Convert 10 000 000 000 000 000 000 000 101 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number 10 000 000 000 000 000 000 000 101(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 10 000 000 000 000 000 000 000 101 ÷ 2 = 5 000 000 000 000 000 000 000 050 + 1;
  • 5 000 000 000 000 000 000 000 050 ÷ 2 = 2 500 000 000 000 000 000 000 025 + 0;
  • 2 500 000 000 000 000 000 000 025 ÷ 2 = 1 250 000 000 000 000 000 000 012 + 1;
  • 1 250 000 000 000 000 000 000 012 ÷ 2 = 625 000 000 000 000 000 000 006 + 0;
  • 625 000 000 000 000 000 000 006 ÷ 2 = 312 500 000 000 000 000 000 003 + 0;
  • 312 500 000 000 000 000 000 003 ÷ 2 = 156 250 000 000 000 000 000 001 + 1;
  • 156 250 000 000 000 000 000 001 ÷ 2 = 78 125 000 000 000 000 000 000 + 1;
  • 78 125 000 000 000 000 000 000 ÷ 2 = 39 062 500 000 000 000 000 000 + 0;
  • 39 062 500 000 000 000 000 000 ÷ 2 = 19 531 250 000 000 000 000 000 + 0;
  • 19 531 250 000 000 000 000 000 ÷ 2 = 9 765 625 000 000 000 000 000 + 0;
  • 9 765 625 000 000 000 000 000 ÷ 2 = 4 882 812 500 000 000 000 000 + 0;
  • 4 882 812 500 000 000 000 000 ÷ 2 = 2 441 406 250 000 000 000 000 + 0;
  • 2 441 406 250 000 000 000 000 ÷ 2 = 1 220 703 125 000 000 000 000 + 0;
  • 1 220 703 125 000 000 000 000 ÷ 2 = 610 351 562 500 000 000 000 + 0;
  • 610 351 562 500 000 000 000 ÷ 2 = 305 175 781 250 000 000 000 + 0;
  • 305 175 781 250 000 000 000 ÷ 2 = 152 587 890 625 000 000 000 + 0;
  • 152 587 890 625 000 000 000 ÷ 2 = 76 293 945 312 500 000 000 + 0;
  • 76 293 945 312 500 000 000 ÷ 2 = 38 146 972 656 250 000 000 + 0;
  • 38 146 972 656 250 000 000 ÷ 2 = 19 073 486 328 125 000 000 + 0;
  • 19 073 486 328 125 000 000 ÷ 2 = 9 536 743 164 062 500 000 + 0;
  • 9 536 743 164 062 500 000 ÷ 2 = 4 768 371 582 031 250 000 + 0;
  • 4 768 371 582 031 250 000 ÷ 2 = 2 384 185 791 015 625 000 + 0;
  • 2 384 185 791 015 625 000 ÷ 2 = 1 192 092 895 507 812 500 + 0;
  • 1 192 092 895 507 812 500 ÷ 2 = 596 046 447 753 906 250 + 0;
  • 596 046 447 753 906 250 ÷ 2 = 298 023 223 876 953 125 + 0;
  • 298 023 223 876 953 125 ÷ 2 = 149 011 611 938 476 562 + 1;
  • 149 011 611 938 476 562 ÷ 2 = 74 505 805 969 238 281 + 0;
  • 74 505 805 969 238 281 ÷ 2 = 37 252 902 984 619 140 + 1;
  • 37 252 902 984 619 140 ÷ 2 = 18 626 451 492 309 570 + 0;
  • 18 626 451 492 309 570 ÷ 2 = 9 313 225 746 154 785 + 0;
  • 9 313 225 746 154 785 ÷ 2 = 4 656 612 873 077 392 + 1;
  • 4 656 612 873 077 392 ÷ 2 = 2 328 306 436 538 696 + 0;
  • 2 328 306 436 538 696 ÷ 2 = 1 164 153 218 269 348 + 0;
  • 1 164 153 218 269 348 ÷ 2 = 582 076 609 134 674 + 0;
  • 582 076 609 134 674 ÷ 2 = 291 038 304 567 337 + 0;
  • 291 038 304 567 337 ÷ 2 = 145 519 152 283 668 + 1;
  • 145 519 152 283 668 ÷ 2 = 72 759 576 141 834 + 0;
  • 72 759 576 141 834 ÷ 2 = 36 379 788 070 917 + 0;
  • 36 379 788 070 917 ÷ 2 = 18 189 894 035 458 + 1;
  • 18 189 894 035 458 ÷ 2 = 9 094 947 017 729 + 0;
  • 9 094 947 017 729 ÷ 2 = 4 547 473 508 864 + 1;
  • 4 547 473 508 864 ÷ 2 = 2 273 736 754 432 + 0;
  • 2 273 736 754 432 ÷ 2 = 1 136 868 377 216 + 0;
  • 1 136 868 377 216 ÷ 2 = 568 434 188 608 + 0;
  • 568 434 188 608 ÷ 2 = 284 217 094 304 + 0;
  • 284 217 094 304 ÷ 2 = 142 108 547 152 + 0;
  • 142 108 547 152 ÷ 2 = 71 054 273 576 + 0;
  • 71 054 273 576 ÷ 2 = 35 527 136 788 + 0;
  • 35 527 136 788 ÷ 2 = 17 763 568 394 + 0;
  • 17 763 568 394 ÷ 2 = 8 881 784 197 + 0;
  • 8 881 784 197 ÷ 2 = 4 440 892 098 + 1;
  • 4 440 892 098 ÷ 2 = 2 220 446 049 + 0;
  • 2 220 446 049 ÷ 2 = 1 110 223 024 + 1;
  • 1 110 223 024 ÷ 2 = 555 111 512 + 0;
  • 555 111 512 ÷ 2 = 277 555 756 + 0;
  • 277 555 756 ÷ 2 = 138 777 878 + 0;
  • 138 777 878 ÷ 2 = 69 388 939 + 0;
  • 69 388 939 ÷ 2 = 34 694 469 + 1;
  • 34 694 469 ÷ 2 = 17 347 234 + 1;
  • 17 347 234 ÷ 2 = 8 673 617 + 0;
  • 8 673 617 ÷ 2 = 4 336 808 + 1;
  • 4 336 808 ÷ 2 = 2 168 404 + 0;
  • 2 168 404 ÷ 2 = 1 084 202 + 0;
  • 1 084 202 ÷ 2 = 542 101 + 0;
  • 542 101 ÷ 2 = 271 050 + 1;
  • 271 050 ÷ 2 = 135 525 + 0;
  • 135 525 ÷ 2 = 67 762 + 1;
  • 67 762 ÷ 2 = 33 881 + 0;
  • 33 881 ÷ 2 = 16 940 + 1;
  • 16 940 ÷ 2 = 8 470 + 0;
  • 8 470 ÷ 2 = 4 235 + 0;
  • 4 235 ÷ 2 = 2 117 + 1;
  • 2 117 ÷ 2 = 1 058 + 1;
  • 1 058 ÷ 2 = 529 + 0;
  • 529 ÷ 2 = 264 + 1;
  • 264 ÷ 2 = 132 + 0;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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

