1.834 139 868 826 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.834 139 868 826 89(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
1.834 139 868 826 89(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.834 139 868 826 89.

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) 0.834 139 868 826 89 × 2 = 1 + 0.668 279 737 653 78;
  • 2) 0.668 279 737 653 78 × 2 = 1 + 0.336 559 475 307 56;
  • 3) 0.336 559 475 307 56 × 2 = 0 + 0.673 118 950 615 12;
  • 4) 0.673 118 950 615 12 × 2 = 1 + 0.346 237 901 230 24;
  • 5) 0.346 237 901 230 24 × 2 = 0 + 0.692 475 802 460 48;
  • 6) 0.692 475 802 460 48 × 2 = 1 + 0.384 951 604 920 96;
  • 7) 0.384 951 604 920 96 × 2 = 0 + 0.769 903 209 841 92;
  • 8) 0.769 903 209 841 92 × 2 = 1 + 0.539 806 419 683 84;
  • 9) 0.539 806 419 683 84 × 2 = 1 + 0.079 612 839 367 68;
  • 10) 0.079 612 839 367 68 × 2 = 0 + 0.159 225 678 735 36;
  • 11) 0.159 225 678 735 36 × 2 = 0 + 0.318 451 357 470 72;
  • 12) 0.318 451 357 470 72 × 2 = 0 + 0.636 902 714 941 44;
  • 13) 0.636 902 714 941 44 × 2 = 1 + 0.273 805 429 882 88;
  • 14) 0.273 805 429 882 88 × 2 = 0 + 0.547 610 859 765 76;
  • 15) 0.547 610 859 765 76 × 2 = 1 + 0.095 221 719 531 52;
  • 16) 0.095 221 719 531 52 × 2 = 0 + 0.190 443 439 063 04;
  • 17) 0.190 443 439 063 04 × 2 = 0 + 0.380 886 878 126 08;
  • 18) 0.380 886 878 126 08 × 2 = 0 + 0.761 773 756 252 16;
  • 19) 0.761 773 756 252 16 × 2 = 1 + 0.523 547 512 504 32;
  • 20) 0.523 547 512 504 32 × 2 = 1 + 0.047 095 025 008 64;
  • 21) 0.047 095 025 008 64 × 2 = 0 + 0.094 190 050 017 28;
  • 22) 0.094 190 050 017 28 × 2 = 0 + 0.188 380 100 034 56;
  • 23) 0.188 380 100 034 56 × 2 = 0 + 0.376 760 200 069 12;
  • 24) 0.376 760 200 069 12 × 2 = 0 + 0.753 520 400 138 24;
  • 25) 0.753 520 400 138 24 × 2 = 1 + 0.507 040 800 276 48;
  • 26) 0.507 040 800 276 48 × 2 = 1 + 0.014 081 600 552 96;
  • 27) 0.014 081 600 552 96 × 2 = 0 + 0.028 163 201 105 92;
  • 28) 0.028 163 201 105 92 × 2 = 0 + 0.056 326 402 211 84;
  • 29) 0.056 326 402 211 84 × 2 = 0 + 0.112 652 804 423 68;
  • 30) 0.112 652 804 423 68 × 2 = 0 + 0.225 305 608 847 36;
  • 31) 0.225 305 608 847 36 × 2 = 0 + 0.450 611 217 694 72;
  • 32) 0.450 611 217 694 72 × 2 = 0 + 0.901 222 435 389 44;
  • 33) 0.901 222 435 389 44 × 2 = 1 + 0.802 444 870 778 88;
  • 34) 0.802 444 870 778 88 × 2 = 1 + 0.604 889 741 557 76;
  • 35) 0.604 889 741 557 76 × 2 = 1 + 0.209 779 483 115 52;
  • 36) 0.209 779 483 115 52 × 2 = 0 + 0.419 558 966 231 04;
  • 37) 0.419 558 966 231 04 × 2 = 0 + 0.839 117 932 462 08;
  • 38) 0.839 117 932 462 08 × 2 = 1 + 0.678 235 864 924 16;
  • 39) 0.678 235 864 924 16 × 2 = 1 + 0.356 471 729 848 32;
  • 40) 0.356 471 729 848 32 × 2 = 0 + 0.712 943 459 696 64;
  • 41) 0.712 943 459 696 64 × 2 = 1 + 0.425 886 919 393 28;
  • 42) 0.425 886 919 393 28 × 2 = 0 + 0.851 773 838 786 56;
  • 43) 0.851 773 838 786 56 × 2 = 1 + 0.703 547 677 573 12;
  • 44) 0.703 547 677 573 12 × 2 = 1 + 0.407 095 355 146 24;
  • 45) 0.407 095 355 146 24 × 2 = 0 + 0.814 190 710 292 48;
  • 46) 0.814 190 710 292 48 × 2 = 1 + 0.628 381 420 584 96;
  • 47) 0.628 381 420 584 96 × 2 = 1 + 0.256 762 841 169 92;
  • 48) 0.256 762 841 169 92 × 2 = 0 + 0.513 525 682 339 84;
  • 49) 0.513 525 682 339 84 × 2 = 1 + 0.027 051 364 679 68;
  • 50) 0.027 051 364 679 68 × 2 = 0 + 0.054 102 729 359 36;
  • 51) 0.054 102 729 359 36 × 2 = 0 + 0.108 205 458 718 72;
  • 52) 0.108 205 458 718 72 × 2 = 0 + 0.216 410 917 437 44;
  • 53) 0.216 410 917 437 44 × 2 = 0 + 0.432 821 834 874 88;

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:


0.834 139 868 826 89(10) =

0.1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0(2)

5. Positive number before normalization:

1.834 139 868 826 89(10) =

1.1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.834 139 868 826 89(10) =

1.1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0(2) =

1.1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0(2) × 20


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): 0

Mantissa (not normalized):
1.1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000 0 =

1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 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) =
011 1111 1111

Mantissa (52 bits) =
1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 1000


Decimal number 1.834 139 868 826 89 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1101 0101 1000 1010 0011 0000 1100 0000 1110 0110 1011 0110 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