1.797 693 134 861 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 861 92(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
1.797 693 134 861 92(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: 0.797 693 134 861 92.

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) 0.797 693 134 861 92 × 2 = 1 + 0.595 386 269 723 84;
  • 2) 0.595 386 269 723 84 × 2 = 1 + 0.190 772 539 447 68;
  • 3) 0.190 772 539 447 68 × 2 = 0 + 0.381 545 078 895 36;
  • 4) 0.381 545 078 895 36 × 2 = 0 + 0.763 090 157 790 72;
  • 5) 0.763 090 157 790 72 × 2 = 1 + 0.526 180 315 581 44;
  • 6) 0.526 180 315 581 44 × 2 = 1 + 0.052 360 631 162 88;
  • 7) 0.052 360 631 162 88 × 2 = 0 + 0.104 721 262 325 76;
  • 8) 0.104 721 262 325 76 × 2 = 0 + 0.209 442 524 651 52;
  • 9) 0.209 442 524 651 52 × 2 = 0 + 0.418 885 049 303 04;
  • 10) 0.418 885 049 303 04 × 2 = 0 + 0.837 770 098 606 08;
  • 11) 0.837 770 098 606 08 × 2 = 1 + 0.675 540 197 212 16;
  • 12) 0.675 540 197 212 16 × 2 = 1 + 0.351 080 394 424 32;
  • 13) 0.351 080 394 424 32 × 2 = 0 + 0.702 160 788 848 64;
  • 14) 0.702 160 788 848 64 × 2 = 1 + 0.404 321 577 697 28;
  • 15) 0.404 321 577 697 28 × 2 = 0 + 0.808 643 155 394 56;
  • 16) 0.808 643 155 394 56 × 2 = 1 + 0.617 286 310 789 12;
  • 17) 0.617 286 310 789 12 × 2 = 1 + 0.234 572 621 578 24;
  • 18) 0.234 572 621 578 24 × 2 = 0 + 0.469 145 243 156 48;
  • 19) 0.469 145 243 156 48 × 2 = 0 + 0.938 290 486 312 96;
  • 20) 0.938 290 486 312 96 × 2 = 1 + 0.876 580 972 625 92;
  • 21) 0.876 580 972 625 92 × 2 = 1 + 0.753 161 945 251 84;
  • 22) 0.753 161 945 251 84 × 2 = 1 + 0.506 323 890 503 68;
  • 23) 0.506 323 890 503 68 × 2 = 1 + 0.012 647 781 007 36;
  • 24) 0.012 647 781 007 36 × 2 = 0 + 0.025 295 562 014 72;
  • 25) 0.025 295 562 014 72 × 2 = 0 + 0.050 591 124 029 44;
  • 26) 0.050 591 124 029 44 × 2 = 0 + 0.101 182 248 058 88;
  • 27) 0.101 182 248 058 88 × 2 = 0 + 0.202 364 496 117 76;
  • 28) 0.202 364 496 117 76 × 2 = 0 + 0.404 728 992 235 52;
  • 29) 0.404 728 992 235 52 × 2 = 0 + 0.809 457 984 471 04;
  • 30) 0.809 457 984 471 04 × 2 = 1 + 0.618 915 968 942 08;
  • 31) 0.618 915 968 942 08 × 2 = 1 + 0.237 831 937 884 16;
  • 32) 0.237 831 937 884 16 × 2 = 0 + 0.475 663 875 768 32;
  • 33) 0.475 663 875 768 32 × 2 = 0 + 0.951 327 751 536 64;
  • 34) 0.951 327 751 536 64 × 2 = 1 + 0.902 655 503 073 28;
  • 35) 0.902 655 503 073 28 × 2 = 1 + 0.805 311 006 146 56;
  • 36) 0.805 311 006 146 56 × 2 = 1 + 0.610 622 012 293 12;
  • 37) 0.610 622 012 293 12 × 2 = 1 + 0.221 244 024 586 24;
  • 38) 0.221 244 024 586 24 × 2 = 0 + 0.442 488 049 172 48;
  • 39) 0.442 488 049 172 48 × 2 = 0 + 0.884 976 098 344 96;
  • 40) 0.884 976 098 344 96 × 2 = 1 + 0.769 952 196 689 92;
  • 41) 0.769 952 196 689 92 × 2 = 1 + 0.539 904 393 379 84;
  • 42) 0.539 904 393 379 84 × 2 = 1 + 0.079 808 786 759 68;
  • 43) 0.079 808 786 759 68 × 2 = 0 + 0.159 617 573 519 36;
  • 44) 0.159 617 573 519 36 × 2 = 0 + 0.319 235 147 038 72;
  • 45) 0.319 235 147 038 72 × 2 = 0 + 0.638 470 294 077 44;
  • 46) 0.638 470 294 077 44 × 2 = 1 + 0.276 940 588 154 88;
  • 47) 0.276 940 588 154 88 × 2 = 0 + 0.553 881 176 309 76;
  • 48) 0.553 881 176 309 76 × 2 = 1 + 0.107 762 352 619 52;
  • 49) 0.107 762 352 619 52 × 2 = 0 + 0.215 524 705 239 04;
  • 50) 0.215 524 705 239 04 × 2 = 0 + 0.431 049 410 478 08;
  • 51) 0.431 049 410 478 08 × 2 = 0 + 0.862 098 820 956 16;
  • 52) 0.862 098 820 956 16 × 2 = 1 + 0.724 197 641 912 32;
  • 53) 0.724 197 641 912 32 × 2 = 1 + 0.448 395 283 824 64;

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:


0.797 693 134 861 92(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1100 0101 0001 1(2)

5. Positive number before normalization:

1.797 693 134 861 92(10) =

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


1.797 693 134 861 92(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1100 0101 0001 1(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1001 1100 0101 0001 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 1001 1100 0101 0001 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 1001 1100 0101 0001 1 =

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


Decimal number 1.797 693 134 861 92 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 1100 0101 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