0.000 000 000 000 55 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 55(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
0.000 000 000 000 55(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 55.

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) 0.000 000 000 000 55 × 2 = 0 + 0.000 000 000 001 1;
  • 2) 0.000 000 000 001 1 × 2 = 0 + 0.000 000 000 002 2;
  • 3) 0.000 000 000 002 2 × 2 = 0 + 0.000 000 000 004 4;
  • 4) 0.000 000 000 004 4 × 2 = 0 + 0.000 000 000 008 8;
  • 5) 0.000 000 000 008 8 × 2 = 0 + 0.000 000 000 017 6;
  • 6) 0.000 000 000 017 6 × 2 = 0 + 0.000 000 000 035 2;
  • 7) 0.000 000 000 035 2 × 2 = 0 + 0.000 000 000 070 4;
  • 8) 0.000 000 000 070 4 × 2 = 0 + 0.000 000 000 140 8;
  • 9) 0.000 000 000 140 8 × 2 = 0 + 0.000 000 000 281 6;
  • 10) 0.000 000 000 281 6 × 2 = 0 + 0.000 000 000 563 2;
  • 11) 0.000 000 000 563 2 × 2 = 0 + 0.000 000 001 126 4;
  • 12) 0.000 000 001 126 4 × 2 = 0 + 0.000 000 002 252 8;
  • 13) 0.000 000 002 252 8 × 2 = 0 + 0.000 000 004 505 6;
  • 14) 0.000 000 004 505 6 × 2 = 0 + 0.000 000 009 011 2;
  • 15) 0.000 000 009 011 2 × 2 = 0 + 0.000 000 018 022 4;
  • 16) 0.000 000 018 022 4 × 2 = 0 + 0.000 000 036 044 8;
  • 17) 0.000 000 036 044 8 × 2 = 0 + 0.000 000 072 089 6;
  • 18) 0.000 000 072 089 6 × 2 = 0 + 0.000 000 144 179 2;
  • 19) 0.000 000 144 179 2 × 2 = 0 + 0.000 000 288 358 4;
  • 20) 0.000 000 288 358 4 × 2 = 0 + 0.000 000 576 716 8;
  • 21) 0.000 000 576 716 8 × 2 = 0 + 0.000 001 153 433 6;
  • 22) 0.000 001 153 433 6 × 2 = 0 + 0.000 002 306 867 2;
  • 23) 0.000 002 306 867 2 × 2 = 0 + 0.000 004 613 734 4;
  • 24) 0.000 004 613 734 4 × 2 = 0 + 0.000 009 227 468 8;
  • 25) 0.000 009 227 468 8 × 2 = 0 + 0.000 018 454 937 6;
  • 26) 0.000 018 454 937 6 × 2 = 0 + 0.000 036 909 875 2;
  • 27) 0.000 036 909 875 2 × 2 = 0 + 0.000 073 819 750 4;
  • 28) 0.000 073 819 750 4 × 2 = 0 + 0.000 147 639 500 8;
  • 29) 0.000 147 639 500 8 × 2 = 0 + 0.000 295 279 001 6;
  • 30) 0.000 295 279 001 6 × 2 = 0 + 0.000 590 558 003 2;
  • 31) 0.000 590 558 003 2 × 2 = 0 + 0.001 181 116 006 4;
  • 32) 0.001 181 116 006 4 × 2 = 0 + 0.002 362 232 012 8;
  • 33) 0.002 362 232 012 8 × 2 = 0 + 0.004 724 464 025 6;
  • 34) 0.004 724 464 025 6 × 2 = 0 + 0.009 448 928 051 2;
  • 35) 0.009 448 928 051 2 × 2 = 0 + 0.018 897 856 102 4;
  • 36) 0.018 897 856 102 4 × 2 = 0 + 0.037 795 712 204 8;
  • 37) 0.037 795 712 204 8 × 2 = 0 + 0.075 591 424 409 6;
  • 38) 0.075 591 424 409 6 × 2 = 0 + 0.151 182 848 819 2;
  • 39) 0.151 182 848 819 2 × 2 = 0 + 0.302 365 697 638 4;
  • 40) 0.302 365 697 638 4 × 2 = 0 + 0.604 731 395 276 8;
  • 41) 0.604 731 395 276 8 × 2 = 1 + 0.209 462 790 553 6;
  • 42) 0.209 462 790 553 6 × 2 = 0 + 0.418 925 581 107 2;
  • 43) 0.418 925 581 107 2 × 2 = 0 + 0.837 851 162 214 4;
  • 44) 0.837 851 162 214 4 × 2 = 1 + 0.675 702 324 428 8;
  • 45) 0.675 702 324 428 8 × 2 = 1 + 0.351 404 648 857 6;
  • 46) 0.351 404 648 857 6 × 2 = 0 + 0.702 809 297 715 2;
  • 47) 0.702 809 297 715 2 × 2 = 1 + 0.405 618 595 430 4;
  • 48) 0.405 618 595 430 4 × 2 = 0 + 0.811 237 190 860 8;
  • 49) 0.811 237 190 860 8 × 2 = 1 + 0.622 474 381 721 6;
  • 50) 0.622 474 381 721 6 × 2 = 1 + 0.244 948 763 443 2;
  • 51) 0.244 948 763 443 2 × 2 = 0 + 0.489 897 526 886 4;
  • 52) 0.489 897 526 886 4 × 2 = 0 + 0.979 795 053 772 8;
  • 53) 0.979 795 053 772 8 × 2 = 1 + 0.959 590 107 545 6;
  • 54) 0.959 590 107 545 6 × 2 = 1 + 0.919 180 215 091 2;
  • 55) 0.919 180 215 091 2 × 2 = 1 + 0.838 360 430 182 4;
  • 56) 0.838 360 430 182 4 × 2 = 1 + 0.676 720 860 364 8;
  • 57) 0.676 720 860 364 8 × 2 = 1 + 0.353 441 720 729 6;
  • 58) 0.353 441 720 729 6 × 2 = 0 + 0.706 883 441 459 2;
  • 59) 0.706 883 441 459 2 × 2 = 1 + 0.413 766 882 918 4;
