1.797 693 134 862 03 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 862 03(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
1.797 693 134 862 03(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: 0.797 693 134 862 03.

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) 0.797 693 134 862 03 × 2 = 1 + 0.595 386 269 724 06;
  • 2) 0.595 386 269 724 06 × 2 = 1 + 0.190 772 539 448 12;
  • 3) 0.190 772 539 448 12 × 2 = 0 + 0.381 545 078 896 24;
  • 4) 0.381 545 078 896 24 × 2 = 0 + 0.763 090 157 792 48;
  • 5) 0.763 090 157 792 48 × 2 = 1 + 0.526 180 315 584 96;
  • 6) 0.526 180 315 584 96 × 2 = 1 + 0.052 360 631 169 92;
  • 7) 0.052 360 631 169 92 × 2 = 0 + 0.104 721 262 339 84;
  • 8) 0.104 721 262 339 84 × 2 = 0 + 0.209 442 524 679 68;
  • 9) 0.209 442 524 679 68 × 2 = 0 + 0.418 885 049 359 36;
  • 10) 0.418 885 049 359 36 × 2 = 0 + 0.837 770 098 718 72;
  • 11) 0.837 770 098 718 72 × 2 = 1 + 0.675 540 197 437 44;
  • 12) 0.675 540 197 437 44 × 2 = 1 + 0.351 080 394 874 88;
  • 13) 0.351 080 394 874 88 × 2 = 0 + 0.702 160 789 749 76;
  • 14) 0.702 160 789 749 76 × 2 = 1 + 0.404 321 579 499 52;
  • 15) 0.404 321 579 499 52 × 2 = 0 + 0.808 643 158 999 04;
  • 16) 0.808 643 158 999 04 × 2 = 1 + 0.617 286 317 998 08;
  • 17) 0.617 286 317 998 08 × 2 = 1 + 0.234 572 635 996 16;
  • 18) 0.234 572 635 996 16 × 2 = 0 + 0.469 145 271 992 32;
  • 19) 0.469 145 271 992 32 × 2 = 0 + 0.938 290 543 984 64;
  • 20) 0.938 290 543 984 64 × 2 = 1 + 0.876 581 087 969 28;
  • 21) 0.876 581 087 969 28 × 2 = 1 + 0.753 162 175 938 56;
  • 22) 0.753 162 175 938 56 × 2 = 1 + 0.506 324 351 877 12;
  • 23) 0.506 324 351 877 12 × 2 = 1 + 0.012 648 703 754 24;
  • 24) 0.012 648 703 754 24 × 2 = 0 + 0.025 297 407 508 48;
  • 25) 0.025 297 407 508 48 × 2 = 0 + 0.050 594 815 016 96;
  • 26) 0.050 594 815 016 96 × 2 = 0 + 0.101 189 630 033 92;
  • 27) 0.101 189 630 033 92 × 2 = 0 + 0.202 379 260 067 84;
  • 28) 0.202 379 260 067 84 × 2 = 0 + 0.404 758 520 135 68;
  • 29) 0.404 758 520 135 68 × 2 = 0 + 0.809 517 040 271 36;
  • 30) 0.809 517 040 271 36 × 2 = 1 + 0.619 034 080 542 72;
  • 31) 0.619 034 080 542 72 × 2 = 1 + 0.238 068 161 085 44;
  • 32) 0.238 068 161 085 44 × 2 = 0 + 0.476 136 322 170 88;
  • 33) 0.476 136 322 170 88 × 2 = 0 + 0.952 272 644 341 76;
  • 34) 0.952 272 644 341 76 × 2 = 1 + 0.904 545 288 683 52;
  • 35) 0.904 545 288 683 52 × 2 = 1 + 0.809 090 577 367 04;
  • 36) 0.809 090 577 367 04 × 2 = 1 + 0.618 181 154 734 08;
  • 37) 0.618 181 154 734 08 × 2 = 1 + 0.236 362 309 468 16;
  • 38) 0.236 362 309 468 16 × 2 = 0 + 0.472 724 618 936 32;
  • 39) 0.472 724 618 936 32 × 2 = 0 + 0.945 449 237 872 64;
  • 40) 0.945 449 237 872 64 × 2 = 1 + 0.890 898 475 745 28;
  • 41) 0.890 898 475 745 28 × 2 = 1 + 0.781 796 951 490 56;
  • 42) 0.781 796 951 490 56 × 2 = 1 + 0.563 593 902 981 12;
  • 43) 0.563 593 902 981 12 × 2 = 1 + 0.127 187 805 962 24;
  • 44) 0.127 187 805 962 24 × 2 = 0 + 0.254 375 611 924 48;
  • 45) 0.254 375 611 924 48 × 2 = 0 + 0.508 751 223 848 96;
  • 46) 0.508 751 223 848 96 × 2 = 1 + 0.017 502 447 697 92;
  • 47) 0.017 502 447 697 92 × 2 = 0 + 0.035 004 895 395 84;
  • 48) 0.035 004 895 395 84 × 2 = 0 + 0.070 009 790 791 68;
  • 49) 0.070 009 790 791 68 × 2 = 0 + 0.140 019 581 583 36;
  • 50) 0.140 019 581 583 36 × 2 = 0 + 0.280 039 163 166 72;
  • 51) 0.280 039 163 166 72 × 2 = 0 + 0.560 078 326 333 44;
  • 52) 0.560 078 326 333 44 × 2 = 1 + 0.120 156 652 666 88;
  • 53) 0.120 156 652 666 88 × 2 = 0 + 0.240 313 305 333 76;

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:


0.797 693 134 862 03(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 0(2)

5. Positive number before normalization:

1.797 693 134 862 03(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 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:


1.797 693 134 862 03(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 0(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 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):
1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 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. 1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001 0 =

1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001


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) =
1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001


Decimal number 1.797 693 134 862 03 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1110 0100 0001

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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