Decimal to 64 Bit IEEE 754 Binary: Convert Number 1.797 693 134 862 26 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 1.797 693 134 862 26(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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 26.

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 26 × 2 = 1 + 0.595 386 269 724 52;
  • 2) 0.595 386 269 724 52 × 2 = 1 + 0.190 772 539 449 04;
  • 3) 0.190 772 539 449 04 × 2 = 0 + 0.381 545 078 898 08;
  • 4) 0.381 545 078 898 08 × 2 = 0 + 0.763 090 157 796 16;
  • 5) 0.763 090 157 796 16 × 2 = 1 + 0.526 180 315 592 32;
  • 6) 0.526 180 315 592 32 × 2 = 1 + 0.052 360 631 184 64;
  • 7) 0.052 360 631 184 64 × 2 = 0 + 0.104 721 262 369 28;
  • 8) 0.104 721 262 369 28 × 2 = 0 + 0.209 442 524 738 56;
  • 9) 0.209 442 524 738 56 × 2 = 0 + 0.418 885 049 477 12;
  • 10) 0.418 885 049 477 12 × 2 = 0 + 0.837 770 098 954 24;
  • 11) 0.837 770 098 954 24 × 2 = 1 + 0.675 540 197 908 48;
  • 12) 0.675 540 197 908 48 × 2 = 1 + 0.351 080 395 816 96;
  • 13) 0.351 080 395 816 96 × 2 = 0 + 0.702 160 791 633 92;
  • 14) 0.702 160 791 633 92 × 2 = 1 + 0.404 321 583 267 84;
  • 15) 0.404 321 583 267 84 × 2 = 0 + 0.808 643 166 535 68;
  • 16) 0.808 643 166 535 68 × 2 = 1 + 0.617 286 333 071 36;
  • 17) 0.617 286 333 071 36 × 2 = 1 + 0.234 572 666 142 72;
  • 18) 0.234 572 666 142 72 × 2 = 0 + 0.469 145 332 285 44;
  • 19) 0.469 145 332 285 44 × 2 = 0 + 0.938 290 664 570 88;
  • 20) 0.938 290 664 570 88 × 2 = 1 + 0.876 581 329 141 76;
  • 21) 0.876 581 329 141 76 × 2 = 1 + 0.753 162 658 283 52;
  • 22) 0.753 162 658 283 52 × 2 = 1 + 0.506 325 316 567 04;
  • 23) 0.506 325 316 567 04 × 2 = 1 + 0.012 650 633 134 08;
  • 24) 0.012 650 633 134 08 × 2 = 0 + 0.025 301 266 268 16;
  • 25) 0.025 301 266 268 16 × 2 = 0 + 0.050 602 532 536 32;
  • 26) 0.050 602 532 536 32 × 2 = 0 + 0.101 205 065 072 64;
  • 27) 0.101 205 065 072 64 × 2 = 0 + 0.202 410 130 145 28;
  • 28) 0.202 410 130 145 28 × 2 = 0 + 0.404 820 260 290 56;
  • 29) 0.404 820 260 290 56 × 2 = 0 + 0.809 640 520 581 12;
  • 30) 0.809 640 520 581 12 × 2 = 1 + 0.619 281 041 162 24;
  • 31) 0.619 281 041 162 24 × 2 = 1 + 0.238 562 082 324 48;
  • 32) 0.238 562 082 324 48 × 2 = 0 + 0.477 124 164 648 96;
  • 33) 0.477 124 164 648 96 × 2 = 0 + 0.954 248 329 297 92;
  • 34) 0.954 248 329 297 92 × 2 = 1 + 0.908 496 658 595 84;
  • 35) 0.908 496 658 595 84 × 2 = 1 + 0.816 993 317 191 68;
  • 36) 0.816 993 317 191 68 × 2 = 1 + 0.633 986 634 383 36;
  • 37) 0.633 986 634 383 36 × 2 = 1 + 0.267 973 268 766 72;
  • 38) 0.267 973 268 766 72 × 2 = 0 + 0.535 946 537 533 44;
  • 39) 0.535 946 537 533 44 × 2 = 1 + 0.071 893 075 066 88;
  • 40) 0.071 893 075 066 88 × 2 = 0 + 0.143 786 150 133 76;
  • 41) 0.143 786 150 133 76 × 2 = 0 + 0.287 572 300 267 52;
  • 42) 0.287 572 300 267 52 × 2 = 0 + 0.575 144 600 535 04;
  • 43) 0.575 144 600 535 04 × 2 = 1 + 0.150 289 201 070 08;
  • 44) 0.150 289 201 070 08 × 2 = 0 + 0.300 578 402 140 16;
  • 45) 0.300 578 402 140 16 × 2 = 0 + 0.601 156 804 280 32;
  • 46) 0.601 156 804 280 32 × 2 = 1 + 0.202 313 608 560 64;
  • 47) 0.202 313 608 560 64 × 2 = 0 + 0.404 627 217 121 28;
  • 48) 0.404 627 217 121 28 × 2 = 0 + 0.809 254 434 242 56;
  • 49) 0.809 254 434 242 56 × 2 = 1 + 0.618 508 868 485 12;
  • 50) 0.618 508 868 485 12 × 2 = 1 + 0.237 017 736 970 24;
  • 51) 0.237 017 736 970 24 × 2 = 0 + 0.474 035 473 940 48;
  • 52) 0.474 035 473 940 48 × 2 = 0 + 0.948 070 947 880 96;
  • 53) 0.948 070 947 880 96 × 2 = 1 + 0.896 141 895 761 92;

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 26(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100 1(2)

5. Positive number before normalization:

1.797 693 134 862 26(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100 1(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 26(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100 1(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100 1(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 1010 0010 0100 1100 1


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 1010 0010 0100 1100 1 =

1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100


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 1010 0010 0100 1100


The base ten decimal number 1.797 693 134 862 26 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0010 0100 1100

» Decimal number in base ten 1.797 693 134 863 18 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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