3.141 592 653 689 769 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 769 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
3.141 592 653 689 769 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: 0.141 592 653 689 769 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) 0.141 592 653 689 769 4 × 2 = 0 + 0.283 185 307 379 538 8;
  • 2) 0.283 185 307 379 538 8 × 2 = 0 + 0.566 370 614 759 077 6;
  • 3) 0.566 370 614 759 077 6 × 2 = 1 + 0.132 741 229 518 155 2;
  • 4) 0.132 741 229 518 155 2 × 2 = 0 + 0.265 482 459 036 310 4;
  • 5) 0.265 482 459 036 310 4 × 2 = 0 + 0.530 964 918 072 620 8;
  • 6) 0.530 964 918 072 620 8 × 2 = 1 + 0.061 929 836 145 241 6;
  • 7) 0.061 929 836 145 241 6 × 2 = 0 + 0.123 859 672 290 483 2;
  • 8) 0.123 859 672 290 483 2 × 2 = 0 + 0.247 719 344 580 966 4;
  • 9) 0.247 719 344 580 966 4 × 2 = 0 + 0.495 438 689 161 932 8;
  • 10) 0.495 438 689 161 932 8 × 2 = 0 + 0.990 877 378 323 865 6;
  • 11) 0.990 877 378 323 865 6 × 2 = 1 + 0.981 754 756 647 731 2;
  • 12) 0.981 754 756 647 731 2 × 2 = 1 + 0.963 509 513 295 462 4;
  • 13) 0.963 509 513 295 462 4 × 2 = 1 + 0.927 019 026 590 924 8;
  • 14) 0.927 019 026 590 924 8 × 2 = 1 + 0.854 038 053 181 849 6;
  • 15) 0.854 038 053 181 849 6 × 2 = 1 + 0.708 076 106 363 699 2;
  • 16) 0.708 076 106 363 699 2 × 2 = 1 + 0.416 152 212 727 398 4;
  • 17) 0.416 152 212 727 398 4 × 2 = 0 + 0.832 304 425 454 796 8;
  • 18) 0.832 304 425 454 796 8 × 2 = 1 + 0.664 608 850 909 593 6;
  • 19) 0.664 608 850 909 593 6 × 2 = 1 + 0.329 217 701 819 187 2;
  • 20) 0.329 217 701 819 187 2 × 2 = 0 + 0.658 435 403 638 374 4;
  • 21) 0.658 435 403 638 374 4 × 2 = 1 + 0.316 870 807 276 748 8;
  • 22) 0.316 870 807 276 748 8 × 2 = 0 + 0.633 741 614 553 497 6;
  • 23) 0.633 741 614 553 497 6 × 2 = 1 + 0.267 483 229 106 995 2;
  • 24) 0.267 483 229 106 995 2 × 2 = 0 + 0.534 966 458 213 990 4;
  • 25) 0.534 966 458 213 990 4 × 2 = 1 + 0.069 932 916 427 980 8;
  • 26) 0.069 932 916 427 980 8 × 2 = 0 + 0.139 865 832 855 961 6;
  • 27) 0.139 865 832 855 961 6 × 2 = 0 + 0.279 731 665 711 923 2;
  • 28) 0.279 731 665 711 923 2 × 2 = 0 + 0.559 463 331 423 846 4;
  • 29) 0.559 463 331 423 846 4 × 2 = 1 + 0.118 926 662 847 692 8;
  • 30) 0.118 926 662 847 692 8 × 2 = 0 + 0.237 853 325 695 385 6;
  • 31) 0.237 853 325 695 385 6 × 2 = 0 + 0.475 706 651 390 771 2;
  • 32) 0.475 706 651 390 771 2 × 2 = 0 + 0.951 413 302 781 542 4;
  • 33) 0.951 413 302 781 542 4 × 2 = 1 + 0.902 826 605 563 084 8;
  • 34) 0.902 826 605 563 084 8 × 2 = 1 + 0.805 653 211 126 169 6;
  • 35) 0.805 653 211 126 169 6 × 2 = 1 + 0.611 306 422 252 339 2;
  • 36) 0.611 306 422 252 339 2 × 2 = 1 + 0.222 612 844 504 678 4;
  • 37) 0.222 612 844 504 678 4 × 2 = 0 + 0.445 225 689 009 356 8;
  • 38) 0.445 225 689 009 356 8 × 2 = 0 + 0.890 451 378 018 713 6;
  • 39) 0.890 451 378 018 713 6 × 2 = 1 + 0.780 902 756 037 427 2;
  • 40) 0.780 902 756 037 427 2 × 2 = 1 + 0.561 805 512 074 854 4;
  • 41) 0.561 805 512 074 854 4 × 2 = 1 + 0.123 611 024 149 708 8;
  • 42) 0.123 611 024 149 708 8 × 2 = 0 + 0.247 222 048 299 417 6;
  • 43) 0.247 222 048 299 417 6 × 2 = 0 + 0.494 444 096 598 835 2;
  • 44) 0.494 444 096 598 835 2 × 2 = 0 + 0.988 888 193 197 670 4;
  • 45) 0.988 888 193 197 670 4 × 2 = 1 + 0.977 776 386 395 340 8;
  • 46) 0.977 776 386 395 340 8 × 2 = 1 + 0.955 552 772 790 681 6;
  • 47) 0.955 552 772 790 681 6 × 2 = 1 + 0.911 105 545 581 363 2;
  • 48) 0.911 105 545 581 363 2 × 2 = 1 + 0.822 211 091 162 726 4;
  • 49) 0.822 211 091 162 726 4 × 2 = 1 + 0.644 422 182 325 452 8;
  • 50) 0.644 422 182 325 452 8 × 2 = 1 + 0.288 844 364 650 905 6;
  • 51) 0.288 844 364 650 905 6 × 2 = 0 + 0.577 688 729 301 811 2;
  • 52) 0.577 688 729 301 811 2 × 2 = 1 + 0.155 377 458 603 622 4;
  • 53) 0.155 377 458 603 622 4 × 2 = 0 + 0.310 754 917 207 244 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:


0.141 592 653 689 769 4(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1000 1111 1101 0(2)

5. Positive number before normalization:

3.141 592 653 689 769 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1000 1111 1101 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:


3.141 592 653 689 769 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1000 1111 1101 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1000 1111 1101 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 0111 1110 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 0111 1110 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 0111 1110 10 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 0111 1110


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 0111 1110


Decimal number 3.141 592 653 689 769 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 0111 1110

» Convert decimal 3.141 592 653 689 764 5 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