-0.000 282 024 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 024 5(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
-0.000 282 024 5(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:

|-0.000 282 024 5| = 0.000 282 024 5


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: 0.000 282 024 5.

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) 0.000 282 024 5 × 2 = 0 + 0.000 564 049;
  • 2) 0.000 564 049 × 2 = 0 + 0.001 128 098;
  • 3) 0.001 128 098 × 2 = 0 + 0.002 256 196;
  • 4) 0.002 256 196 × 2 = 0 + 0.004 512 392;
  • 5) 0.004 512 392 × 2 = 0 + 0.009 024 784;
  • 6) 0.009 024 784 × 2 = 0 + 0.018 049 568;
  • 7) 0.018 049 568 × 2 = 0 + 0.036 099 136;
  • 8) 0.036 099 136 × 2 = 0 + 0.072 198 272;
  • 9) 0.072 198 272 × 2 = 0 + 0.144 396 544;
  • 10) 0.144 396 544 × 2 = 0 + 0.288 793 088;
  • 11) 0.288 793 088 × 2 = 0 + 0.577 586 176;
  • 12) 0.577 586 176 × 2 = 1 + 0.155 172 352;
  • 13) 0.155 172 352 × 2 = 0 + 0.310 344 704;
  • 14) 0.310 344 704 × 2 = 0 + 0.620 689 408;
  • 15) 0.620 689 408 × 2 = 1 + 0.241 378 816;
  • 16) 0.241 378 816 × 2 = 0 + 0.482 757 632;
  • 17) 0.482 757 632 × 2 = 0 + 0.965 515 264;
  • 18) 0.965 515 264 × 2 = 1 + 0.931 030 528;
  • 19) 0.931 030 528 × 2 = 1 + 0.862 061 056;
  • 20) 0.862 061 056 × 2 = 1 + 0.724 122 112;
  • 21) 0.724 122 112 × 2 = 1 + 0.448 244 224;
  • 22) 0.448 244 224 × 2 = 0 + 0.896 488 448;
  • 23) 0.896 488 448 × 2 = 1 + 0.792 976 896;
  • 24) 0.792 976 896 × 2 = 1 + 0.585 953 792;
  • 25) 0.585 953 792 × 2 = 1 + 0.171 907 584;
  • 26) 0.171 907 584 × 2 = 0 + 0.343 815 168;
  • 27) 0.343 815 168 × 2 = 0 + 0.687 630 336;
  • 28) 0.687 630 336 × 2 = 1 + 0.375 260 672;
  • 29) 0.375 260 672 × 2 = 0 + 0.750 521 344;
  • 30) 0.750 521 344 × 2 = 1 + 0.501 042 688;
  • 31) 0.501 042 688 × 2 = 1 + 0.002 085 376;
  • 32) 0.002 085 376 × 2 = 0 + 0.004 170 752;
  • 33) 0.004 170 752 × 2 = 0 + 0.008 341 504;
  • 34) 0.008 341 504 × 2 = 0 + 0.016 683 008;
  • 35) 0.016 683 008 × 2 = 0 + 0.033 366 016;
  • 36) 0.033 366 016 × 2 = 0 + 0.066 732 032;
  • 37) 0.066 732 032 × 2 = 0 + 0.133 464 064;
  • 38) 0.133 464 064 × 2 = 0 + 0.266 928 128;
  • 39) 0.266 928 128 × 2 = 0 + 0.533 856 256;
  • 40) 0.533 856 256 × 2 = 1 + 0.067 712 512;
  • 41) 0.067 712 512 × 2 = 0 + 0.135 425 024;
  • 42) 0.135 425 024 × 2 = 0 + 0.270 850 048;
  • 43) 0.270 850 048 × 2 = 0 + 0.541 700 096;
  • 44) 0.541 700 096 × 2 = 1 + 0.083 400 192;
  • 45) 0.083 400 192 × 2 = 0 + 0.166 800 384;
  • 46) 0.166 800 384 × 2 = 0 + 0.333 600 768;
  • 47) 0.333 600 768 × 2 = 0 + 0.667 201 536;
  • 48) 0.667 201 536 × 2 = 1 + 0.334 403 072;
  • 49) 0.334 403 072 × 2 = 0 + 0.668 806 144;
  • 50) 0.668 806 144 × 2 = 1 + 0.337 612 288;
  • 51) 0.337 612 288 × 2 = 0 + 0.675 224 576;
  • 52) 0.675 224 576 × 2 = 1 + 0.350 449 152;
  • 53) 0.350 449 152 × 2 = 0 + 0.700 898 304;
  • 54) 0.700 898 304 × 2 = 1 + 0.401 796 608;
  • 55) 0.401 796 608 × 2 = 0 + 0.803 593 216;
  • 56) 0.803 593 216 × 2 = 1 + 0.607 186 432;
  • 57) 0.607 186 432 × 2 = 1 + 0.214 372 864;
  • 58) 0.214 372 864 × 2 = 0 + 0.428 745 728;
  • 59) 0.428 745 728 × 2 = 0 + 0.857 491 456;
  • 60) 0.857 491 456 × 2 = 1 + 0.714 982 912;
  • 61) 0.714 982 912 × 2 = 1 + 0.429 965 824;
  • 62) 0.429 965 824 × 2 = 0 + 0.859 931 648;
  • 63) 0.859 931 648 × 2 = 1 + 0.719 863 296;
  • 64) 0.719 863 296 × 2 = 1 + 0.439 726 592;

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:


0.000 282 024 5(10) =

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

6. Positive number before normalization:

0.000 282 024 5(10) =

0.0000 0000 0001 0010 0111 1011 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011(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:


0.000 282 024 5(10) =

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

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

1.0010 0111 1011 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011(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 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011


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 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011 =

0010 0111 1011 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011


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 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011


Decimal number -0.000 282 024 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1001 0110 0000 0001 0001 0001 0101 0101 1001 1011

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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