
  • 34) 0.044 037 917 192 086 490 185 728 × 2 = 0 + 0.088 075 834 384 172 980 371 456;
  • 35) 0.088 075 834 384 172 980 371 456 × 2 = 0 + 0.176 151 668 768 345 960 742 912;
  • 36) 0.176 151 668 768 345 960 742 912 × 2 = 0 + 0.352 303 337 536 691 921 485 824;
  • 37) 0.352 303 337 536 691 921 485 824 × 2 = 0 + 0.704 606 675 073 383 842 971 648;
  • 38) 0.704 606 675 073 383 842 971 648 × 2 = 1 + 0.409 213 350 146 767 685 943 296;
  • 39) 0.409 213 350 146 767 685 943 296 × 2 = 0 + 0.818 426 700 293 535 371 886 592;
  • 40) 0.818 426 700 293 535 371 886 592 × 2 = 1 + 0.636 853 400 587 070 743 773 184;
  • 41) 0.636 853 400 587 070 743 773 184 × 2 = 1 + 0.273 706 801 174 141 487 546 368;
  • 42) 0.273 706 801 174 141 487 546 368 × 2 = 0 + 0.547 413 602 348 282 975 092 736;
  • 43) 0.547 413 602 348 282 975 092 736 × 2 = 1 + 0.094 827 204 696 565 950 185 472;
  • 44) 0.094 827 204 696 565 950 185 472 × 2 = 0 + 0.189 654 409 393 131 900 370 944;
  • 45) 0.189 654 409 393 131 900 370 944 × 2 = 0 + 0.379 308 818 786 263 800 741 888;
  • 46) 0.379 308 818 786 263 800 741 888 × 2 = 0 + 0.758 617 637 572 527 601 483 776;
  • 47) 0.758 617 637 572 527 601 483 776 × 2 = 1 + 0.517 235 275 145 055 202 967 552;
  • 48) 0.517 235 275 145 055 202 967 552 × 2 = 1 + 0.034 470 550 290 110 405 935 104;
  • 49) 0.034 470 550 290 110 405 935 104 × 2 = 0 + 0.068 941 100 580 220 811 870 208;
  • 50) 0.068 941 100 580 220 811 870 208 × 2 = 0 + 0.137 882 201 160 441 623 740 416;
  • 51) 0.137 882 201 160 441 623 740 416 × 2 = 0 + 0.275 764 402 320 883 247 480 832;
  • 52) 0.275 764 402 320 883 247 480 832 × 2 = 0 + 0.551 528 804 641 766 494 961 664;
  • 53) 0.551 528 804 641 766 494 961 664 × 2 = 1 + 0.103 057 609 283 532 989 923 328;

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:


0.141 592 653 589 793 238 461 959(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 461 959(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:


3.141 592 653 589 793 238 461 959(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 3.141 592 653 589 793 238 461 959 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