1.797 693 134 863 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 863 09(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
1.797 693 134 863 09(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: 0.797 693 134 863 09.

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) 0.797 693 134 863 09 × 2 = 1 + 0.595 386 269 726 18;
  • 2) 0.595 386 269 726 18 × 2 = 1 + 0.190 772 539 452 36;
  • 3) 0.190 772 539 452 36 × 2 = 0 + 0.381 545 078 904 72;
  • 4) 0.381 545 078 904 72 × 2 = 0 + 0.763 090 157 809 44;
  • 5) 0.763 090 157 809 44 × 2 = 1 + 0.526 180 315 618 88;
  • 6) 0.526 180 315 618 88 × 2 = 1 + 0.052 360 631 237 76;
  • 7) 0.052 360 631 237 76 × 2 = 0 + 0.104 721 262 475 52;
  • 8) 0.104 721 262 475 52 × 2 = 0 + 0.209 442 524 951 04;
  • 9) 0.209 442 524 951 04 × 2 = 0 + 0.418 885 049 902 08;
  • 10) 0.418 885 049 902 08 × 2 = 0 + 0.837 770 099 804 16;
  • 11) 0.837 770 099 804 16 × 2 = 1 + 0.675 540 199 608 32;
  • 12) 0.675 540 199 608 32 × 2 = 1 + 0.351 080 399 216 64;
  • 13) 0.351 080 399 216 64 × 2 = 0 + 0.702 160 798 433 28;
  • 14) 0.702 160 798 433 28 × 2 = 1 + 0.404 321 596 866 56;
  • 15) 0.404 321 596 866 56 × 2 = 0 + 0.808 643 193 733 12;
  • 16) 0.808 643 193 733 12 × 2 = 1 + 0.617 286 387 466 24;
  • 17) 0.617 286 387 466 24 × 2 = 1 + 0.234 572 774 932 48;
  • 18) 0.234 572 774 932 48 × 2 = 0 + 0.469 145 549 864 96;
  • 19) 0.469 145 549 864 96 × 2 = 0 + 0.938 291 099 729 92;
  • 20) 0.938 291 099 729 92 × 2 = 1 + 0.876 582 199 459 84;
  • 21) 0.876 582 199 459 84 × 2 = 1 + 0.753 164 398 919 68;
  • 22) 0.753 164 398 919 68 × 2 = 1 + 0.506 328 797 839 36;
  • 23) 0.506 328 797 839 36 × 2 = 1 + 0.012 657 595 678 72;
  • 24) 0.012 657 595 678 72 × 2 = 0 + 0.025 315 191 357 44;
  • 25) 0.025 315 191 357 44 × 2 = 0 + 0.050 630 382 714 88;
  • 26) 0.050 630 382 714 88 × 2 = 0 + 0.101 260 765 429 76;
  • 27) 0.101 260 765 429 76 × 2 = 0 + 0.202 521 530 859 52;
  • 28) 0.202 521 530 859 52 × 2 = 0 + 0.405 043 061 719 04;
  • 29) 0.405 043 061 719 04 × 2 = 0 + 0.810 086 123 438 08;
  • 30) 0.810 086 123 438 08 × 2 = 1 + 0.620 172 246 876 16;
  • 31) 0.620 172 246 876 16 × 2 = 1 + 0.240 344 493 752 32;
  • 32) 0.240 344 493 752 32 × 2 = 0 + 0.480 688 987 504 64;
  • 33) 0.480 688 987 504 64 × 2 = 0 + 0.961 377 975 009 28;
  • 34) 0.961 377 975 009 28 × 2 = 1 + 0.922 755 950 018 56;
  • 35) 0.922 755 950 018 56 × 2 = 1 + 0.845 511 900 037 12;
  • 36) 0.845 511 900 037 12 × 2 = 1 + 0.691 023 800 074 24;
  • 37) 0.691 023 800 074 24 × 2 = 1 + 0.382 047 600 148 48;
  • 38) 0.382 047 600 148 48 × 2 = 0 + 0.764 095 200 296 96;
  • 39) 0.764 095 200 296 96 × 2 = 1 + 0.528 190 400 593 92;
  • 40) 0.528 190 400 593 92 × 2 = 1 + 0.056 380 801 187 84;
  • 41) 0.056 380 801 187 84 × 2 = 0 + 0.112 761 602 375 68;
  • 42) 0.112 761 602 375 68 × 2 = 0 + 0.225 523 204 751 36;
  • 43) 0.225 523 204 751 36 × 2 = 0 + 0.451 046 409 502 72;
  • 44) 0.451 046 409 502 72 × 2 = 0 + 0.902 092 819 005 44;
  • 45) 0.902 092 819 005 44 × 2 = 1 + 0.804 185 638 010 88;
  • 46) 0.804 185 638 010 88 × 2 = 1 + 0.608 371 276 021 76;
  • 47) 0.608 371 276 021 76 × 2 = 1 + 0.216 742 552 043 52;
  • 48) 0.216 742 552 043 52 × 2 = 0 + 0.433 485 104 087 04;
  • 49) 0.433 485 104 087 04 × 2 = 0 + 0.866 970 208 174 08;
  • 50) 0.866 970 208 174 08 × 2 = 1 + 0.733 940 416 348 16;
  • 51) 0.733 940 416 348 16 × 2 = 1 + 0.467 880 832 696 32;
  • 52) 0.467 880 832 696 32 × 2 = 0 + 0.935 761 665 392 64;
  • 53) 0.935 761 665 392 64 × 2 = 1 + 0.871 523 330 785 28;

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:


0.797 693 134 863 09(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1011 0000 1110 0110 1(2)

5. Positive number before normalization:

1.797 693 134 863 09(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1011 0000 1110 0110 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:


1.797 693 134 863 09(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1011 0000 1110 0110 1(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1011 0000 1110 0110 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 1011 0000 1110 0110 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 1011 0000 1110 0110 1 =

1100 1100 0011 0101 1001 1110 0000 0110 0111 1011 0000 1110 0110


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 1011 0000 1110 0110


Decimal number 1.797 693 134 863 09 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 1011 0000 1110 0110

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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