Convert 0.000 004 123 232 324 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 0.000 004 123 232 324(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 004 123 232 324.

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) 0.000 004 123 232 324 × 2 = 0 + 0.000 008 246 464 648;
  • 2) 0.000 008 246 464 648 × 2 = 0 + 0.000 016 492 929 296;
  • 3) 0.000 016 492 929 296 × 2 = 0 + 0.000 032 985 858 592;
  • 4) 0.000 032 985 858 592 × 2 = 0 + 0.000 065 971 717 184;
  • 5) 0.000 065 971 717 184 × 2 = 0 + 0.000 131 943 434 368;
  • 6) 0.000 131 943 434 368 × 2 = 0 + 0.000 263 886 868 736;
  • 7) 0.000 263 886 868 736 × 2 = 0 + 0.000 527 773 737 472;
  • 8) 0.000 527 773 737 472 × 2 = 0 + 0.001 055 547 474 944;
  • 9) 0.001 055 547 474 944 × 2 = 0 + 0.002 111 094 949 888;
  • 10) 0.002 111 094 949 888 × 2 = 0 + 0.004 222 189 899 776;
  • 11) 0.004 222 189 899 776 × 2 = 0 + 0.008 444 379 799 552;
  • 12) 0.008 444 379 799 552 × 2 = 0 + 0.016 888 759 599 104;
  • 13) 0.016 888 759 599 104 × 2 = 0 + 0.033 777 519 198 208;
  • 14) 0.033 777 519 198 208 × 2 = 0 + 0.067 555 038 396 416;
  • 15) 0.067 555 038 396 416 × 2 = 0 + 0.135 110 076 792 832;
  • 16) 0.135 110 076 792 832 × 2 = 0 + 0.270 220 153 585 664;
  • 17) 0.270 220 153 585 664 × 2 = 0 + 0.540 440 307 171 328;
  • 18) 0.540 440 307 171 328 × 2 = 1 + 0.080 880 614 342 656;
  • 19) 0.080 880 614 342 656 × 2 = 0 + 0.161 761 228 685 312;
  • 20) 0.161 761 228 685 312 × 2 = 0 + 0.323 522 457 370 624;
  • 21) 0.323 522 457 370 624 × 2 = 0 + 0.647 044 914 741 248;
  • 22) 0.647 044 914 741 248 × 2 = 1 + 0.294 089 829 482 496;
  • 23) 0.294 089 829 482 496 × 2 = 0 + 0.588 179 658 964 992;
  • 24) 0.588 179 658 964 992 × 2 = 1 + 0.176 359 317 929 984;
  • 25) 0.176 359 317 929 984 × 2 = 0 + 0.352 718 635 859 968;
  • 26) 0.352 718 635 859 968 × 2 = 0 + 0.705 437 271 719 936;
  • 27) 0.705 437 271 719 936 × 2 = 1 + 0.410 874 543 439 872;
  • 28) 0.410 874 543 439 872 × 2 = 0 + 0.821 749 086 879 744;
  • 29) 0.821 749 086 879 744 × 2 = 1 + 0.643 498 173 759 488;
  • 30) 0.643 498 173 759 488 × 2 = 1 + 0.286 996 347 518 976;
  • 31) 0.286 996 347 518 976 × 2 = 0 + 0.573 992 695 037 952;
  • 32) 0.573 992 695 037 952 × 2 = 1 + 0.147 985 390 075 904;
  • 33) 0.147 985 390 075 904 × 2 = 0 + 0.295 970 780 151 808;
  • 34) 0.295 970 780 151 808 × 2 = 0 + 0.591 941 560 303 616;
  • 35) 0.591 941 560 303 616 × 2 = 1 + 0.183 883 120 607 232;
  • 36) 0.183 883 120 607 232 × 2 = 0 + 0.367 766 241 214 464;
  • 37) 0.367 766 241 214 464 × 2 = 0 + 0.735 532 482 428 928;
  • 38) 0.735 532 482 428 928 × 2 = 1 + 0.471 064 964 857 856;
  • 39) 0.471 064 964 857 856 × 2 = 0 + 0.942 129 929 715 712;
  • 40) 0.942 129 929 715 712 × 2 = 1 + 0.884 259 859 431 424;
  • 41) 0.884 259 859 431 424 × 2 = 1 + 0.768 519 718 862 848;
  • 42) 0.768 519 718 862 848 × 2 = 1 + 0.537 039 437 725 696;
  • 43) 0.537 039 437 725 696 × 2 = 1 + 0.074 078 875 451 392;
  • 44) 0.074 078 875 451 392 × 2 = 0 + 0.148 157 750 902 784;
  • 45) 0.148 157 750 902 784 × 2 = 0 + 0.296 315 501 805 568;
  • 46) 0.296 315 501 805 568 × 2 = 0 + 0.592 631 003 611 136;
  • 47) 0.592 631 003 611 136 × 2 = 1 + 0.185 262 007 222 272;
  • 48) 0.185 262 007 222 272 × 2 = 0 + 0.370 524 014 444 544;
  • 49) 0.370 524 014 444 544 × 2 = 0 + 0.741 048 028 889 088;
  • 50) 0.741 048 028 889 088 × 2 = 1 + 0.482 096 057 778 176;
  • 51) 0.482 096 057 778 176 × 2 = 0 + 0.964 192 115 556 352;
  • 52) 0.964 192 115 556 352 × 2 = 1 + 0.928 384 231 112 704;
  • 53) 0.928 384 231 112 704 × 2 = 1 + 0.856 768 462 225 408;
  • 54) 0.856 768 462 225 408 × 2 = 1 + 0.713 536 924 450 816;
  • 55) 0.713 536 924 450 816 × 2 = 1 + 0.427 073 848 901 632;
  • 56) 0.427 073 848 901 632 × 2 = 0 + 0.854 147 697 803 264;
  • 57) 0.854 147 697 803 264 × 2 = 1 + 0.708 295 395 606 528;
  • 58) 0.708 295 395 606 528 × 2 = 1 + 0.416 590 791 213 056;
  • 59) 0.416 590 791 213 056 × 2 = 0 + 0.833 181 582 426 112;
  • 60) 0.833 181 582 426 112 × 2 = 1 + 0.666 363 164 852 224;
  • 61) 0.666 363 164 852 224 × 2 = 1 + 0.332 726 329 704 448;
  • 62) 0.332 726 329 704 448 × 2 = 0 + 0.665 452 659 408 896;
  • 63) 0.665 452 659 408 896 × 2 = 1 + 0.330 905 318 817 792;
  • 64) 0.330 905 318 817 792 × 2 = 0 + 0.661 810 637 635 584;
  • 65) 0.661 810 637 635 584 × 2 = 1 + 0.323 621 275 271 168;
  • 66) 0.323 621 275 271 168 × 2 = 0 + 0.647 242 550 542 336;
  • 67) 0.647 242 550 542 336 × 2 = 1 + 0.294 485 101 084 672;
  • 68) 0.294 485 101 084 672 × 2 = 0 + 0.588 970 202 169 344;
  • 69) 0.588 970 202 169 344 × 2 = 1 + 0.177 940 404 338 688;
  • 70) 0.177 940 404 338 688 × 2 = 0 + 0.355 880 808 677 376;

