Convert 265.105 000 000 000 1 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

265.105 000 000 000 1(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 265.
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;
  • 265 ÷ 2 = 132 + 1;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.

265(10) =


1 0000 1001(2)


3. Convert to the binary (base 2) the fractional part: 0.105 000 000 000 1.

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) 0.105 000 000 000 1 × 2 = 0 + 0.210 000 000 000 2;
  • 2) 0.210 000 000 000 2 × 2 = 0 + 0.420 000 000 000 4;
  • 3) 0.420 000 000 000 4 × 2 = 0 + 0.840 000 000 000 8;
  • 4) 0.840 000 000 000 8 × 2 = 1 + 0.680 000 000 001 6;
  • 5) 0.680 000 000 001 6 × 2 = 1 + 0.360 000 000 003 2;
  • 6) 0.360 000 000 003 2 × 2 = 0 + 0.720 000 000 006 4;
  • 7) 0.720 000 000 006 4 × 2 = 1 + 0.440 000 000 012 8;
  • 8) 0.440 000 000 012 8 × 2 = 0 + 0.880 000 000 025 6;
  • 9) 0.880 000 000 025 6 × 2 = 1 + 0.760 000 000 051 2;
  • 10) 0.760 000 000 051 2 × 2 = 1 + 0.520 000 000 102 4;
  • 11) 0.520 000 000 102 4 × 2 = 1 + 0.040 000 000 204 8;
  • 12) 0.040 000 000 204 8 × 2 = 0 + 0.080 000 000 409 6;
  • 13) 0.080 000 000 409 6 × 2 = 0 + 0.160 000 000 819 2;
  • 14) 0.160 000 000 819 2 × 2 = 0 + 0.320 000 001 638 4;
  • 15) 0.320 000 001 638 4 × 2 = 0 + 0.640 000 003 276 8;
  • 16) 0.640 000 003 276 8 × 2 = 1 + 0.280 000 006 553 6;
  • 17) 0.280 000 006 553 6 × 2 = 0 + 0.560 000 013 107 2;
  • 18) 0.560 000 013 107 2 × 2 = 1 + 0.120 000 026 214 4;
  • 19) 0.120 000 026 214 4 × 2 = 0 + 0.240 000 052 428 8;
  • 20) 0.240 000 052 428 8 × 2 = 0 + 0.480 000 104 857 6;
  • 21) 0.480 000 104 857 6 × 2 = 0 + 0.960 000 209 715 2;
  • 22) 0.960 000 209 715 2 × 2 = 1 + 0.920 000 419 430 4;
  • 23) 0.920 000 419 430 4 × 2 = 1 + 0.840 000 838 860 8;
  • 24) 0.840 000 838 860 8 × 2 = 1 + 0.680 001 677 721 6;
  • 25) 0.680 001 677 721 6 × 2 = 1 + 0.360 003 355 443 2;
  • 26) 0.360 003 355 443 2 × 2 = 0 + 0.720 006 710 886 4;
  • 27) 0.720 006 710 886 4 × 2 = 1 + 0.440 013 421 772 8;
  • 28) 0.440 013 421 772 8 × 2 = 0 + 0.880 026 843 545 6;
  • 29) 0.880 026 843 545 6 × 2 = 1 + 0.760 053 687 091 2;
  • 30) 0.760 053 687 091 2 × 2 = 1 + 0.520 107 374 182 4;
  • 31) 0.520 107 374 182 4 × 2 = 1 + 0.040 214 748 364 8;
  • 32) 0.040 214 748 364 8 × 2 = 0 + 0.080 429 496 729 6;
  • 33) 0.080 429 496 729 6 × 2 = 0 + 0.160 858 993 459 2;
  • 34) 0.160 858 993 459 2 × 2 = 0 + 0.321 717 986 918 4;
  • 35) 0.321 717 986 918 4 × 2 = 0 + 0.643 435 973 836 8;
  • 36) 0.643 435 973 836 8 × 2 = 1 + 0.286 871 947 673 6;
  • 37) 0.286 871 947 673 6 × 2 = 0 + 0.573 743 895 347 2;
  • 38) 0.573 743 895 347 2 × 2 = 1 + 0.147 487 790 694 4;
  • 39) 0.147 487 790 694 4 × 2 = 0 + 0.294 975 581 388 8;
  • 40) 0.294 975 581 388 8 × 2 = 0 + 0.589 951 162 777 6;
  • 41) 0.589 951 162 777 6 × 2 = 1 + 0.179 902 325 555 2;
  • 42) 0.179 902 325 555 2 × 2 = 0 + 0.359 804 651 110 4;
  • 43) 0.359 804 651 110 4 × 2 = 0 + 0.719 609 302 220 8;
  • 44) 0.719 609 302 220 8 × 2 = 1 + 0.439 218 604 441 6;
  • 45) 0.439 218 604 441 6 × 2 = 0 + 0.878 437 208 883 2;
  • 46) 0.878 437 208 883 2 × 2 = 1 + 0.756 874 417 766 4;
  • 47) 0.756 874 417 766 4 × 2 = 1 + 0.513 748 835 532 8;
  • 48) 0.513 748 835 532 8 × 2 = 1 + 0.027 497 671 065 6;
  • 49) 0.027 497 671 065 6 × 2 = 0 + 0.054 995 342 131 2;
  • 50) 0.054 995 342 131 2 × 2 = 0 + 0.109 990 684 262 4;
  • 51) 0.109 990 684 262 4 × 2 = 0 + 0.219 981 368 524 8;
  • 52) 0.219 981 368 524 8 × 2 = 0 + 0.439 962 737 049 6;
  • 53) 0.439 962 737 049 6 × 2 = 0 + 0.879 925 474 099 2;

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...)


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:

0.105 000 000 000 1(10) =


0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0(2)


5. Positive number before normalization:

265.105 000 000 000 1(10) =


1 0000 1001.0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left so that only one non zero digit remains to the left of it:

265.105 000 000 000 1(10) =


1 0000 1001.0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0(2) =


1 0000 1001.0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0(2) × 20 =


1.0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0(2) × 28


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): 8


Mantissa (not normalized):
1.0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0111 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


8 + 2(11-1) - 1 =


(8 + 1 023)(10) =


1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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) =


1031(10) =


100 0000 0111(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. 0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001 0 1110 0000 =


0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 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 0111


Mantissa (52 bits) =
0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001


Number 265.105 000 000 000 1 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0111 - 0000 1001 0001 1010 1110 0001 0100 0111 1010 1110 0001 0100 1001

(64 bits IEEE 754)

More operations of this kind:

265.105 = ? ... 265.105 000 000 000 2 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

265.105 000 000 000 1 to 64 bit double precision IEEE 754 binary floating point = ? Dec 02 22:58 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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