0.707 106 781 186 546 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.707 106 781 186 546 43(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
0.707 106 781 186 546 43(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.707 106 781 186 546 43.

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) 0.707 106 781 186 546 43 × 2 = 1 + 0.414 213 562 373 092 86;
  • 2) 0.414 213 562 373 092 86 × 2 = 0 + 0.828 427 124 746 185 72;
  • 3) 0.828 427 124 746 185 72 × 2 = 1 + 0.656 854 249 492 371 44;
  • 4) 0.656 854 249 492 371 44 × 2 = 1 + 0.313 708 498 984 742 88;
  • 5) 0.313 708 498 984 742 88 × 2 = 0 + 0.627 416 997 969 485 76;
  • 6) 0.627 416 997 969 485 76 × 2 = 1 + 0.254 833 995 938 971 52;
  • 7) 0.254 833 995 938 971 52 × 2 = 0 + 0.509 667 991 877 943 04;
  • 8) 0.509 667 991 877 943 04 × 2 = 1 + 0.019 335 983 755 886 08;
  • 9) 0.019 335 983 755 886 08 × 2 = 0 + 0.038 671 967 511 772 16;
  • 10) 0.038 671 967 511 772 16 × 2 = 0 + 0.077 343 935 023 544 32;
  • 11) 0.077 343 935 023 544 32 × 2 = 0 + 0.154 687 870 047 088 64;
  • 12) 0.154 687 870 047 088 64 × 2 = 0 + 0.309 375 740 094 177 28;
  • 13) 0.309 375 740 094 177 28 × 2 = 0 + 0.618 751 480 188 354 56;
  • 14) 0.618 751 480 188 354 56 × 2 = 1 + 0.237 502 960 376 709 12;
  • 15) 0.237 502 960 376 709 12 × 2 = 0 + 0.475 005 920 753 418 24;
  • 16) 0.475 005 920 753 418 24 × 2 = 0 + 0.950 011 841 506 836 48;
  • 17) 0.950 011 841 506 836 48 × 2 = 1 + 0.900 023 683 013 672 96;
  • 18) 0.900 023 683 013 672 96 × 2 = 1 + 0.800 047 366 027 345 92;
  • 19) 0.800 047 366 027 345 92 × 2 = 1 + 0.600 094 732 054 691 84;
  • 20) 0.600 094 732 054 691 84 × 2 = 1 + 0.200 189 464 109 383 68;
  • 21) 0.200 189 464 109 383 68 × 2 = 0 + 0.400 378 928 218 767 36;
  • 22) 0.400 378 928 218 767 36 × 2 = 0 + 0.800 757 856 437 534 72;
  • 23) 0.800 757 856 437 534 72 × 2 = 1 + 0.601 515 712 875 069 44;
  • 24) 0.601 515 712 875 069 44 × 2 = 1 + 0.203 031 425 750 138 88;
  • 25) 0.203 031 425 750 138 88 × 2 = 0 + 0.406 062 851 500 277 76;
  • 26) 0.406 062 851 500 277 76 × 2 = 0 + 0.812 125 703 000 555 52;
  • 27) 0.812 125 703 000 555 52 × 2 = 1 + 0.624 251 406 001 111 04;
  • 28) 0.624 251 406 001 111 04 × 2 = 1 + 0.248 502 812 002 222 08;
  • 29) 0.248 502 812 002 222 08 × 2 = 0 + 0.497 005 624 004 444 16;
  • 30) 0.497 005 624 004 444 16 × 2 = 0 + 0.994 011 248 008 888 32;
  • 31) 0.994 011 248 008 888 32 × 2 = 1 + 0.988 022 496 017 776 64;
  • 32) 0.988 022 496 017 776 64 × 2 = 1 + 0.976 044 992 035 553 28;
  • 33) 0.976 044 992 035 553 28 × 2 = 1 + 0.952 089 984 071 106 56;
  • 34) 0.952 089 984 071 106 56 × 2 = 1 + 0.904 179 968 142 213 12;
  • 35) 0.904 179 968 142 213 12 × 2 = 1 + 0.808 359 936 284 426 24;
  • 36) 0.808 359 936 284 426 24 × 2 = 1 + 0.616 719 872 568 852 48;
  • 37) 0.616 719 872 568 852 48 × 2 = 1 + 0.233 439 745 137 704 96;
  • 38) 0.233 439 745 137 704 96 × 2 = 0 + 0.466 879 490 275 409 92;
  • 39) 0.466 879 490 275 409 92 × 2 = 0 + 0.933 758 980 550 819 84;
  • 40) 0.933 758 980 550 819 84 × 2 = 1 + 0.867 517 961 101 639 68;
  • 41) 0.867 517 961 101 639 68 × 2 = 1 + 0.735 035 922 203 279 36;
  • 42) 0.735 035 922 203 279 36 × 2 = 1 + 0.470 071 844 406 558 72;
  • 43) 0.470 071 844 406 558 72 × 2 = 0 + 0.940 143 688 813 117 44;
  • 44) 0.940 143 688 813 117 44 × 2 = 1 + 0.880 287 377 626 234 88;
  • 45) 0.880 287 377 626 234 88 × 2 = 1 + 0.760 574 755 252 469 76;
  • 46) 0.760 574 755 252 469 76 × 2 = 1 + 0.521 149 510 504 939 52;
  • 47) 0.521 149 510 504 939 52 × 2 = 1 + 0.042 299 021 009 879 04;
  • 48) 0.042 299 021 009 879 04 × 2 = 0 + 0.084 598 042 019 758 08;
  • 49) 0.084 598 042 019 758 08 × 2 = 0 + 0.169 196 084 039 516 16;
  • 50) 0.169 196 084 039 516 16 × 2 = 0 + 0.338 392 168 079 032 32;
  • 51) 0.338 392 168 079 032 32 × 2 = 0 + 0.676 784 336 158 064 64;
  • 52) 0.676 784 336 158 064 64 × 2 = 1 + 0.353 568 672 316 129 28;
  • 53) 0.353 568 672 316 129 28 × 2 = 0 + 0.707 137 344 632 258 56;

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:


0.707 106 781 186 546 43(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0001 0(2)

5. Positive number before normalization:

0.707 106 781 186 546 43(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.707 106 781 186 546 43(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0001 0(2) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0001 0(2) × 20 =

1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010(2) × 2-1


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):
1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010


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) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1022(10) =

011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010 =

0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010


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) =
011 1111 1110

Mantissa (52 bits) =
0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010


Decimal number 0.707 106 781 186 546 43 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 0010

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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