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...)


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:

0.000 004 123 232 324(10) =


0.0000 0000 0000 0000 0100 0101 0010 1101 0010 0101 1110 0010 0101 1110 1101 1010 1010 10(2)


5. Positive number before normalization:

0.000 004 123 232 324(10) =


0.0000 0000 0000 0000 0100 0101 0010 1101 0010 0101 1110 0010 0101 1110 1101 1010 1010 10(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right so that only one non zero digit remains to the left of it:

0.000 004 123 232 324(10) =


0.0000 0000 0000 0000 0100 0101 0010 1101 0010 0101 1110 0010 0101 1110 1101 1010 1010 10(2) =


0.0000 0000 0000 0000 0100 0101 0010 1101 0010 0101 1110 0010 0101 1110 1101 1010 1010 10(2) × 20 =


1.0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010(2) × 2-18


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): -18


Mantissa (not normalized):
1.0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-18 + 2(11-1) - 1 =


(-18 + 1 023)(10) =


1 005(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;
  • 1 005 ÷ 2 = 502 + 1;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =


1005(10) =


011 1110 1101(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. 0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010 =


0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010


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 1110 1101


Mantissa (52 bits) =
0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010


Conclusion:

Number 0.000 004 123 232 324 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 011 1110 1101 - 0001 0100 1011 0100 1001 0111 1000 1001 0111 1011 0110 1010 1010

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 0

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

0.000 004 123 232 323 = ? ... 0.000 004 123 232 325 = ?


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