-0.014 657 790 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.014 657 790 46(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 46(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 46| = 0.014 657 790 46


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 46.

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 46 × 2 = 0 + 0.029 315 580 92;
  • 2) 0.029 315 580 92 × 2 = 0 + 0.058 631 161 84;
  • 3) 0.058 631 161 84 × 2 = 0 + 0.117 262 323 68;
  • 4) 0.117 262 323 68 × 2 = 0 + 0.234 524 647 36;
  • 5) 0.234 524 647 36 × 2 = 0 + 0.469 049 294 72;
  • 6) 0.469 049 294 72 × 2 = 0 + 0.938 098 589 44;
  • 7) 0.938 098 589 44 × 2 = 1 + 0.876 197 178 88;
  • 8) 0.876 197 178 88 × 2 = 1 + 0.752 394 357 76;
  • 9) 0.752 394 357 76 × 2 = 1 + 0.504 788 715 52;
  • 10) 0.504 788 715 52 × 2 = 1 + 0.009 577 431 04;
  • 11) 0.009 577 431 04 × 2 = 0 + 0.019 154 862 08;
  • 12) 0.019 154 862 08 × 2 = 0 + 0.038 309 724 16;
  • 13) 0.038 309 724 16 × 2 = 0 + 0.076 619 448 32;
  • 14) 0.076 619 448 32 × 2 = 0 + 0.153 238 896 64;
  • 15) 0.153 238 896 64 × 2 = 0 + 0.306 477 793 28;
  • 16) 0.306 477 793 28 × 2 = 0 + 0.612 955 586 56;
  • 17) 0.612 955 586 56 × 2 = 1 + 0.225 911 173 12;
  • 18) 0.225 911 173 12 × 2 = 0 + 0.451 822 346 24;
  • 19) 0.451 822 346 24 × 2 = 0 + 0.903 644 692 48;
  • 20) 0.903 644 692 48 × 2 = 1 + 0.807 289 384 96;
  • 21) 0.807 289 384 96 × 2 = 1 + 0.614 578 769 92;
  • 22) 0.614 578 769 92 × 2 = 1 + 0.229 157 539 84;
  • 23) 0.229 157 539 84 × 2 = 0 + 0.458 315 079 68;
  • 24) 0.458 315 079 68 × 2 = 0 + 0.916 630 159 36;
  • 25) 0.916 630 159 36 × 2 = 1 + 0.833 260 318 72;
  • 26) 0.833 260 318 72 × 2 = 1 + 0.666 520 637 44;
  • 27) 0.666 520 637 44 × 2 = 1 + 0.333 041 274 88;
  • 28) 0.333 041 274 88 × 2 = 0 + 0.666 082 549 76;
  • 29) 0.666 082 549 76 × 2 = 1 + 0.332 165 099 52;
  • 30) 0.332 165 099 52 × 2 = 0 + 0.664 330 199 04;
  • 31) 0.664 330 199 04 × 2 = 1 + 0.328 660 398 08;
  • 32) 0.328 660 398 08 × 2 = 0 + 0.657 320 796 16;
  • 33) 0.657 320 796 16 × 2 = 1 + 0.314 641 592 32;
  • 34) 0.314 641 592 32 × 2 = 0 + 0.629 283 184 64;
  • 35) 0.629 283 184 64 × 2 = 1 + 0.258 566 369 28;
  • 36) 0.258 566 369 28 × 2 = 0 + 0.517 132 738 56;
  • 37) 0.517 132 738 56 × 2 = 1 + 0.034 265 477 12;
  • 38) 0.034 265 477 12 × 2 = 0 + 0.068 530 954 24;
  • 39) 0.068 530 954 24 × 2 = 0 + 0.137 061 908 48;
  • 40) 0.137 061 908 48 × 2 = 0 + 0.274 123 816 96;
  • 41) 0.274 123 816 96 × 2 = 0 + 0.548 247 633 92;
  • 42) 0.548 247 633 92 × 2 = 1 + 0.096 495 267 84;
  • 43) 0.096 495 267 84 × 2 = 0 + 0.192 990 535 68;
  • 44) 0.192 990 535 68 × 2 = 0 + 0.385 981 071 36;
  • 45) 0.385 981 071 36 × 2 = 0 + 0.771 962 142 72;
  • 46) 0.771 962 142 72 × 2 = 1 + 0.543 924 285 44;
  • 47) 0.543 924 285 44 × 2 = 1 + 0.087 848 570 88;
  • 48) 0.087 848 570 88 × 2 = 0 + 0.175 697 141 76;
  • 49) 0.175 697 141 76 × 2 = 0 + 0.351 394 283 52;
  • 50) 0.351 394 283 52 × 2 = 0 + 0.702 788 567 04;
  • 51) 0.702 788 567 04 × 2 = 1 + 0.405 577 134 08;
  • 52) 0.405 577 134 08 × 2 = 0 + 0.811 154 268 16;
  • 53) 0.811 154 268 16 × 2 = 1 + 0.622 308 536 32;
  • 54) 0.622 308 536 32 × 2 = 1 + 0.244 617 072 64;
  • 55) 0.244 617 072 64 × 2 = 0 + 0.489 234 145 28;
  • 56) 0.489 234 145 28 × 2 = 0 + 0.978 468 290 56;
  • 57) 0.978 468 290 56 × 2 = 1 + 0.956 936 581 12;
  • 58) 0.956 936 581 12 × 2 = 1 + 0.913 873 162 24;
  • 59) 0.913 873 162 24 × 2 = 1 + 0.827 746 324 48;

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 46(10) =

0.0000 0011 1100 0000 1001 1100 1110 1010 1010 1000 0100 0110 0010 1100 111(2)

6. Positive number before normalization:

0.014 657 790 46(10) =

0.0000 0011 1100 0000 1001 1100 1110 1010 1010 1000 0100 0110 0010 1100 111(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 46(10) =

0.0000 0011 1100 0000 1001 1100 1110 1010 1010 1000 0100 0110 0010 1100 111(2) =

0.0000 0011 1100 0000 1001 1100 1110 1010 1010 1000 0100 0110 0010 1100 111(2) × 20 =

1.1110 0000 0100 1110 0111 0101 0101 0100 0010 0011 0001 0110 0111(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 0101 0101 0100 0010 0011 0001 0110 0111


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 0101 0101 0100 0010 0011 0001 0110 0111 =

1110 0000 0100 1110 0111 0101 0101 0100 0010 0011 0001 0110 0111


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 0101 0101 0100 0010 0011 0001 0110 0111


Decimal number -0.014 657 790 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1110 0000 0100 1110 0111 0101 0101 0100 0010 0011 0001 0110 0111

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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