-0.000 282 006 422 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 422(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
-0.000 282 006 422(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:

|-0.000 282 006 422| = 0.000 282 006 422


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: 0.000 282 006 422.

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) 0.000 282 006 422 × 2 = 0 + 0.000 564 012 844;
  • 2) 0.000 564 012 844 × 2 = 0 + 0.001 128 025 688;
  • 3) 0.001 128 025 688 × 2 = 0 + 0.002 256 051 376;
  • 4) 0.002 256 051 376 × 2 = 0 + 0.004 512 102 752;
  • 5) 0.004 512 102 752 × 2 = 0 + 0.009 024 205 504;
  • 6) 0.009 024 205 504 × 2 = 0 + 0.018 048 411 008;
  • 7) 0.018 048 411 008 × 2 = 0 + 0.036 096 822 016;
  • 8) 0.036 096 822 016 × 2 = 0 + 0.072 193 644 032;
  • 9) 0.072 193 644 032 × 2 = 0 + 0.144 387 288 064;
  • 10) 0.144 387 288 064 × 2 = 0 + 0.288 774 576 128;
  • 11) 0.288 774 576 128 × 2 = 0 + 0.577 549 152 256;
  • 12) 0.577 549 152 256 × 2 = 1 + 0.155 098 304 512;
  • 13) 0.155 098 304 512 × 2 = 0 + 0.310 196 609 024;
  • 14) 0.310 196 609 024 × 2 = 0 + 0.620 393 218 048;
  • 15) 0.620 393 218 048 × 2 = 1 + 0.240 786 436 096;
  • 16) 0.240 786 436 096 × 2 = 0 + 0.481 572 872 192;
  • 17) 0.481 572 872 192 × 2 = 0 + 0.963 145 744 384;
  • 18) 0.963 145 744 384 × 2 = 1 + 0.926 291 488 768;
  • 19) 0.926 291 488 768 × 2 = 1 + 0.852 582 977 536;
  • 20) 0.852 582 977 536 × 2 = 1 + 0.705 165 955 072;
  • 21) 0.705 165 955 072 × 2 = 1 + 0.410 331 910 144;
  • 22) 0.410 331 910 144 × 2 = 0 + 0.820 663 820 288;
  • 23) 0.820 663 820 288 × 2 = 1 + 0.641 327 640 576;
  • 24) 0.641 327 640 576 × 2 = 1 + 0.282 655 281 152;
  • 25) 0.282 655 281 152 × 2 = 0 + 0.565 310 562 304;
  • 26) 0.565 310 562 304 × 2 = 1 + 0.130 621 124 608;
  • 27) 0.130 621 124 608 × 2 = 0 + 0.261 242 249 216;
  • 28) 0.261 242 249 216 × 2 = 0 + 0.522 484 498 432;
  • 29) 0.522 484 498 432 × 2 = 1 + 0.044 968 996 864;
  • 30) 0.044 968 996 864 × 2 = 0 + 0.089 937 993 728;
  • 31) 0.089 937 993 728 × 2 = 0 + 0.179 875 987 456;
  • 32) 0.179 875 987 456 × 2 = 0 + 0.359 751 974 912;
  • 33) 0.359 751 974 912 × 2 = 0 + 0.719 503 949 824;
  • 34) 0.719 503 949 824 × 2 = 1 + 0.439 007 899 648;
  • 35) 0.439 007 899 648 × 2 = 0 + 0.878 015 799 296;
  • 36) 0.878 015 799 296 × 2 = 1 + 0.756 031 598 592;
  • 37) 0.756 031 598 592 × 2 = 1 + 0.512 063 197 184;
  • 38) 0.512 063 197 184 × 2 = 1 + 0.024 126 394 368;
  • 39) 0.024 126 394 368 × 2 = 0 + 0.048 252 788 736;
  • 40) 0.048 252 788 736 × 2 = 0 + 0.096 505 577 472;
  • 41) 0.096 505 577 472 × 2 = 0 + 0.193 011 154 944;
  • 42) 0.193 011 154 944 × 2 = 0 + 0.386 022 309 888;
  • 43) 0.386 022 309 888 × 2 = 0 + 0.772 044 619 776;
  • 44) 0.772 044 619 776 × 2 = 1 + 0.544 089 239 552;
  • 45) 0.544 089 239 552 × 2 = 1 + 0.088 178 479 104;
  • 46) 0.088 178 479 104 × 2 = 0 + 0.176 356 958 208;
  • 47) 0.176 356 958 208 × 2 = 0 + 0.352 713 916 416;
  • 48) 0.352 713 916 416 × 2 = 0 + 0.705 427 832 832;
  • 49) 0.705 427 832 832 × 2 = 1 + 0.410 855 665 664;
  • 50) 0.410 855 665 664 × 2 = 0 + 0.821 711 331 328;
  • 51) 0.821 711 331 328 × 2 = 1 + 0.643 422 662 656;
  • 52) 0.643 422 662 656 × 2 = 1 + 0.286 845 325 312;
  • 53) 0.286 845 325 312 × 2 = 0 + 0.573 690 650 624;
  • 54) 0.573 690 650 624 × 2 = 1 + 0.147 381 301 248;
  • 55) 0.147 381 301 248 × 2 = 0 + 0.294 762 602 496;
  • 56) 0.294 762 602 496 × 2 = 0 + 0.589 525 204 992;
  • 57) 0.589 525 204 992 × 2 = 1 + 0.179 050 409 984;
  • 58) 0.179 050 409 984 × 2 = 0 + 0.358 100 819 968;
  • 59) 0.358 100 819 968 × 2 = 0 + 0.716 201 639 936;
  • 60) 0.716 201 639 936 × 2 = 1 + 0.432 403 279 872;
  • 61) 0.432 403 279 872 × 2 = 0 + 0.864 806 559 744;
  • 62) 0.864 806 559 744 × 2 = 1 + 0.729 613 119 488;
  • 63) 0.729 613 119 488 × 2 = 1 + 0.459 226 238 976;
  • 64) 0.459 226 238 976 × 2 = 0 + 0.918 452 477 952;

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:


0.000 282 006 422(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110(2)

6. Positive number before normalization:

0.000 282 006 422(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110(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:


0.000 282 006 422(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110(2) × 20 =

1.0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110(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 1000 0101 1100 0001 1000 1011 0100 1001 0110


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 1000 0101 1100 0001 1000 1011 0100 1001 0110 =

0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110


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 1000 0101 1100 0001 1000 1011 0100 1001 0110


Decimal number -0.000 282 006 422 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0101 1100 0001 1000 1011 0100 1001 0110

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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