  • 60) 0.413 766 882 918 4 × 2 = 0 + 0.827 533 765 836 8;
  • 61) 0.827 533 765 836 8 × 2 = 1 + 0.655 067 531 673 6;
  • 62) 0.655 067 531 673 6 × 2 = 1 + 0.310 135 063 347 2;
  • 63) 0.310 135 063 347 2 × 2 = 0 + 0.620 270 126 694 4;
  • 64) 0.620 270 126 694 4 × 2 = 1 + 0.240 540 253 388 8;
  • 65) 0.240 540 253 388 8 × 2 = 0 + 0.481 080 506 777 6;
  • 66) 0.481 080 506 777 6 × 2 = 0 + 0.962 161 013 555 2;
  • 67) 0.962 161 013 555 2 × 2 = 1 + 0.924 322 027 110 4;
  • 68) 0.924 322 027 110 4 × 2 = 1 + 0.848 644 054 220 8;
  • 69) 0.848 644 054 220 8 × 2 = 1 + 0.697 288 108 441 6;
  • 70) 0.697 288 108 441 6 × 2 = 1 + 0.394 576 216 883 2;
  • 71) 0.394 576 216 883 2 × 2 = 0 + 0.789 152 433 766 4;
  • 72) 0.789 152 433 766 4 × 2 = 1 + 0.578 304 867 532 8;
  • 73) 0.578 304 867 532 8 × 2 = 1 + 0.156 609 735 065 6;
  • 74) 0.156 609 735 065 6 × 2 = 0 + 0.313 219 470 131 2;
  • 75) 0.313 219 470 131 2 × 2 = 0 + 0.626 438 940 262 4;
  • 76) 0.626 438 940 262 4 × 2 = 1 + 0.252 877 880 524 8;
  • 77) 0.252 877 880 524 8 × 2 = 0 + 0.505 755 761 049 6;
  • 78) 0.505 755 761 049 6 × 2 = 1 + 0.011 511 522 099 2;
  • 79) 0.011 511 522 099 2 × 2 = 0 + 0.023 023 044 198 4;
  • 80) 0.023 023 044 198 4 × 2 = 0 + 0.046 046 088 396 8;
  • 81) 0.046 046 088 396 8 × 2 = 0 + 0.092 092 176 793 6;
  • 82) 0.092 092 176 793 6 × 2 = 0 + 0.184 184 353 587 2;
  • 83) 0.184 184 353 587 2 × 2 = 0 + 0.368 368 707 174 4;
  • 84) 0.368 368 707 174 4 × 2 = 0 + 0.736 737 414 348 8;
  • 85) 0.736 737 414 348 8 × 2 = 1 + 0.473 474 828 697 6;
  • 86) 0.473 474 828 697 6 × 2 = 0 + 0.946 949 657 395 2;
  • 87) 0.946 949 657 395 2 × 2 = 1 + 0.893 899 314 790 4;
  • 88) 0.893 899 314 790 4 × 2 = 1 + 0.787 798 629 580 8;
  • 89) 0.787 798 629 580 8 × 2 = 1 + 0.575 597 259 161 6;
  • 90) 0.575 597 259 161 6 × 2 = 1 + 0.151 194 518 323 2;
  • 91) 0.151 194 518 323 2 × 2 = 0 + 0.302 389 036 646 4;
  • 92) 0.302 389 036 646 4 × 2 = 0 + 0.604 778 073 292 8;
  • 93) 0.604 778 073 292 8 × 2 = 1 + 0.209 556 146 585 6;

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).


4. 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:


0.000 000 000 000 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1010 1100 1111 1010 1101 0011 1101 1001 0100 0000 1011 1100 1(2)

5. Positive number before normalization:

0.000 000 000 000 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1010 1100 1111 1010 1101 0011 1101 1001 0100 0000 1011 1100 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 41 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1010 1100 1111 1010 1101 0011 1101 1001 0100 0000 1011 1100 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1010 1100 1111 1010 1101 0011 1101 1001 0100 0000 1011 1100 1(2) × 20 =

1.0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 1001(2) × 2-41


7. 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 0 (a positive number)

Exponent (unadjusted): -41

Mantissa (not normalized):
1.0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-41 + 2(11-1) - 1 =

(-41 + 1 023)(10) =

982(10)


9. 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;
  • 982 ÷ 2 = 491 + 0;
  • 491 ÷ 2 = 245 + 1;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

982(10) =

011 1101 0110(2)


11. 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. 0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 1001 =

0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0110

Mantissa (52 bits) =
0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 1001


Decimal number 0.000 000 000 000 55 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0110 - 0011 0101 1001 1111 0101 1010 0111 1011 0010 1000 0001 0111 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