-0.014 657 790 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.014 657 790 93(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
-0.014 657 790 93(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. Start with the positive version of the number:

|-0.014 657 790 93| = 0.014 657 790 93


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.014 657 790 93.

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) 0.014 657 790 93 × 2 = 0 + 0.029 315 581 86;
  • 2) 0.029 315 581 86 × 2 = 0 + 0.058 631 163 72;
  • 3) 0.058 631 163 72 × 2 = 0 + 0.117 262 327 44;
  • 4) 0.117 262 327 44 × 2 = 0 + 0.234 524 654 88;
  • 5) 0.234 524 654 88 × 2 = 0 + 0.469 049 309 76;
  • 6) 0.469 049 309 76 × 2 = 0 + 0.938 098 619 52;
  • 7) 0.938 098 619 52 × 2 = 1 + 0.876 197 239 04;
  • 8) 0.876 197 239 04 × 2 = 1 + 0.752 394 478 08;
  • 9) 0.752 394 478 08 × 2 = 1 + 0.504 788 956 16;
  • 10) 0.504 788 956 16 × 2 = 1 + 0.009 577 912 32;
  • 11) 0.009 577 912 32 × 2 = 0 + 0.019 155 824 64;
  • 12) 0.019 155 824 64 × 2 = 0 + 0.038 311 649 28;
  • 13) 0.038 311 649 28 × 2 = 0 + 0.076 623 298 56;
  • 14) 0.076 623 298 56 × 2 = 0 + 0.153 246 597 12;
  • 15) 0.153 246 597 12 × 2 = 0 + 0.306 493 194 24;
  • 16) 0.306 493 194 24 × 2 = 0 + 0.612 986 388 48;
  • 17) 0.612 986 388 48 × 2 = 1 + 0.225 972 776 96;
  • 18) 0.225 972 776 96 × 2 = 0 + 0.451 945 553 92;
  • 19) 0.451 945 553 92 × 2 = 0 + 0.903 891 107 84;
  • 20) 0.903 891 107 84 × 2 = 1 + 0.807 782 215 68;
  • 21) 0.807 782 215 68 × 2 = 1 + 0.615 564 431 36;
  • 22) 0.615 564 431 36 × 2 = 1 + 0.231 128 862 72;
  • 23) 0.231 128 862 72 × 2 = 0 + 0.462 257 725 44;
  • 24) 0.462 257 725 44 × 2 = 0 + 0.924 515 450 88;
  • 25) 0.924 515 450 88 × 2 = 1 + 0.849 030 901 76;
  • 26) 0.849 030 901 76 × 2 = 1 + 0.698 061 803 52;
  • 27) 0.698 061 803 52 × 2 = 1 + 0.396 123 607 04;
  • 28) 0.396 123 607 04 × 2 = 0 + 0.792 247 214 08;
  • 29) 0.792 247 214 08 × 2 = 1 + 0.584 494 428 16;
  • 30) 0.584 494 428 16 × 2 = 1 + 0.168 988 856 32;
  • 31) 0.168 988 856 32 × 2 = 0 + 0.337 977 712 64;
  • 32) 0.337 977 712 64 × 2 = 0 + 0.675 955 425 28;
  • 33) 0.675 955 425 28 × 2 = 1 + 0.351 910 850 56;
  • 34) 0.351 910 850 56 × 2 = 0 + 0.703 821 701 12;
  • 35) 0.703 821 701 12 × 2 = 1 + 0.407 643 402 24;
  • 36) 0.407 643 402 24 × 2 = 0 + 0.815 286 804 48;
  • 37) 0.815 286 804 48 × 2 = 1 + 0.630 573 608 96;
  • 38) 0.630 573 608 96 × 2 = 1 + 0.261 147 217 92;
  • 39) 0.261 147 217 92 × 2 = 0 + 0.522 294 435 84;
  • 40) 0.522 294 435 84 × 2 = 1 + 0.044 588 871 68;
  • 41) 0.044 588 871 68 × 2 = 0 + 0.089 177 743 36;
  • 42) 0.089 177 743 36 × 2 = 0 + 0.178 355 486 72;
  • 43) 0.178 355 486 72 × 2 = 0 + 0.356 710 973 44;
  • 44) 0.356 710 973 44 × 2 = 0 + 0.713 421 946 88;
  • 45) 0.713 421 946 88 × 2 = 1 + 0.426 843 893 76;
  • 46) 0.426 843 893 76 × 2 = 0 + 0.853 687 787 52;
  • 47) 0.853 687 787 52 × 2 = 1 + 0.707 375 575 04;
  • 48) 0.707 375 575 04 × 2 = 1 + 0.414 751 150 08;
  • 49) 0.414 751 150 08 × 2 = 0 + 0.829 502 300 16;
  • 50) 0.829 502 300 16 × 2 = 1 + 0.659 004 600 32;
  • 51) 0.659 004 600 32 × 2 = 1 + 0.318 009 200 64;
  • 52) 0.318 009 200 64 × 2 = 0 + 0.636 018 401 28;
  • 53) 0.636 018 401 28 × 2 = 1 + 0.272 036 802 56;
  • 54) 0.272 036 802 56 × 2 = 0 + 0.544 073 605 12;
  • 55) 0.544 073 605 12 × 2 = 1 + 0.088 147 210 24;
  • 56) 0.088 147 210 24 × 2 = 0 + 0.176 294 420 48;
  • 57) 0.176 294 420 48 × 2 = 0 + 0.352 588 840 96;
  • 58) 0.352 588 840 96 × 2 = 0 + 0.705 177 681 92;
  • 59) 0.705 177 681 92 × 2 = 1 + 0.410 355 363 84;

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).


5. 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:


0.014 657 790 93(10) =

0.0000 0011 1100 0000 1001 1100 1110 1100 1010 1101 0000 1011 0110 1010 001(2)

6. Positive number before normalization:

0.014 657 790 93(10) =

0.0000 0011 1100 0000 1001 1100 1110 1100 1010 1101 0000 1011 0110 1010 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.014 657 790 93(10) =

0.0000 0011 1100 0000 1001 1100 1110 1100 1010 1101 0000 1011 0110 1010 001(2) =

0.0000 0011 1100 0000 1001 1100 1110 1100 1010 1101 0000 1011 0110 1010 001(2) × 20 =

1.1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 0101 0001(2) × 2-7


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): -7

Mantissa (not normalized):
1.1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 0101 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(10)


10. 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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1016(10) =

011 1111 1000(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 0101 0001 =

1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 0101 0001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1000

Mantissa (52 bits) =
1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 0101 0001


Decimal number -0.014 657 790 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1110 0000 0100 1110 0111 0110 0101 0110 1000 0101 1011 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