-0.000 282 016 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 016(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
-0.000 282 016(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:

|-0.000 282 016| = 0.000 282 016


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: 0.000 282 016.

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) 0.000 282 016 × 2 = 0 + 0.000 564 032;
  • 2) 0.000 564 032 × 2 = 0 + 0.001 128 064;
  • 3) 0.001 128 064 × 2 = 0 + 0.002 256 128;
  • 4) 0.002 256 128 × 2 = 0 + 0.004 512 256;
  • 5) 0.004 512 256 × 2 = 0 + 0.009 024 512;
  • 6) 0.009 024 512 × 2 = 0 + 0.018 049 024;
  • 7) 0.018 049 024 × 2 = 0 + 0.036 098 048;
  • 8) 0.036 098 048 × 2 = 0 + 0.072 196 096;
  • 9) 0.072 196 096 × 2 = 0 + 0.144 392 192;
  • 10) 0.144 392 192 × 2 = 0 + 0.288 784 384;
  • 11) 0.288 784 384 × 2 = 0 + 0.577 568 768;
  • 12) 0.577 568 768 × 2 = 1 + 0.155 137 536;
  • 13) 0.155 137 536 × 2 = 0 + 0.310 275 072;
  • 14) 0.310 275 072 × 2 = 0 + 0.620 550 144;
  • 15) 0.620 550 144 × 2 = 1 + 0.241 100 288;
  • 16) 0.241 100 288 × 2 = 0 + 0.482 200 576;
  • 17) 0.482 200 576 × 2 = 0 + 0.964 401 152;
  • 18) 0.964 401 152 × 2 = 1 + 0.928 802 304;
  • 19) 0.928 802 304 × 2 = 1 + 0.857 604 608;
  • 20) 0.857 604 608 × 2 = 1 + 0.715 209 216;
  • 21) 0.715 209 216 × 2 = 1 + 0.430 418 432;
  • 22) 0.430 418 432 × 2 = 0 + 0.860 836 864;
  • 23) 0.860 836 864 × 2 = 1 + 0.721 673 728;
  • 24) 0.721 673 728 × 2 = 1 + 0.443 347 456;
  • 25) 0.443 347 456 × 2 = 0 + 0.886 694 912;
  • 26) 0.886 694 912 × 2 = 1 + 0.773 389 824;
  • 27) 0.773 389 824 × 2 = 1 + 0.546 779 648;
  • 28) 0.546 779 648 × 2 = 1 + 0.093 559 296;
  • 29) 0.093 559 296 × 2 = 0 + 0.187 118 592;
  • 30) 0.187 118 592 × 2 = 0 + 0.374 237 184;
  • 31) 0.374 237 184 × 2 = 0 + 0.748 474 368;
  • 32) 0.748 474 368 × 2 = 1 + 0.496 948 736;
  • 33) 0.496 948 736 × 2 = 0 + 0.993 897 472;
  • 34) 0.993 897 472 × 2 = 1 + 0.987 794 944;
  • 35) 0.987 794 944 × 2 = 1 + 0.975 589 888;
  • 36) 0.975 589 888 × 2 = 1 + 0.951 179 776;
  • 37) 0.951 179 776 × 2 = 1 + 0.902 359 552;
  • 38) 0.902 359 552 × 2 = 1 + 0.804 719 104;
  • 39) 0.804 719 104 × 2 = 1 + 0.609 438 208;
  • 40) 0.609 438 208 × 2 = 1 + 0.218 876 416;
  • 41) 0.218 876 416 × 2 = 0 + 0.437 752 832;
  • 42) 0.437 752 832 × 2 = 0 + 0.875 505 664;
  • 43) 0.875 505 664 × 2 = 1 + 0.751 011 328;
  • 44) 0.751 011 328 × 2 = 1 + 0.502 022 656;
  • 45) 0.502 022 656 × 2 = 1 + 0.004 045 312;
  • 46) 0.004 045 312 × 2 = 0 + 0.008 090 624;
  • 47) 0.008 090 624 × 2 = 0 + 0.016 181 248;
  • 48) 0.016 181 248 × 2 = 0 + 0.032 362 496;
  • 49) 0.032 362 496 × 2 = 0 + 0.064 724 992;
  • 50) 0.064 724 992 × 2 = 0 + 0.129 449 984;
  • 51) 0.129 449 984 × 2 = 0 + 0.258 899 968;
  • 52) 0.258 899 968 × 2 = 0 + 0.517 799 936;
  • 53) 0.517 799 936 × 2 = 1 + 0.035 599 872;
  • 54) 0.035 599 872 × 2 = 0 + 0.071 199 744;
  • 55) 0.071 199 744 × 2 = 0 + 0.142 399 488;
  • 56) 0.142 399 488 × 2 = 0 + 0.284 798 976;
  • 57) 0.284 798 976 × 2 = 0 + 0.569 597 952;
  • 58) 0.569 597 952 × 2 = 1 + 0.139 195 904;
  • 59) 0.139 195 904 × 2 = 0 + 0.278 391 808;
  • 60) 0.278 391 808 × 2 = 0 + 0.556 783 616;
  • 61) 0.556 783 616 × 2 = 1 + 0.113 567 232;
  • 62) 0.113 567 232 × 2 = 0 + 0.227 134 464;
  • 63) 0.227 134 464 × 2 = 0 + 0.454 268 928;
  • 64) 0.454 268 928 × 2 = 0 + 0.908 537 856;

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:


0.000 282 016(10) =

0.0000 0000 0001 0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000(2)

6. Positive number before normalization:

0.000 282 016(10) =

0.0000 0000 0001 0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000(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:


0.000 282 016(10) =

0.0000 0000 0001 0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000(2) =

0.0000 0000 0001 0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000(2) × 20 =

1.0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000(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 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000


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 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000 =

0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 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 0011

Mantissa (52 bits) =
0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 1000


Decimal number -0.000 282 016 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0111 0001 0111 1111 0011 1000 0000 1000 0100 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