-0.000 282 021 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 021(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
-0.000 282 021(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:

|-0.000 282 021| = 0.000 282 021


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: 0.000 282 021.

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) 0.000 282 021 × 2 = 0 + 0.000 564 042;
  • 2) 0.000 564 042 × 2 = 0 + 0.001 128 084;
  • 3) 0.001 128 084 × 2 = 0 + 0.002 256 168;
  • 4) 0.002 256 168 × 2 = 0 + 0.004 512 336;
  • 5) 0.004 512 336 × 2 = 0 + 0.009 024 672;
  • 6) 0.009 024 672 × 2 = 0 + 0.018 049 344;
  • 7) 0.018 049 344 × 2 = 0 + 0.036 098 688;
  • 8) 0.036 098 688 × 2 = 0 + 0.072 197 376;
  • 9) 0.072 197 376 × 2 = 0 + 0.144 394 752;
  • 10) 0.144 394 752 × 2 = 0 + 0.288 789 504;
  • 11) 0.288 789 504 × 2 = 0 + 0.577 579 008;
  • 12) 0.577 579 008 × 2 = 1 + 0.155 158 016;
  • 13) 0.155 158 016 × 2 = 0 + 0.310 316 032;
  • 14) 0.310 316 032 × 2 = 0 + 0.620 632 064;
  • 15) 0.620 632 064 × 2 = 1 + 0.241 264 128;
  • 16) 0.241 264 128 × 2 = 0 + 0.482 528 256;
  • 17) 0.482 528 256 × 2 = 0 + 0.965 056 512;
  • 18) 0.965 056 512 × 2 = 1 + 0.930 113 024;
  • 19) 0.930 113 024 × 2 = 1 + 0.860 226 048;
  • 20) 0.860 226 048 × 2 = 1 + 0.720 452 096;
  • 21) 0.720 452 096 × 2 = 1 + 0.440 904 192;
  • 22) 0.440 904 192 × 2 = 0 + 0.881 808 384;
  • 23) 0.881 808 384 × 2 = 1 + 0.763 616 768;
  • 24) 0.763 616 768 × 2 = 1 + 0.527 233 536;
  • 25) 0.527 233 536 × 2 = 1 + 0.054 467 072;
  • 26) 0.054 467 072 × 2 = 0 + 0.108 934 144;
  • 27) 0.108 934 144 × 2 = 0 + 0.217 868 288;
  • 28) 0.217 868 288 × 2 = 0 + 0.435 736 576;
  • 29) 0.435 736 576 × 2 = 0 + 0.871 473 152;
  • 30) 0.871 473 152 × 2 = 1 + 0.742 946 304;
  • 31) 0.742 946 304 × 2 = 1 + 0.485 892 608;
  • 32) 0.485 892 608 × 2 = 0 + 0.971 785 216;
  • 33) 0.971 785 216 × 2 = 1 + 0.943 570 432;
  • 34) 0.943 570 432 × 2 = 1 + 0.887 140 864;
  • 35) 0.887 140 864 × 2 = 1 + 0.774 281 728;
  • 36) 0.774 281 728 × 2 = 1 + 0.548 563 456;
  • 37) 0.548 563 456 × 2 = 1 + 0.097 126 912;
  • 38) 0.097 126 912 × 2 = 0 + 0.194 253 824;
  • 39) 0.194 253 824 × 2 = 0 + 0.388 507 648;
  • 40) 0.388 507 648 × 2 = 0 + 0.777 015 296;
  • 41) 0.777 015 296 × 2 = 1 + 0.554 030 592;
  • 42) 0.554 030 592 × 2 = 1 + 0.108 061 184;
  • 43) 0.108 061 184 × 2 = 0 + 0.216 122 368;
  • 44) 0.216 122 368 × 2 = 0 + 0.432 244 736;
  • 45) 0.432 244 736 × 2 = 0 + 0.864 489 472;
  • 46) 0.864 489 472 × 2 = 1 + 0.728 978 944;
  • 47) 0.728 978 944 × 2 = 1 + 0.457 957 888;
  • 48) 0.457 957 888 × 2 = 0 + 0.915 915 776;
  • 49) 0.915 915 776 × 2 = 1 + 0.831 831 552;
  • 50) 0.831 831 552 × 2 = 1 + 0.663 663 104;
  • 51) 0.663 663 104 × 2 = 1 + 0.327 326 208;
  • 52) 0.327 326 208 × 2 = 0 + 0.654 652 416;
  • 53) 0.654 652 416 × 2 = 1 + 0.309 304 832;
  • 54) 0.309 304 832 × 2 = 0 + 0.618 609 664;
  • 55) 0.618 609 664 × 2 = 1 + 0.237 219 328;
  • 56) 0.237 219 328 × 2 = 0 + 0.474 438 656;
  • 57) 0.474 438 656 × 2 = 0 + 0.948 877 312;
  • 58) 0.948 877 312 × 2 = 1 + 0.897 754 624;
  • 59) 0.897 754 624 × 2 = 1 + 0.795 509 248;
  • 60) 0.795 509 248 × 2 = 1 + 0.591 018 496;
  • 61) 0.591 018 496 × 2 = 1 + 0.182 036 992;
  • 62) 0.182 036 992 × 2 = 0 + 0.364 073 984;
  • 63) 0.364 073 984 × 2 = 0 + 0.728 147 968;
  • 64) 0.728 147 968 × 2 = 1 + 0.456 295 936;

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:


0.000 282 021(10) =

0.0000 0000 0001 0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 1001(2)

6. Positive number before normalization:

0.000 282 021(10) =

0.0000 0000 0001 0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 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:


0.000 282 021(10) =

0.0000 0000 0001 0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 1001(2) =

0.0000 0000 0001 0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 1001(2) × 20 =

1.0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 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 1000 0110 1111 1000 1100 0110 1110 1010 0111 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 1000 0110 1111 1000 1100 0110 1110 1010 0111 1001 =

0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 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 1000 0110 1111 1000 1100 0110 1110 1010 0111 1001


Decimal number -0.000 282 021 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1000 0110 1111 1000 1100 0110 1110 1010 0111 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