-0.000 282 006 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 42(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
-0.000 282 006 42(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:

|-0.000 282 006 42| = 0.000 282 006 42


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: 0.000 282 006 42.

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) 0.000 282 006 42 × 2 = 0 + 0.000 564 012 84;
  • 2) 0.000 564 012 84 × 2 = 0 + 0.001 128 025 68;
  • 3) 0.001 128 025 68 × 2 = 0 + 0.002 256 051 36;
  • 4) 0.002 256 051 36 × 2 = 0 + 0.004 512 102 72;
  • 5) 0.004 512 102 72 × 2 = 0 + 0.009 024 205 44;
  • 6) 0.009 024 205 44 × 2 = 0 + 0.018 048 410 88;
  • 7) 0.018 048 410 88 × 2 = 0 + 0.036 096 821 76;
  • 8) 0.036 096 821 76 × 2 = 0 + 0.072 193 643 52;
  • 9) 0.072 193 643 52 × 2 = 0 + 0.144 387 287 04;
  • 10) 0.144 387 287 04 × 2 = 0 + 0.288 774 574 08;
  • 11) 0.288 774 574 08 × 2 = 0 + 0.577 549 148 16;
  • 12) 0.577 549 148 16 × 2 = 1 + 0.155 098 296 32;
  • 13) 0.155 098 296 32 × 2 = 0 + 0.310 196 592 64;
  • 14) 0.310 196 592 64 × 2 = 0 + 0.620 393 185 28;
  • 15) 0.620 393 185 28 × 2 = 1 + 0.240 786 370 56;
  • 16) 0.240 786 370 56 × 2 = 0 + 0.481 572 741 12;
  • 17) 0.481 572 741 12 × 2 = 0 + 0.963 145 482 24;
  • 18) 0.963 145 482 24 × 2 = 1 + 0.926 290 964 48;
  • 19) 0.926 290 964 48 × 2 = 1 + 0.852 581 928 96;
  • 20) 0.852 581 928 96 × 2 = 1 + 0.705 163 857 92;
  • 21) 0.705 163 857 92 × 2 = 1 + 0.410 327 715 84;
  • 22) 0.410 327 715 84 × 2 = 0 + 0.820 655 431 68;
  • 23) 0.820 655 431 68 × 2 = 1 + 0.641 310 863 36;
  • 24) 0.641 310 863 36 × 2 = 1 + 0.282 621 726 72;
  • 25) 0.282 621 726 72 × 2 = 0 + 0.565 243 453 44;
  • 26) 0.565 243 453 44 × 2 = 1 + 0.130 486 906 88;
  • 27) 0.130 486 906 88 × 2 = 0 + 0.260 973 813 76;
  • 28) 0.260 973 813 76 × 2 = 0 + 0.521 947 627 52;
  • 29) 0.521 947 627 52 × 2 = 1 + 0.043 895 255 04;
  • 30) 0.043 895 255 04 × 2 = 0 + 0.087 790 510 08;
  • 31) 0.087 790 510 08 × 2 = 0 + 0.175 581 020 16;
  • 32) 0.175 581 020 16 × 2 = 0 + 0.351 162 040 32;
  • 33) 0.351 162 040 32 × 2 = 0 + 0.702 324 080 64;
  • 34) 0.702 324 080 64 × 2 = 1 + 0.404 648 161 28;
  • 35) 0.404 648 161 28 × 2 = 0 + 0.809 296 322 56;
  • 36) 0.809 296 322 56 × 2 = 1 + 0.618 592 645 12;
  • 37) 0.618 592 645 12 × 2 = 1 + 0.237 185 290 24;
  • 38) 0.237 185 290 24 × 2 = 0 + 0.474 370 580 48;
  • 39) 0.474 370 580 48 × 2 = 0 + 0.948 741 160 96;
  • 40) 0.948 741 160 96 × 2 = 1 + 0.897 482 321 92;
  • 41) 0.897 482 321 92 × 2 = 1 + 0.794 964 643 84;
  • 42) 0.794 964 643 84 × 2 = 1 + 0.589 929 287 68;
  • 43) 0.589 929 287 68 × 2 = 1 + 0.179 858 575 36;
  • 44) 0.179 858 575 36 × 2 = 0 + 0.359 717 150 72;
  • 45) 0.359 717 150 72 × 2 = 0 + 0.719 434 301 44;
  • 46) 0.719 434 301 44 × 2 = 1 + 0.438 868 602 88;
  • 47) 0.438 868 602 88 × 2 = 0 + 0.877 737 205 76;
  • 48) 0.877 737 205 76 × 2 = 1 + 0.755 474 411 52;
  • 49) 0.755 474 411 52 × 2 = 1 + 0.510 948 823 04;
  • 50) 0.510 948 823 04 × 2 = 1 + 0.021 897 646 08;
  • 51) 0.021 897 646 08 × 2 = 0 + 0.043 795 292 16;
  • 52) 0.043 795 292 16 × 2 = 0 + 0.087 590 584 32;
  • 53) 0.087 590 584 32 × 2 = 0 + 0.175 181 168 64;
  • 54) 0.175 181 168 64 × 2 = 0 + 0.350 362 337 28;
  • 55) 0.350 362 337 28 × 2 = 0 + 0.700 724 674 56;
  • 56) 0.700 724 674 56 × 2 = 1 + 0.401 449 349 12;
  • 57) 0.401 449 349 12 × 2 = 0 + 0.802 898 698 24;
  • 58) 0.802 898 698 24 × 2 = 1 + 0.605 797 396 48;
  • 59) 0.605 797 396 48 × 2 = 1 + 0.211 594 792 96;
  • 60) 0.211 594 792 96 × 2 = 0 + 0.423 189 585 92;
  • 61) 0.423 189 585 92 × 2 = 0 + 0.846 379 171 84;
  • 62) 0.846 379 171 84 × 2 = 1 + 0.692 758 343 68;
  • 63) 0.692 758 343 68 × 2 = 1 + 0.385 516 687 36;
  • 64) 0.385 516 687 36 × 2 = 0 + 0.771 033 374 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:


0.000 282 006 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110(2)

6. Positive number before normalization:

0.000 282 006 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 006 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110(2) × 20 =

1.0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110 =

0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 0110


Decimal number -0.000 282 006 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0101 1001 1110 0101 1100 0001 0110 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