3.141 592 653 689 781 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 781 4(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
3.141 592 653 689 781 4(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: 0.141 592 653 689 781 4.

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) 0.141 592 653 689 781 4 × 2 = 0 + 0.283 185 307 379 562 8;
  • 2) 0.283 185 307 379 562 8 × 2 = 0 + 0.566 370 614 759 125 6;
  • 3) 0.566 370 614 759 125 6 × 2 = 1 + 0.132 741 229 518 251 2;
  • 4) 0.132 741 229 518 251 2 × 2 = 0 + 0.265 482 459 036 502 4;
  • 5) 0.265 482 459 036 502 4 × 2 = 0 + 0.530 964 918 073 004 8;
  • 6) 0.530 964 918 073 004 8 × 2 = 1 + 0.061 929 836 146 009 6;
  • 7) 0.061 929 836 146 009 6 × 2 = 0 + 0.123 859 672 292 019 2;
  • 8) 0.123 859 672 292 019 2 × 2 = 0 + 0.247 719 344 584 038 4;
  • 9) 0.247 719 344 584 038 4 × 2 = 0 + 0.495 438 689 168 076 8;
  • 10) 0.495 438 689 168 076 8 × 2 = 0 + 0.990 877 378 336 153 6;
  • 11) 0.990 877 378 336 153 6 × 2 = 1 + 0.981 754 756 672 307 2;
  • 12) 0.981 754 756 672 307 2 × 2 = 1 + 0.963 509 513 344 614 4;
  • 13) 0.963 509 513 344 614 4 × 2 = 1 + 0.927 019 026 689 228 8;
  • 14) 0.927 019 026 689 228 8 × 2 = 1 + 0.854 038 053 378 457 6;
  • 15) 0.854 038 053 378 457 6 × 2 = 1 + 0.708 076 106 756 915 2;
  • 16) 0.708 076 106 756 915 2 × 2 = 1 + 0.416 152 213 513 830 4;
  • 17) 0.416 152 213 513 830 4 × 2 = 0 + 0.832 304 427 027 660 8;
  • 18) 0.832 304 427 027 660 8 × 2 = 1 + 0.664 608 854 055 321 6;
  • 19) 0.664 608 854 055 321 6 × 2 = 1 + 0.329 217 708 110 643 2;
  • 20) 0.329 217 708 110 643 2 × 2 = 0 + 0.658 435 416 221 286 4;
  • 21) 0.658 435 416 221 286 4 × 2 = 1 + 0.316 870 832 442 572 8;
  • 22) 0.316 870 832 442 572 8 × 2 = 0 + 0.633 741 664 885 145 6;
  • 23) 0.633 741 664 885 145 6 × 2 = 1 + 0.267 483 329 770 291 2;
  • 24) 0.267 483 329 770 291 2 × 2 = 0 + 0.534 966 659 540 582 4;
  • 25) 0.534 966 659 540 582 4 × 2 = 1 + 0.069 933 319 081 164 8;
  • 26) 0.069 933 319 081 164 8 × 2 = 0 + 0.139 866 638 162 329 6;
  • 27) 0.139 866 638 162 329 6 × 2 = 0 + 0.279 733 276 324 659 2;
  • 28) 0.279 733 276 324 659 2 × 2 = 0 + 0.559 466 552 649 318 4;
  • 29) 0.559 466 552 649 318 4 × 2 = 1 + 0.118 933 105 298 636 8;
  • 30) 0.118 933 105 298 636 8 × 2 = 0 + 0.237 866 210 597 273 6;
  • 31) 0.237 866 210 597 273 6 × 2 = 0 + 0.475 732 421 194 547 2;
  • 32) 0.475 732 421 194 547 2 × 2 = 0 + 0.951 464 842 389 094 4;
  • 33) 0.951 464 842 389 094 4 × 2 = 1 + 0.902 929 684 778 188 8;
  • 34) 0.902 929 684 778 188 8 × 2 = 1 + 0.805 859 369 556 377 6;
  • 35) 0.805 859 369 556 377 6 × 2 = 1 + 0.611 718 739 112 755 2;
  • 36) 0.611 718 739 112 755 2 × 2 = 1 + 0.223 437 478 225 510 4;
  • 37) 0.223 437 478 225 510 4 × 2 = 0 + 0.446 874 956 451 020 8;
  • 38) 0.446 874 956 451 020 8 × 2 = 0 + 0.893 749 912 902 041 6;
  • 39) 0.893 749 912 902 041 6 × 2 = 1 + 0.787 499 825 804 083 2;
  • 40) 0.787 499 825 804 083 2 × 2 = 1 + 0.574 999 651 608 166 4;
  • 41) 0.574 999 651 608 166 4 × 2 = 1 + 0.149 999 303 216 332 8;
  • 42) 0.149 999 303 216 332 8 × 2 = 0 + 0.299 998 606 432 665 6;
  • 43) 0.299 998 606 432 665 6 × 2 = 0 + 0.599 997 212 865 331 2;
  • 44) 0.599 997 212 865 331 2 × 2 = 1 + 0.199 994 425 730 662 4;
  • 45) 0.199 994 425 730 662 4 × 2 = 0 + 0.399 988 851 461 324 8;
  • 46) 0.399 988 851 461 324 8 × 2 = 0 + 0.799 977 702 922 649 6;
  • 47) 0.799 977 702 922 649 6 × 2 = 1 + 0.599 955 405 845 299 2;
  • 48) 0.599 955 405 845 299 2 × 2 = 1 + 0.199 910 811 690 598 4;
  • 49) 0.199 910 811 690 598 4 × 2 = 0 + 0.399 821 623 381 196 8;
  • 50) 0.399 821 623 381 196 8 × 2 = 0 + 0.799 643 246 762 393 6;
  • 51) 0.799 643 246 762 393 6 × 2 = 1 + 0.599 286 493 524 787 2;
  • 52) 0.599 286 493 524 787 2 × 2 = 1 + 0.198 572 987 049 574 4;
  • 53) 0.198 572 987 049 574 4 × 2 = 0 + 0.397 145 974 099 148 8;

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:


0.141 592 653 689 781 4(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0011 0011 0(2)

5. Positive number before normalization:

3.141 592 653 689 781 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0011 0011 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:


3.141 592 653 689 781 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0011 0011 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0011 0011 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1001 1001 10(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 0111 1001 1100 1001 1001 10


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 0111 1001 1100 1001 1001 10 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1001 1001


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 0111 1001 1100 1001 1001


Decimal number 3.141 592 653 689 781 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1001 1001

» Convert decimal 3.141 592 653 689 783 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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