-0.057 015 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.057 015 4(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
-0.057 015 4(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:

|-0.057 015 4| = 0.057 015 4


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: 0.057 015 4.

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) 0.057 015 4 × 2 = 0 + 0.114 030 8;
  • 2) 0.114 030 8 × 2 = 0 + 0.228 061 6;
  • 3) 0.228 061 6 × 2 = 0 + 0.456 123 2;
  • 4) 0.456 123 2 × 2 = 0 + 0.912 246 4;
  • 5) 0.912 246 4 × 2 = 1 + 0.824 492 8;
  • 6) 0.824 492 8 × 2 = 1 + 0.648 985 6;
  • 7) 0.648 985 6 × 2 = 1 + 0.297 971 2;
  • 8) 0.297 971 2 × 2 = 0 + 0.595 942 4;
  • 9) 0.595 942 4 × 2 = 1 + 0.191 884 8;
  • 10) 0.191 884 8 × 2 = 0 + 0.383 769 6;
  • 11) 0.383 769 6 × 2 = 0 + 0.767 539 2;
  • 12) 0.767 539 2 × 2 = 1 + 0.535 078 4;
  • 13) 0.535 078 4 × 2 = 1 + 0.070 156 8;
  • 14) 0.070 156 8 × 2 = 0 + 0.140 313 6;
  • 15) 0.140 313 6 × 2 = 0 + 0.280 627 2;
  • 16) 0.280 627 2 × 2 = 0 + 0.561 254 4;
  • 17) 0.561 254 4 × 2 = 1 + 0.122 508 8;
  • 18) 0.122 508 8 × 2 = 0 + 0.245 017 6;
  • 19) 0.245 017 6 × 2 = 0 + 0.490 035 2;
  • 20) 0.490 035 2 × 2 = 0 + 0.980 070 4;
  • 21) 0.980 070 4 × 2 = 1 + 0.960 140 8;
  • 22) 0.960 140 8 × 2 = 1 + 0.920 281 6;
  • 23) 0.920 281 6 × 2 = 1 + 0.840 563 2;
  • 24) 0.840 563 2 × 2 = 1 + 0.681 126 4;
  • 25) 0.681 126 4 × 2 = 1 + 0.362 252 8;
  • 26) 0.362 252 8 × 2 = 0 + 0.724 505 6;
  • 27) 0.724 505 6 × 2 = 1 + 0.449 011 2;
  • 28) 0.449 011 2 × 2 = 0 + 0.898 022 4;
  • 29) 0.898 022 4 × 2 = 1 + 0.796 044 8;
  • 30) 0.796 044 8 × 2 = 1 + 0.592 089 6;
  • 31) 0.592 089 6 × 2 = 1 + 0.184 179 2;
  • 32) 0.184 179 2 × 2 = 0 + 0.368 358 4;
  • 33) 0.368 358 4 × 2 = 0 + 0.736 716 8;
  • 34) 0.736 716 8 × 2 = 1 + 0.473 433 6;
  • 35) 0.473 433 6 × 2 = 0 + 0.946 867 2;
  • 36) 0.946 867 2 × 2 = 1 + 0.893 734 4;
  • 37) 0.893 734 4 × 2 = 1 + 0.787 468 8;
  • 38) 0.787 468 8 × 2 = 1 + 0.574 937 6;
  • 39) 0.574 937 6 × 2 = 1 + 0.149 875 2;
  • 40) 0.149 875 2 × 2 = 0 + 0.299 750 4;
  • 41) 0.299 750 4 × 2 = 0 + 0.599 500 8;
  • 42) 0.599 500 8 × 2 = 1 + 0.199 001 6;
  • 43) 0.199 001 6 × 2 = 0 + 0.398 003 2;
  • 44) 0.398 003 2 × 2 = 0 + 0.796 006 4;
  • 45) 0.796 006 4 × 2 = 1 + 0.592 012 8;
  • 46) 0.592 012 8 × 2 = 1 + 0.184 025 6;
  • 47) 0.184 025 6 × 2 = 0 + 0.368 051 2;
  • 48) 0.368 051 2 × 2 = 0 + 0.736 102 4;
  • 49) 0.736 102 4 × 2 = 1 + 0.472 204 8;
  • 50) 0.472 204 8 × 2 = 0 + 0.944 409 6;
  • 51) 0.944 409 6 × 2 = 1 + 0.888 819 2;
  • 52) 0.888 819 2 × 2 = 1 + 0.777 638 4;
  • 53) 0.777 638 4 × 2 = 1 + 0.555 276 8;
  • 54) 0.555 276 8 × 2 = 1 + 0.110 553 6;
  • 55) 0.110 553 6 × 2 = 0 + 0.221 107 2;
  • 56) 0.221 107 2 × 2 = 0 + 0.442 214 4;
  • 57) 0.442 214 4 × 2 = 0 + 0.884 428 8;

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:


0.057 015 4(10) =

0.0000 1110 1001 1000 1000 1111 1010 1110 0101 1110 0100 1100 1011 1100 0(2)

6. Positive number before normalization:

0.057 015 4(10) =

0.0000 1110 1001 1000 1000 1111 1010 1110 0101 1110 0100 1100 1011 1100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.057 015 4(10) =

0.0000 1110 1001 1000 1000 1111 1010 1110 0101 1110 0100 1100 1011 1100 0(2) =

0.0000 1110 1001 1000 1000 1111 1010 1110 0101 1110 0100 1100 1011 1100 0(2) × 20 =

1.1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 1000(2) × 2-5


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): -5

Mantissa (not normalized):
1.1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 1000 =

1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 1000


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 1010

Mantissa (52 bits) =
1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 1000


Decimal number -0.057 015 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1101 0011 0001 0001 1111 0101 1100 1011 1100 1001 1001 0111 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