10 000 000 000 000 000 000 000 101(10) =


1000 0100 0101 1001 0101 0001 0110 0001 0100 0000 0001 0100 1000 0100 1010 0000 0000 0000 0000 0110 0101(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 83 positions to the left so that only one non zero digit remains to the left of it:

10 000 000 000 000 000 000 000 101(10) =


1000 0100 0101 1001 0101 0001 0110 0001 0100 0000 0001 0100 1000 0100 1010 0000 0000 0000 0000 0110 0101(2) =


1000 0100 0101 1001 0101 0001 0110 0001 0100 0000 0001 0100 1000 0100 1010 0000 0000 0000 0000 0110 0101(2) × 20 =


1.0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000 1001 0100 0000 0000 0000 0000 1100 101(2) × 283


4. 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): 83


Mantissa (not normalized):
1.0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000 1001 0100 0000 0000 0000 0000 1100 101


5. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


83 + 2(11-1) - 1 =


(83 + 1 023)(10) =


1 106(10)


6. 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;
  • 1 106 ÷ 2 = 553 + 0;
  • 553 ÷ 2 = 276 + 1;
  • 276 ÷ 2 = 138 + 0;
  • 138 ÷ 2 = 69 + 0;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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

Exponent (adjusted) =


1106(10) =


100 0101 0010(2)


8. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000 100 1010 0000 0000 0000 0000 0110 0101 =


0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000


9. 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) =
100 0101 0010


Mantissa (52 bits) =
0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000


Conclusion:

Number 10 000 000 000 000 000 000 000 101 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 100 0101 0010 - 0000 1000 1011 0010 1010 0010 1100 0010 1000 0000 0010 1001 0000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 1

      58
    • 0

      57
    • 1

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

10 000 000 000 000 000 000 000 100 = ? ... 10 000 000 000 000 000 000 000 102 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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