-0.000 282 006 238 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 238(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
-0.000 282 006 238(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:

|-0.000 282 006 238| = 0.000 282 006 238


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: 0.000 282 006 238.

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) 0.000 282 006 238 × 2 = 0 + 0.000 564 012 476;
  • 2) 0.000 564 012 476 × 2 = 0 + 0.001 128 024 952;
  • 3) 0.001 128 024 952 × 2 = 0 + 0.002 256 049 904;
  • 4) 0.002 256 049 904 × 2 = 0 + 0.004 512 099 808;
  • 5) 0.004 512 099 808 × 2 = 0 + 0.009 024 199 616;
  • 6) 0.009 024 199 616 × 2 = 0 + 0.018 048 399 232;
  • 7) 0.018 048 399 232 × 2 = 0 + 0.036 096 798 464;
  • 8) 0.036 096 798 464 × 2 = 0 + 0.072 193 596 928;
  • 9) 0.072 193 596 928 × 2 = 0 + 0.144 387 193 856;
  • 10) 0.144 387 193 856 × 2 = 0 + 0.288 774 387 712;
  • 11) 0.288 774 387 712 × 2 = 0 + 0.577 548 775 424;
  • 12) 0.577 548 775 424 × 2 = 1 + 0.155 097 550 848;
  • 13) 0.155 097 550 848 × 2 = 0 + 0.310 195 101 696;
  • 14) 0.310 195 101 696 × 2 = 0 + 0.620 390 203 392;
  • 15) 0.620 390 203 392 × 2 = 1 + 0.240 780 406 784;
  • 16) 0.240 780 406 784 × 2 = 0 + 0.481 560 813 568;
  • 17) 0.481 560 813 568 × 2 = 0 + 0.963 121 627 136;
  • 18) 0.963 121 627 136 × 2 = 1 + 0.926 243 254 272;
  • 19) 0.926 243 254 272 × 2 = 1 + 0.852 486 508 544;
  • 20) 0.852 486 508 544 × 2 = 1 + 0.704 973 017 088;
  • 21) 0.704 973 017 088 × 2 = 1 + 0.409 946 034 176;
  • 22) 0.409 946 034 176 × 2 = 0 + 0.819 892 068 352;
  • 23) 0.819 892 068 352 × 2 = 1 + 0.639 784 136 704;
  • 24) 0.639 784 136 704 × 2 = 1 + 0.279 568 273 408;
  • 25) 0.279 568 273 408 × 2 = 0 + 0.559 136 546 816;
  • 26) 0.559 136 546 816 × 2 = 1 + 0.118 273 093 632;
  • 27) 0.118 273 093 632 × 2 = 0 + 0.236 546 187 264;
  • 28) 0.236 546 187 264 × 2 = 0 + 0.473 092 374 528;
  • 29) 0.473 092 374 528 × 2 = 0 + 0.946 184 749 056;
  • 30) 0.946 184 749 056 × 2 = 1 + 0.892 369 498 112;
  • 31) 0.892 369 498 112 × 2 = 1 + 0.784 738 996 224;
  • 32) 0.784 738 996 224 × 2 = 1 + 0.569 477 992 448;
  • 33) 0.569 477 992 448 × 2 = 1 + 0.138 955 984 896;
  • 34) 0.138 955 984 896 × 2 = 0 + 0.277 911 969 792;
  • 35) 0.277 911 969 792 × 2 = 0 + 0.555 823 939 584;
  • 36) 0.555 823 939 584 × 2 = 1 + 0.111 647 879 168;
  • 37) 0.111 647 879 168 × 2 = 0 + 0.223 295 758 336;
  • 38) 0.223 295 758 336 × 2 = 0 + 0.446 591 516 672;
  • 39) 0.446 591 516 672 × 2 = 0 + 0.893 183 033 344;
  • 40) 0.893 183 033 344 × 2 = 1 + 0.786 366 066 688;
  • 41) 0.786 366 066 688 × 2 = 1 + 0.572 732 133 376;
  • 42) 0.572 732 133 376 × 2 = 1 + 0.145 464 266 752;
  • 43) 0.145 464 266 752 × 2 = 0 + 0.290 928 533 504;
  • 44) 0.290 928 533 504 × 2 = 0 + 0.581 857 067 008;
  • 45) 0.581 857 067 008 × 2 = 1 + 0.163 714 134 016;
  • 46) 0.163 714 134 016 × 2 = 0 + 0.327 428 268 032;
  • 47) 0.327 428 268 032 × 2 = 0 + 0.654 856 536 064;
  • 48) 0.654 856 536 064 × 2 = 1 + 0.309 713 072 128;
  • 49) 0.309 713 072 128 × 2 = 0 + 0.619 426 144 256;
  • 50) 0.619 426 144 256 × 2 = 1 + 0.238 852 288 512;
  • 51) 0.238 852 288 512 × 2 = 0 + 0.477 704 577 024;
  • 52) 0.477 704 577 024 × 2 = 0 + 0.955 409 154 048;
  • 53) 0.955 409 154 048 × 2 = 1 + 0.910 818 308 096;
  • 54) 0.910 818 308 096 × 2 = 1 + 0.821 636 616 192;
  • 55) 0.821 636 616 192 × 2 = 1 + 0.643 273 232 384;
  • 56) 0.643 273 232 384 × 2 = 1 + 0.286 546 464 768;
  • 57) 0.286 546 464 768 × 2 = 0 + 0.573 092 929 536;
  • 58) 0.573 092 929 536 × 2 = 1 + 0.146 185 859 072;
  • 59) 0.146 185 859 072 × 2 = 0 + 0.292 371 718 144;
  • 60) 0.292 371 718 144 × 2 = 0 + 0.584 743 436 288;
  • 61) 0.584 743 436 288 × 2 = 1 + 0.169 486 872 576;
  • 62) 0.169 486 872 576 × 2 = 0 + 0.338 973 745 152;
  • 63) 0.338 973 745 152 × 2 = 0 + 0.677 947 490 304;
  • 64) 0.677 947 490 304 × 2 = 1 + 0.355 894 980 608;

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:


0.000 282 006 238(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 1001(2)

6. Positive number before normalization:

0.000 282 006 238(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 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:


0.000 282 006 238(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 1001(2) × 20 =

1.0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 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 0100 0111 1001 0001 1100 1001 0100 1111 0100 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 0100 0111 1001 0001 1100 1001 0100 1111 0100 1001 =

0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 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 0100 0111 1001 0001 1100 1001 0100 1111 0100 1001


Decimal number -0.000 282 006 238 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1001 0001 1100 1001 0100 1111 0100 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