2.718 281 828 458 974 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 458 974(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
2.718 281 828 458 974(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.718 281 828 458 974.

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) 0.718 281 828 458 974 × 2 = 1 + 0.436 563 656 917 948;
  • 2) 0.436 563 656 917 948 × 2 = 0 + 0.873 127 313 835 896;
  • 3) 0.873 127 313 835 896 × 2 = 1 + 0.746 254 627 671 792;
  • 4) 0.746 254 627 671 792 × 2 = 1 + 0.492 509 255 343 584;
  • 5) 0.492 509 255 343 584 × 2 = 0 + 0.985 018 510 687 168;
  • 6) 0.985 018 510 687 168 × 2 = 1 + 0.970 037 021 374 336;
  • 7) 0.970 037 021 374 336 × 2 = 1 + 0.940 074 042 748 672;
  • 8) 0.940 074 042 748 672 × 2 = 1 + 0.880 148 085 497 344;
  • 9) 0.880 148 085 497 344 × 2 = 1 + 0.760 296 170 994 688;
  • 10) 0.760 296 170 994 688 × 2 = 1 + 0.520 592 341 989 376;
  • 11) 0.520 592 341 989 376 × 2 = 1 + 0.041 184 683 978 752;
  • 12) 0.041 184 683 978 752 × 2 = 0 + 0.082 369 367 957 504;
  • 13) 0.082 369 367 957 504 × 2 = 0 + 0.164 738 735 915 008;
  • 14) 0.164 738 735 915 008 × 2 = 0 + 0.329 477 471 830 016;
  • 15) 0.329 477 471 830 016 × 2 = 0 + 0.658 954 943 660 032;
  • 16) 0.658 954 943 660 032 × 2 = 1 + 0.317 909 887 320 064;
  • 17) 0.317 909 887 320 064 × 2 = 0 + 0.635 819 774 640 128;
  • 18) 0.635 819 774 640 128 × 2 = 1 + 0.271 639 549 280 256;
  • 19) 0.271 639 549 280 256 × 2 = 0 + 0.543 279 098 560 512;
  • 20) 0.543 279 098 560 512 × 2 = 1 + 0.086 558 197 121 024;
  • 21) 0.086 558 197 121 024 × 2 = 0 + 0.173 116 394 242 048;
  • 22) 0.173 116 394 242 048 × 2 = 0 + 0.346 232 788 484 096;
  • 23) 0.346 232 788 484 096 × 2 = 0 + 0.692 465 576 968 192;
  • 24) 0.692 465 576 968 192 × 2 = 1 + 0.384 931 153 936 384;
  • 25) 0.384 931 153 936 384 × 2 = 0 + 0.769 862 307 872 768;
  • 26) 0.769 862 307 872 768 × 2 = 1 + 0.539 724 615 745 536;
  • 27) 0.539 724 615 745 536 × 2 = 1 + 0.079 449 231 491 072;
  • 28) 0.079 449 231 491 072 × 2 = 0 + 0.158 898 462 982 144;
  • 29) 0.158 898 462 982 144 × 2 = 0 + 0.317 796 925 964 288;
  • 30) 0.317 796 925 964 288 × 2 = 0 + 0.635 593 851 928 576;
  • 31) 0.635 593 851 928 576 × 2 = 1 + 0.271 187 703 857 152;
  • 32) 0.271 187 703 857 152 × 2 = 0 + 0.542 375 407 714 304;
  • 33) 0.542 375 407 714 304 × 2 = 1 + 0.084 750 815 428 608;
  • 34) 0.084 750 815 428 608 × 2 = 0 + 0.169 501 630 857 216;
  • 35) 0.169 501 630 857 216 × 2 = 0 + 0.339 003 261 714 432;
  • 36) 0.339 003 261 714 432 × 2 = 0 + 0.678 006 523 428 864;
  • 37) 0.678 006 523 428 864 × 2 = 1 + 0.356 013 046 857 728;
  • 38) 0.356 013 046 857 728 × 2 = 0 + 0.712 026 093 715 456;
  • 39) 0.712 026 093 715 456 × 2 = 1 + 0.424 052 187 430 912;
  • 40) 0.424 052 187 430 912 × 2 = 0 + 0.848 104 374 861 824;
  • 41) 0.848 104 374 861 824 × 2 = 1 + 0.696 208 749 723 648;
  • 42) 0.696 208 749 723 648 × 2 = 1 + 0.392 417 499 447 296;
  • 43) 0.392 417 499 447 296 × 2 = 0 + 0.784 834 998 894 592;
  • 44) 0.784 834 998 894 592 × 2 = 1 + 0.569 669 997 789 184;
  • 45) 0.569 669 997 789 184 × 2 = 1 + 0.139 339 995 578 368;
  • 46) 0.139 339 995 578 368 × 2 = 0 + 0.278 679 991 156 736;
  • 47) 0.278 679 991 156 736 × 2 = 0 + 0.557 359 982 313 472;
  • 48) 0.557 359 982 313 472 × 2 = 1 + 0.114 719 964 626 944;
  • 49) 0.114 719 964 626 944 × 2 = 0 + 0.229 439 929 253 888;
  • 50) 0.229 439 929 253 888 × 2 = 0 + 0.458 879 858 507 776;
  • 51) 0.458 879 858 507 776 × 2 = 0 + 0.917 759 717 015 552;
  • 52) 0.917 759 717 015 552 × 2 = 1 + 0.835 519 434 031 104;
  • 53) 0.835 519 434 031 104 × 2 = 1 + 0.671 038 868 062 208;

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:


0.718 281 828 458 974(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1101 1001 0001 1(2)

5. Positive number before normalization:

2.718 281 828 458 974(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1101 1001 0001 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:


2.718 281 828 458 974(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1101 1001 0001 1(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1101 1001 0001 1(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 1000 11(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):
1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 1000 11


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. 0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 1000 11 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 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) =
0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 1000


Decimal number 2.718 281 828 458 974 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0110 1100 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