-0.026 154 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.026 154 38(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
-0.026 154 38(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:

|-0.026 154 38| = 0.026 154 38


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: 0.026 154 38.

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) 0.026 154 38 × 2 = 0 + 0.052 308 76;
  • 2) 0.052 308 76 × 2 = 0 + 0.104 617 52;
  • 3) 0.104 617 52 × 2 = 0 + 0.209 235 04;
  • 4) 0.209 235 04 × 2 = 0 + 0.418 470 08;
  • 5) 0.418 470 08 × 2 = 0 + 0.836 940 16;
  • 6) 0.836 940 16 × 2 = 1 + 0.673 880 32;
  • 7) 0.673 880 32 × 2 = 1 + 0.347 760 64;
  • 8) 0.347 760 64 × 2 = 0 + 0.695 521 28;
  • 9) 0.695 521 28 × 2 = 1 + 0.391 042 56;
  • 10) 0.391 042 56 × 2 = 0 + 0.782 085 12;
  • 11) 0.782 085 12 × 2 = 1 + 0.564 170 24;
  • 12) 0.564 170 24 × 2 = 1 + 0.128 340 48;
  • 13) 0.128 340 48 × 2 = 0 + 0.256 680 96;
  • 14) 0.256 680 96 × 2 = 0 + 0.513 361 92;
  • 15) 0.513 361 92 × 2 = 1 + 0.026 723 84;
  • 16) 0.026 723 84 × 2 = 0 + 0.053 447 68;
  • 17) 0.053 447 68 × 2 = 0 + 0.106 895 36;
  • 18) 0.106 895 36 × 2 = 0 + 0.213 790 72;
  • 19) 0.213 790 72 × 2 = 0 + 0.427 581 44;
  • 20) 0.427 581 44 × 2 = 0 + 0.855 162 88;
  • 21) 0.855 162 88 × 2 = 1 + 0.710 325 76;
  • 22) 0.710 325 76 × 2 = 1 + 0.420 651 52;
  • 23) 0.420 651 52 × 2 = 0 + 0.841 303 04;
  • 24) 0.841 303 04 × 2 = 1 + 0.682 606 08;
  • 25) 0.682 606 08 × 2 = 1 + 0.365 212 16;
  • 26) 0.365 212 16 × 2 = 0 + 0.730 424 32;
  • 27) 0.730 424 32 × 2 = 1 + 0.460 848 64;
  • 28) 0.460 848 64 × 2 = 0 + 0.921 697 28;
  • 29) 0.921 697 28 × 2 = 1 + 0.843 394 56;
  • 30) 0.843 394 56 × 2 = 1 + 0.686 789 12;
  • 31) 0.686 789 12 × 2 = 1 + 0.373 578 24;
  • 32) 0.373 578 24 × 2 = 0 + 0.747 156 48;
  • 33) 0.747 156 48 × 2 = 1 + 0.494 312 96;
  • 34) 0.494 312 96 × 2 = 0 + 0.988 625 92;
  • 35) 0.988 625 92 × 2 = 1 + 0.977 251 84;
  • 36) 0.977 251 84 × 2 = 1 + 0.954 503 68;
  • 37) 0.954 503 68 × 2 = 1 + 0.909 007 36;
  • 38) 0.909 007 36 × 2 = 1 + 0.818 014 72;
  • 39) 0.818 014 72 × 2 = 1 + 0.636 029 44;
  • 40) 0.636 029 44 × 2 = 1 + 0.272 058 88;
  • 41) 0.272 058 88 × 2 = 0 + 0.544 117 76;
  • 42) 0.544 117 76 × 2 = 1 + 0.088 235 52;
  • 43) 0.088 235 52 × 2 = 0 + 0.176 471 04;
  • 44) 0.176 471 04 × 2 = 0 + 0.352 942 08;
  • 45) 0.352 942 08 × 2 = 0 + 0.705 884 16;
  • 46) 0.705 884 16 × 2 = 1 + 0.411 768 32;
  • 47) 0.411 768 32 × 2 = 0 + 0.823 536 64;
  • 48) 0.823 536 64 × 2 = 1 + 0.647 073 28;
  • 49) 0.647 073 28 × 2 = 1 + 0.294 146 56;
  • 50) 0.294 146 56 × 2 = 0 + 0.588 293 12;
  • 51) 0.588 293 12 × 2 = 1 + 0.176 586 24;
  • 52) 0.176 586 24 × 2 = 0 + 0.353 172 48;
  • 53) 0.353 172 48 × 2 = 0 + 0.706 344 96;
  • 54) 0.706 344 96 × 2 = 1 + 0.412 689 92;
  • 55) 0.412 689 92 × 2 = 0 + 0.825 379 84;
  • 56) 0.825 379 84 × 2 = 1 + 0.650 759 68;
  • 57) 0.650 759 68 × 2 = 1 + 0.301 519 36;
  • 58) 0.301 519 36 × 2 = 0 + 0.603 038 72;

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:


0.026 154 38(10) =

0.0000 0110 1011 0010 0000 1101 1010 1110 1011 1111 0100 0101 1010 0101 10(2)

6. Positive number before normalization:

0.026 154 38(10) =

0.0000 0110 1011 0010 0000 1101 1010 1110 1011 1111 0100 0101 1010 0101 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.026 154 38(10) =

0.0000 0110 1011 0010 0000 1101 1010 1110 1011 1111 0100 0101 1010 0101 10(2) =

0.0000 0110 1011 0010 0000 1101 1010 1110 1011 1111 0100 0101 1010 0101 10(2) × 20 =

1.1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110(2) × 2-6


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): -6

Mantissa (not normalized):
1.1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110 =

1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110


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 1001

Mantissa (52 bits) =
1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110


Decimal number -0.026 154 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 1010 1100 1000 0011 0110 1011 1010 1111 1101 0001 0110 1001 0110

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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