-0.215 000 000 000 189 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.215 000 000 000 189(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
-0.215 000 000 000 189(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:

|-0.215 000 000 000 189| = 0.215 000 000 000 189


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: 0.215 000 000 000 189.

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) 0.215 000 000 000 189 × 2 = 0 + 0.430 000 000 000 378;
  • 2) 0.430 000 000 000 378 × 2 = 0 + 0.860 000 000 000 756;
  • 3) 0.860 000 000 000 756 × 2 = 1 + 0.720 000 000 001 512;
  • 4) 0.720 000 000 001 512 × 2 = 1 + 0.440 000 000 003 024;
  • 5) 0.440 000 000 003 024 × 2 = 0 + 0.880 000 000 006 048;
  • 6) 0.880 000 000 006 048 × 2 = 1 + 0.760 000 000 012 096;
  • 7) 0.760 000 000 012 096 × 2 = 1 + 0.520 000 000 024 192;
  • 8) 0.520 000 000 024 192 × 2 = 1 + 0.040 000 000 048 384;
  • 9) 0.040 000 000 048 384 × 2 = 0 + 0.080 000 000 096 768;
  • 10) 0.080 000 000 096 768 × 2 = 0 + 0.160 000 000 193 536;
  • 11) 0.160 000 000 193 536 × 2 = 0 + 0.320 000 000 387 072;
  • 12) 0.320 000 000 387 072 × 2 = 0 + 0.640 000 000 774 144;
  • 13) 0.640 000 000 774 144 × 2 = 1 + 0.280 000 001 548 288;
  • 14) 0.280 000 001 548 288 × 2 = 0 + 0.560 000 003 096 576;
  • 15) 0.560 000 003 096 576 × 2 = 1 + 0.120 000 006 193 152;
  • 16) 0.120 000 006 193 152 × 2 = 0 + 0.240 000 012 386 304;
  • 17) 0.240 000 012 386 304 × 2 = 0 + 0.480 000 024 772 608;
  • 18) 0.480 000 024 772 608 × 2 = 0 + 0.960 000 049 545 216;
  • 19) 0.960 000 049 545 216 × 2 = 1 + 0.920 000 099 090 432;
  • 20) 0.920 000 099 090 432 × 2 = 1 + 0.840 000 198 180 864;
  • 21) 0.840 000 198 180 864 × 2 = 1 + 0.680 000 396 361 728;
  • 22) 0.680 000 396 361 728 × 2 = 1 + 0.360 000 792 723 456;
  • 23) 0.360 000 792 723 456 × 2 = 0 + 0.720 001 585 446 912;
  • 24) 0.720 001 585 446 912 × 2 = 1 + 0.440 003 170 893 824;
  • 25) 0.440 003 170 893 824 × 2 = 0 + 0.880 006 341 787 648;
  • 26) 0.880 006 341 787 648 × 2 = 1 + 0.760 012 683 575 296;
  • 27) 0.760 012 683 575 296 × 2 = 1 + 0.520 025 367 150 592;
  • 28) 0.520 025 367 150 592 × 2 = 1 + 0.040 050 734 301 184;
  • 29) 0.040 050 734 301 184 × 2 = 0 + 0.080 101 468 602 368;
  • 30) 0.080 101 468 602 368 × 2 = 0 + 0.160 202 937 204 736;
  • 31) 0.160 202 937 204 736 × 2 = 0 + 0.320 405 874 409 472;
  • 32) 0.320 405 874 409 472 × 2 = 0 + 0.640 811 748 818 944;
  • 33) 0.640 811 748 818 944 × 2 = 1 + 0.281 623 497 637 888;
  • 34) 0.281 623 497 637 888 × 2 = 0 + 0.563 246 995 275 776;
  • 35) 0.563 246 995 275 776 × 2 = 1 + 0.126 493 990 551 552;
  • 36) 0.126 493 990 551 552 × 2 = 0 + 0.252 987 981 103 104;
  • 37) 0.252 987 981 103 104 × 2 = 0 + 0.505 975 962 206 208;
  • 38) 0.505 975 962 206 208 × 2 = 1 + 0.011 951 924 412 416;
  • 39) 0.011 951 924 412 416 × 2 = 0 + 0.023 903 848 824 832;
  • 40) 0.023 903 848 824 832 × 2 = 0 + 0.047 807 697 649 664;
  • 41) 0.047 807 697 649 664 × 2 = 0 + 0.095 615 395 299 328;
  • 42) 0.095 615 395 299 328 × 2 = 0 + 0.191 230 790 598 656;
  • 43) 0.191 230 790 598 656 × 2 = 0 + 0.382 461 581 197 312;
  • 44) 0.382 461 581 197 312 × 2 = 0 + 0.764 923 162 394 624;
  • 45) 0.764 923 162 394 624 × 2 = 1 + 0.529 846 324 789 248;
  • 46) 0.529 846 324 789 248 × 2 = 1 + 0.059 692 649 578 496;
  • 47) 0.059 692 649 578 496 × 2 = 0 + 0.119 385 299 156 992;
  • 48) 0.119 385 299 156 992 × 2 = 0 + 0.238 770 598 313 984;
  • 49) 0.238 770 598 313 984 × 2 = 0 + 0.477 541 196 627 968;
  • 50) 0.477 541 196 627 968 × 2 = 0 + 0.955 082 393 255 936;
  • 51) 0.955 082 393 255 936 × 2 = 1 + 0.910 164 786 511 872;
  • 52) 0.910 164 786 511 872 × 2 = 1 + 0.820 329 573 023 744;
  • 53) 0.820 329 573 023 744 × 2 = 1 + 0.640 659 146 047 488;
  • 54) 0.640 659 146 047 488 × 2 = 1 + 0.281 318 292 094 976;
  • 55) 0.281 318 292 094 976 × 2 = 0 + 0.562 636 584 189 952;

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:


0.215 000 000 000 189(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0100 0000 1100 0011 110(2)

6. Positive number before normalization:

0.215 000 000 000 189(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0100 0000 1100 0011 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.215 000 000 000 189(10) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0100 0000 1100 0011 110(2) =

0.0011 0111 0000 1010 0011 1101 0111 0000 1010 0100 0000 1100 0011 110(2) × 20 =

1.1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110(2) × 2-3


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): -3

Mantissa (not normalized):
1.1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110 =

1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110


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 1100

Mantissa (52 bits) =
1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110


Decimal number -0.215 000 000 000 189 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1011 1000 0101 0001 1110 1011 1000 0101 0010 0000 0110 0001 1110

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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