1.797 693 134 862 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 862 6(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 6(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 6.

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 6 × 2 = 1 + 0.595 386 269 725 2;
  • 2) 0.595 386 269 725 2 × 2 = 1 + 0.190 772 539 450 4;
  • 3) 0.190 772 539 450 4 × 2 = 0 + 0.381 545 078 900 8;
  • 4) 0.381 545 078 900 8 × 2 = 0 + 0.763 090 157 801 6;
  • 5) 0.763 090 157 801 6 × 2 = 1 + 0.526 180 315 603 2;
  • 6) 0.526 180 315 603 2 × 2 = 1 + 0.052 360 631 206 4;
  • 7) 0.052 360 631 206 4 × 2 = 0 + 0.104 721 262 412 8;
  • 8) 0.104 721 262 412 8 × 2 = 0 + 0.209 442 524 825 6;
  • 9) 0.209 442 524 825 6 × 2 = 0 + 0.418 885 049 651 2;
  • 10) 0.418 885 049 651 2 × 2 = 0 + 0.837 770 099 302 4;
  • 11) 0.837 770 099 302 4 × 2 = 1 + 0.675 540 198 604 8;
  • 12) 0.675 540 198 604 8 × 2 = 1 + 0.351 080 397 209 6;
  • 13) 0.351 080 397 209 6 × 2 = 0 + 0.702 160 794 419 2;
  • 14) 0.702 160 794 419 2 × 2 = 1 + 0.404 321 588 838 4;
  • 15) 0.404 321 588 838 4 × 2 = 0 + 0.808 643 177 676 8;
  • 16) 0.808 643 177 676 8 × 2 = 1 + 0.617 286 355 353 6;
  • 17) 0.617 286 355 353 6 × 2 = 1 + 0.234 572 710 707 2;
  • 18) 0.234 572 710 707 2 × 2 = 0 + 0.469 145 421 414 4;
  • 19) 0.469 145 421 414 4 × 2 = 0 + 0.938 290 842 828 8;
  • 20) 0.938 290 842 828 8 × 2 = 1 + 0.876 581 685 657 6;
  • 21) 0.876 581 685 657 6 × 2 = 1 + 0.753 163 371 315 2;
  • 22) 0.753 163 371 315 2 × 2 = 1 + 0.506 326 742 630 4;
  • 23) 0.506 326 742 630 4 × 2 = 1 + 0.012 653 485 260 8;
  • 24) 0.012 653 485 260 8 × 2 = 0 + 0.025 306 970 521 6;
  • 25) 0.025 306 970 521 6 × 2 = 0 + 0.050 613 941 043 2;
  • 26) 0.050 613 941 043 2 × 2 = 0 + 0.101 227 882 086 4;
  • 27) 0.101 227 882 086 4 × 2 = 0 + 0.202 455 764 172 8;
  • 28) 0.202 455 764 172 8 × 2 = 0 + 0.404 911 528 345 6;
  • 29) 0.404 911 528 345 6 × 2 = 0 + 0.809 823 056 691 2;
  • 30) 0.809 823 056 691 2 × 2 = 1 + 0.619 646 113 382 4;
  • 31) 0.619 646 113 382 4 × 2 = 1 + 0.239 292 226 764 8;
  • 32) 0.239 292 226 764 8 × 2 = 0 + 0.478 584 453 529 6;
  • 33) 0.478 584 453 529 6 × 2 = 0 + 0.957 168 907 059 2;
  • 34) 0.957 168 907 059 2 × 2 = 1 + 0.914 337 814 118 4;
  • 35) 0.914 337 814 118 4 × 2 = 1 + 0.828 675 628 236 8;
  • 36) 0.828 675 628 236 8 × 2 = 1 + 0.657 351 256 473 6;
  • 37) 0.657 351 256 473 6 × 2 = 1 + 0.314 702 512 947 2;
  • 38) 0.314 702 512 947 2 × 2 = 0 + 0.629 405 025 894 4;
  • 39) 0.629 405 025 894 4 × 2 = 1 + 0.258 810 051 788 8;
  • 40) 0.258 810 051 788 8 × 2 = 0 + 0.517 620 103 577 6;
  • 41) 0.517 620 103 577 6 × 2 = 1 + 0.035 240 207 155 2;
  • 42) 0.035 240 207 155 2 × 2 = 0 + 0.070 480 414 310 4;
  • 43) 0.070 480 414 310 4 × 2 = 0 + 0.140 960 828 620 8;
  • 44) 0.140 960 828 620 8 × 2 = 0 + 0.281 921 657 241 6;
  • 45) 0.281 921 657 241 6 × 2 = 0 + 0.563 843 314 483 2;
  • 46) 0.563 843 314 483 2 × 2 = 1 + 0.127 686 628 966 4;
  • 47) 0.127 686 628 966 4 × 2 = 0 + 0.255 373 257 932 8;
  • 48) 0.255 373 257 932 8 × 2 = 0 + 0.510 746 515 865 6;
  • 49) 0.510 746 515 865 6 × 2 = 1 + 0.021 493 031 731 2;
  • 50) 0.021 493 031 731 2 × 2 = 0 + 0.042 986 063 462 4;
  • 51) 0.042 986 063 462 4 × 2 = 0 + 0.085 972 126 924 8;
  • 52) 0.085 972 126 924 8 × 2 = 0 + 0.171 944 253 849 6;
  • 53) 0.171 944 253 849 6 × 2 = 0 + 0.343 888 507 699 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 - 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 6(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 1000 0(2)

5. Positive number before normalization:

1.797 693 134 862 6(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 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:


1.797 693 134 862 6(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 1000 0(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 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):
1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 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. 1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 1000 0 =

1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 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) =
1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 1000 0100 1000


Decimal number 1.797 693 134 862 6 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 1010 1000 0100 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