-0.000 282 524 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 524(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
-0.000 282 524(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:

|-0.000 282 524| = 0.000 282 524


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: 0.000 282 524.

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) 0.000 282 524 × 2 = 0 + 0.000 565 048;
  • 2) 0.000 565 048 × 2 = 0 + 0.001 130 096;
  • 3) 0.001 130 096 × 2 = 0 + 0.002 260 192;
  • 4) 0.002 260 192 × 2 = 0 + 0.004 520 384;
  • 5) 0.004 520 384 × 2 = 0 + 0.009 040 768;
  • 6) 0.009 040 768 × 2 = 0 + 0.018 081 536;
  • 7) 0.018 081 536 × 2 = 0 + 0.036 163 072;
  • 8) 0.036 163 072 × 2 = 0 + 0.072 326 144;
  • 9) 0.072 326 144 × 2 = 0 + 0.144 652 288;
  • 10) 0.144 652 288 × 2 = 0 + 0.289 304 576;
  • 11) 0.289 304 576 × 2 = 0 + 0.578 609 152;
  • 12) 0.578 609 152 × 2 = 1 + 0.157 218 304;
  • 13) 0.157 218 304 × 2 = 0 + 0.314 436 608;
  • 14) 0.314 436 608 × 2 = 0 + 0.628 873 216;
  • 15) 0.628 873 216 × 2 = 1 + 0.257 746 432;
  • 16) 0.257 746 432 × 2 = 0 + 0.515 492 864;
  • 17) 0.515 492 864 × 2 = 1 + 0.030 985 728;
  • 18) 0.030 985 728 × 2 = 0 + 0.061 971 456;
  • 19) 0.061 971 456 × 2 = 0 + 0.123 942 912;
  • 20) 0.123 942 912 × 2 = 0 + 0.247 885 824;
  • 21) 0.247 885 824 × 2 = 0 + 0.495 771 648;
  • 22) 0.495 771 648 × 2 = 0 + 0.991 543 296;
  • 23) 0.991 543 296 × 2 = 1 + 0.983 086 592;
  • 24) 0.983 086 592 × 2 = 1 + 0.966 173 184;
  • 25) 0.966 173 184 × 2 = 1 + 0.932 346 368;
  • 26) 0.932 346 368 × 2 = 1 + 0.864 692 736;
  • 27) 0.864 692 736 × 2 = 1 + 0.729 385 472;
  • 28) 0.729 385 472 × 2 = 1 + 0.458 770 944;
  • 29) 0.458 770 944 × 2 = 0 + 0.917 541 888;
  • 30) 0.917 541 888 × 2 = 1 + 0.835 083 776;
  • 31) 0.835 083 776 × 2 = 1 + 0.670 167 552;
  • 32) 0.670 167 552 × 2 = 1 + 0.340 335 104;
  • 33) 0.340 335 104 × 2 = 0 + 0.680 670 208;
  • 34) 0.680 670 208 × 2 = 1 + 0.361 340 416;
  • 35) 0.361 340 416 × 2 = 0 + 0.722 680 832;
  • 36) 0.722 680 832 × 2 = 1 + 0.445 361 664;
  • 37) 0.445 361 664 × 2 = 0 + 0.890 723 328;
  • 38) 0.890 723 328 × 2 = 1 + 0.781 446 656;
  • 39) 0.781 446 656 × 2 = 1 + 0.562 893 312;
  • 40) 0.562 893 312 × 2 = 1 + 0.125 786 624;
  • 41) 0.125 786 624 × 2 = 0 + 0.251 573 248;
  • 42) 0.251 573 248 × 2 = 0 + 0.503 146 496;
  • 43) 0.503 146 496 × 2 = 1 + 0.006 292 992;
  • 44) 0.006 292 992 × 2 = 0 + 0.012 585 984;
  • 45) 0.012 585 984 × 2 = 0 + 0.025 171 968;
  • 46) 0.025 171 968 × 2 = 0 + 0.050 343 936;
  • 47) 0.050 343 936 × 2 = 0 + 0.100 687 872;
  • 48) 0.100 687 872 × 2 = 0 + 0.201 375 744;
  • 49) 0.201 375 744 × 2 = 0 + 0.402 751 488;
  • 50) 0.402 751 488 × 2 = 0 + 0.805 502 976;
  • 51) 0.805 502 976 × 2 = 1 + 0.611 005 952;
  • 52) 0.611 005 952 × 2 = 1 + 0.222 011 904;
  • 53) 0.222 011 904 × 2 = 0 + 0.444 023 808;
  • 54) 0.444 023 808 × 2 = 0 + 0.888 047 616;
  • 55) 0.888 047 616 × 2 = 1 + 0.776 095 232;
  • 56) 0.776 095 232 × 2 = 1 + 0.552 190 464;
  • 57) 0.552 190 464 × 2 = 1 + 0.104 380 928;
  • 58) 0.104 380 928 × 2 = 0 + 0.208 761 856;
  • 59) 0.208 761 856 × 2 = 0 + 0.417 523 712;
  • 60) 0.417 523 712 × 2 = 0 + 0.835 047 424;
  • 61) 0.835 047 424 × 2 = 1 + 0.670 094 848;
  • 62) 0.670 094 848 × 2 = 1 + 0.340 189 696;
  • 63) 0.340 189 696 × 2 = 0 + 0.680 379 392;
  • 64) 0.680 379 392 × 2 = 1 + 0.360 758 784;

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:


0.000 282 524(10) =

0.0000 0000 0001 0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101(2)

6. Positive number before normalization:

0.000 282 524(10) =

0.0000 0000 0001 0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101(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:


0.000 282 524(10) =

0.0000 0000 0001 0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101(2) =

0.0000 0000 0001 0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101(2) × 20 =

1.0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101(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 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101


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 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101 =

0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101


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 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101


Decimal number -0.000 282 524 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 1111 0111 0101 0111 0010 0000 0011 0011 1000 1101

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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