-0.000 282 006 137 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 137(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
-0.000 282 006 137(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:

|-0.000 282 006 137| = 0.000 282 006 137


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: 0.000 282 006 137.

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) 0.000 282 006 137 × 2 = 0 + 0.000 564 012 274;
  • 2) 0.000 564 012 274 × 2 = 0 + 0.001 128 024 548;
  • 3) 0.001 128 024 548 × 2 = 0 + 0.002 256 049 096;
  • 4) 0.002 256 049 096 × 2 = 0 + 0.004 512 098 192;
  • 5) 0.004 512 098 192 × 2 = 0 + 0.009 024 196 384;
  • 6) 0.009 024 196 384 × 2 = 0 + 0.018 048 392 768;
  • 7) 0.018 048 392 768 × 2 = 0 + 0.036 096 785 536;
  • 8) 0.036 096 785 536 × 2 = 0 + 0.072 193 571 072;
  • 9) 0.072 193 571 072 × 2 = 0 + 0.144 387 142 144;
  • 10) 0.144 387 142 144 × 2 = 0 + 0.288 774 284 288;
  • 11) 0.288 774 284 288 × 2 = 0 + 0.577 548 568 576;
  • 12) 0.577 548 568 576 × 2 = 1 + 0.155 097 137 152;
  • 13) 0.155 097 137 152 × 2 = 0 + 0.310 194 274 304;
  • 14) 0.310 194 274 304 × 2 = 0 + 0.620 388 548 608;
  • 15) 0.620 388 548 608 × 2 = 1 + 0.240 777 097 216;
  • 16) 0.240 777 097 216 × 2 = 0 + 0.481 554 194 432;
  • 17) 0.481 554 194 432 × 2 = 0 + 0.963 108 388 864;
  • 18) 0.963 108 388 864 × 2 = 1 + 0.926 216 777 728;
  • 19) 0.926 216 777 728 × 2 = 1 + 0.852 433 555 456;
  • 20) 0.852 433 555 456 × 2 = 1 + 0.704 867 110 912;
  • 21) 0.704 867 110 912 × 2 = 1 + 0.409 734 221 824;
  • 22) 0.409 734 221 824 × 2 = 0 + 0.819 468 443 648;
  • 23) 0.819 468 443 648 × 2 = 1 + 0.638 936 887 296;
  • 24) 0.638 936 887 296 × 2 = 1 + 0.277 873 774 592;
  • 25) 0.277 873 774 592 × 2 = 0 + 0.555 747 549 184;
  • 26) 0.555 747 549 184 × 2 = 1 + 0.111 495 098 368;
  • 27) 0.111 495 098 368 × 2 = 0 + 0.222 990 196 736;
  • 28) 0.222 990 196 736 × 2 = 0 + 0.445 980 393 472;
  • 29) 0.445 980 393 472 × 2 = 0 + 0.891 960 786 944;
  • 30) 0.891 960 786 944 × 2 = 1 + 0.783 921 573 888;
  • 31) 0.783 921 573 888 × 2 = 1 + 0.567 843 147 776;
  • 32) 0.567 843 147 776 × 2 = 1 + 0.135 686 295 552;
  • 33) 0.135 686 295 552 × 2 = 0 + 0.271 372 591 104;
  • 34) 0.271 372 591 104 × 2 = 0 + 0.542 745 182 208;
  • 35) 0.542 745 182 208 × 2 = 1 + 0.085 490 364 416;
  • 36) 0.085 490 364 416 × 2 = 0 + 0.170 980 728 832;
  • 37) 0.170 980 728 832 × 2 = 0 + 0.341 961 457 664;
  • 38) 0.341 961 457 664 × 2 = 0 + 0.683 922 915 328;
  • 39) 0.683 922 915 328 × 2 = 1 + 0.367 845 830 656;
  • 40) 0.367 845 830 656 × 2 = 0 + 0.735 691 661 312;
  • 41) 0.735 691 661 312 × 2 = 1 + 0.471 383 322 624;
  • 42) 0.471 383 322 624 × 2 = 0 + 0.942 766 645 248;
  • 43) 0.942 766 645 248 × 2 = 1 + 0.885 533 290 496;
  • 44) 0.885 533 290 496 × 2 = 1 + 0.771 066 580 992;
  • 45) 0.771 066 580 992 × 2 = 1 + 0.542 133 161 984;
  • 46) 0.542 133 161 984 × 2 = 1 + 0.084 266 323 968;
  • 47) 0.084 266 323 968 × 2 = 0 + 0.168 532 647 936;
  • 48) 0.168 532 647 936 × 2 = 0 + 0.337 065 295 872;
  • 49) 0.337 065 295 872 × 2 = 0 + 0.674 130 591 744;
  • 50) 0.674 130 591 744 × 2 = 1 + 0.348 261 183 488;
  • 51) 0.348 261 183 488 × 2 = 0 + 0.696 522 366 976;
  • 52) 0.696 522 366 976 × 2 = 1 + 0.393 044 733 952;
  • 53) 0.393 044 733 952 × 2 = 0 + 0.786 089 467 904;
  • 54) 0.786 089 467 904 × 2 = 1 + 0.572 178 935 808;
  • 55) 0.572 178 935 808 × 2 = 1 + 0.144 357 871 616;
  • 56) 0.144 357 871 616 × 2 = 0 + 0.288 715 743 232;
  • 57) 0.288 715 743 232 × 2 = 0 + 0.577 431 486 464;
  • 58) 0.577 431 486 464 × 2 = 1 + 0.154 862 972 928;
  • 59) 0.154 862 972 928 × 2 = 0 + 0.309 725 945 856;
  • 60) 0.309 725 945 856 × 2 = 0 + 0.619 451 891 712;
  • 61) 0.619 451 891 712 × 2 = 1 + 0.238 903 783 424;
  • 62) 0.238 903 783 424 × 2 = 0 + 0.477 807 566 848;
  • 63) 0.477 807 566 848 × 2 = 0 + 0.955 615 133 696;
  • 64) 0.955 615 133 696 × 2 = 1 + 0.911 230 267 392;

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:


0.000 282 006 137(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 0100 1001(2)

6. Positive number before normalization:

0.000 282 006 137(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 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:


0.000 282 006 137(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 0100 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 0100 1001(2) × 20 =

1.0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 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 0010 0010 1011 1100 0101 0110 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 0010 0010 1011 1100 0101 0110 0100 1001 =

0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 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 0010 0010 1011 1100 0101 0110 0100 1001


Decimal number -0.000 282 006 137 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0010 0010 1011 1100 0101 0110 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