-0.000 282 010 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 15(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
-0.000 282 010 15(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:

|-0.000 282 010 15| = 0.000 282 010 15


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: 0.000 282 010 15.

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) 0.000 282 010 15 × 2 = 0 + 0.000 564 020 3;
  • 2) 0.000 564 020 3 × 2 = 0 + 0.001 128 040 6;
  • 3) 0.001 128 040 6 × 2 = 0 + 0.002 256 081 2;
  • 4) 0.002 256 081 2 × 2 = 0 + 0.004 512 162 4;
  • 5) 0.004 512 162 4 × 2 = 0 + 0.009 024 324 8;
  • 6) 0.009 024 324 8 × 2 = 0 + 0.018 048 649 6;
  • 7) 0.018 048 649 6 × 2 = 0 + 0.036 097 299 2;
  • 8) 0.036 097 299 2 × 2 = 0 + 0.072 194 598 4;
  • 9) 0.072 194 598 4 × 2 = 0 + 0.144 389 196 8;
  • 10) 0.144 389 196 8 × 2 = 0 + 0.288 778 393 6;
  • 11) 0.288 778 393 6 × 2 = 0 + 0.577 556 787 2;
  • 12) 0.577 556 787 2 × 2 = 1 + 0.155 113 574 4;
  • 13) 0.155 113 574 4 × 2 = 0 + 0.310 227 148 8;
  • 14) 0.310 227 148 8 × 2 = 0 + 0.620 454 297 6;
  • 15) 0.620 454 297 6 × 2 = 1 + 0.240 908 595 2;
  • 16) 0.240 908 595 2 × 2 = 0 + 0.481 817 190 4;
  • 17) 0.481 817 190 4 × 2 = 0 + 0.963 634 380 8;
  • 18) 0.963 634 380 8 × 2 = 1 + 0.927 268 761 6;
  • 19) 0.927 268 761 6 × 2 = 1 + 0.854 537 523 2;
  • 20) 0.854 537 523 2 × 2 = 1 + 0.709 075 046 4;
  • 21) 0.709 075 046 4 × 2 = 1 + 0.418 150 092 8;
  • 22) 0.418 150 092 8 × 2 = 0 + 0.836 300 185 6;
  • 23) 0.836 300 185 6 × 2 = 1 + 0.672 600 371 2;
  • 24) 0.672 600 371 2 × 2 = 1 + 0.345 200 742 4;
  • 25) 0.345 200 742 4 × 2 = 0 + 0.690 401 484 8;
  • 26) 0.690 401 484 8 × 2 = 1 + 0.380 802 969 6;
  • 27) 0.380 802 969 6 × 2 = 0 + 0.761 605 939 2;
  • 28) 0.761 605 939 2 × 2 = 1 + 0.523 211 878 4;
  • 29) 0.523 211 878 4 × 2 = 1 + 0.046 423 756 8;
  • 30) 0.046 423 756 8 × 2 = 0 + 0.092 847 513 6;
  • 31) 0.092 847 513 6 × 2 = 0 + 0.185 695 027 2;
  • 32) 0.185 695 027 2 × 2 = 0 + 0.371 390 054 4;
  • 33) 0.371 390 054 4 × 2 = 0 + 0.742 780 108 8;
  • 34) 0.742 780 108 8 × 2 = 1 + 0.485 560 217 6;
  • 35) 0.485 560 217 6 × 2 = 0 + 0.971 120 435 2;
  • 36) 0.971 120 435 2 × 2 = 1 + 0.942 240 870 4;
  • 37) 0.942 240 870 4 × 2 = 1 + 0.884 481 740 8;
  • 38) 0.884 481 740 8 × 2 = 1 + 0.768 963 481 6;
  • 39) 0.768 963 481 6 × 2 = 1 + 0.537 926 963 2;
  • 40) 0.537 926 963 2 × 2 = 1 + 0.075 853 926 4;
  • 41) 0.075 853 926 4 × 2 = 0 + 0.151 707 852 8;
  • 42) 0.151 707 852 8 × 2 = 0 + 0.303 415 705 6;
  • 43) 0.303 415 705 6 × 2 = 0 + 0.606 831 411 2;
  • 44) 0.606 831 411 2 × 2 = 1 + 0.213 662 822 4;
  • 45) 0.213 662 822 4 × 2 = 0 + 0.427 325 644 8;
  • 46) 0.427 325 644 8 × 2 = 0 + 0.854 651 289 6;
  • 47) 0.854 651 289 6 × 2 = 1 + 0.709 302 579 2;
  • 48) 0.709 302 579 2 × 2 = 1 + 0.418 605 158 4;
  • 49) 0.418 605 158 4 × 2 = 0 + 0.837 210 316 8;
  • 50) 0.837 210 316 8 × 2 = 1 + 0.674 420 633 6;
  • 51) 0.674 420 633 6 × 2 = 1 + 0.348 841 267 2;
  • 52) 0.348 841 267 2 × 2 = 0 + 0.697 682 534 4;
  • 53) 0.697 682 534 4 × 2 = 1 + 0.395 365 068 8;
  • 54) 0.395 365 068 8 × 2 = 0 + 0.790 730 137 6;
  • 55) 0.790 730 137 6 × 2 = 1 + 0.581 460 275 2;
  • 56) 0.581 460 275 2 × 2 = 1 + 0.162 920 550 4;
  • 57) 0.162 920 550 4 × 2 = 0 + 0.325 841 100 8;
  • 58) 0.325 841 100 8 × 2 = 0 + 0.651 682 201 6;
  • 59) 0.651 682 201 6 × 2 = 1 + 0.303 364 403 2;
  • 60) 0.303 364 403 2 × 2 = 0 + 0.606 728 806 4;
  • 61) 0.606 728 806 4 × 2 = 1 + 0.213 457 612 8;
  • 62) 0.213 457 612 8 × 2 = 0 + 0.426 915 225 6;
  • 63) 0.426 915 225 6 × 2 = 0 + 0.853 830 451 2;
  • 64) 0.853 830 451 2 × 2 = 1 + 0.707 660 902 4;

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:


0.000 282 010 15(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001(2)

6. Positive number before normalization:

0.000 282 010 15(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001(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:


0.000 282 010 15(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001(2) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001(2) × 20 =

1.0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001(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 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001


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 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001 =

0010 0111 1011 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001


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 0101 1000 0101 1111 0001 0011 0110 1011 0010 1001


Decimal number -0.000 282 010 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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