Convert 111 011 010 001.000 114 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

111 011 010 001.000 114(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: 111 011 010 001.
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;
  • 111 011 010 001 ÷ 2 = 55 505 505 000 + 1;
  • 55 505 505 000 ÷ 2 = 27 752 752 500 + 0;
  • 27 752 752 500 ÷ 2 = 13 876 376 250 + 0;
  • 13 876 376 250 ÷ 2 = 6 938 188 125 + 0;
  • 6 938 188 125 ÷ 2 = 3 469 094 062 + 1;
  • 3 469 094 062 ÷ 2 = 1 734 547 031 + 0;
  • 1 734 547 031 ÷ 2 = 867 273 515 + 1;
  • 867 273 515 ÷ 2 = 433 636 757 + 1;
  • 433 636 757 ÷ 2 = 216 818 378 + 1;
  • 216 818 378 ÷ 2 = 108 409 189 + 0;
  • 108 409 189 ÷ 2 = 54 204 594 + 1;
  • 54 204 594 ÷ 2 = 27 102 297 + 0;
  • 27 102 297 ÷ 2 = 13 551 148 + 1;
  • 13 551 148 ÷ 2 = 6 775 574 + 0;
  • 6 775 574 ÷ 2 = 3 387 787 + 0;
  • 3 387 787 ÷ 2 = 1 693 893 + 1;
  • 1 693 893 ÷ 2 = 846 946 + 1;
  • 846 946 ÷ 2 = 423 473 + 0;
  • 423 473 ÷ 2 = 211 736 + 1;
  • 211 736 ÷ 2 = 105 868 + 0;
  • 105 868 ÷ 2 = 52 934 + 0;
  • 52 934 ÷ 2 = 26 467 + 0;
  • 26 467 ÷ 2 = 13 233 + 1;
  • 13 233 ÷ 2 = 6 616 + 1;
  • 6 616 ÷ 2 = 3 308 + 0;
  • 3 308 ÷ 2 = 1 654 + 0;
  • 1 654 ÷ 2 = 827 + 0;
  • 827 ÷ 2 = 413 + 1;
  • 413 ÷ 2 = 206 + 1;
  • 206 ÷ 2 = 103 + 0;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

111 011 010 001(10) =


1 1001 1101 1000 1100 0101 1001 0101 1101 0001(2)


3. Convert to the binary (base 2) the fractional part: 0.000 114.

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) 0.000 114 × 2 = 0 + 0.000 228;
  • 2) 0.000 228 × 2 = 0 + 0.000 456;
  • 3) 0.000 456 × 2 = 0 + 0.000 912;
  • 4) 0.000 912 × 2 = 0 + 0.001 824;
  • 5) 0.001 824 × 2 = 0 + 0.003 648;
  • 6) 0.003 648 × 2 = 0 + 0.007 296;
  • 7) 0.007 296 × 2 = 0 + 0.014 592;
  • 8) 0.014 592 × 2 = 0 + 0.029 184;
  • 9) 0.029 184 × 2 = 0 + 0.058 368;
  • 10) 0.058 368 × 2 = 0 + 0.116 736;
  • 11) 0.116 736 × 2 = 0 + 0.233 472;
  • 12) 0.233 472 × 2 = 0 + 0.466 944;
  • 13) 0.466 944 × 2 = 0 + 0.933 888;
  • 14) 0.933 888 × 2 = 1 + 0.867 776;
  • 15) 0.867 776 × 2 = 1 + 0.735 552;
  • 16) 0.735 552 × 2 = 1 + 0.471 104;
  • 17) 0.471 104 × 2 = 0 + 0.942 208;
  • 18) 0.942 208 × 2 = 1 + 0.884 416;
  • 19) 0.884 416 × 2 = 1 + 0.768 832;
  • 20) 0.768 832 × 2 = 1 + 0.537 664;
  • 21) 0.537 664 × 2 = 1 + 0.075 328;
  • 22) 0.075 328 × 2 = 0 + 0.150 656;
  • 23) 0.150 656 × 2 = 0 + 0.301 312;
  • 24) 0.301 312 × 2 = 0 + 0.602 624;
  • 25) 0.602 624 × 2 = 1 + 0.205 248;
  • 26) 0.205 248 × 2 = 0 + 0.410 496;
  • 27) 0.410 496 × 2 = 0 + 0.820 992;
  • 28) 0.820 992 × 2 = 1 + 0.641 984;
  • 29) 0.641 984 × 2 = 1 + 0.283 968;
  • 30) 0.283 968 × 2 = 0 + 0.567 936;
  • 31) 0.567 936 × 2 = 1 + 0.135 872;
  • 32) 0.135 872 × 2 = 0 + 0.271 744;
  • 33) 0.271 744 × 2 = 0 + 0.543 488;
  • 34) 0.543 488 × 2 = 1 + 0.086 976;
  • 35) 0.086 976 × 2 = 0 + 0.173 952;
  • 36) 0.173 952 × 2 = 0 + 0.347 904;
  • 37) 0.347 904 × 2 = 0 + 0.695 808;
  • 38) 0.695 808 × 2 = 1 + 0.391 616;
  • 39) 0.391 616 × 2 = 0 + 0.783 232;
  • 40) 0.783 232 × 2 = 1 + 0.566 464;
  • 41) 0.566 464 × 2 = 1 + 0.132 928;
  • 42) 0.132 928 × 2 = 0 + 0.265 856;
  • 43) 0.265 856 × 2 = 0 + 0.531 712;
  • 44) 0.531 712 × 2 = 1 + 0.063 424;
  • 45) 0.063 424 × 2 = 0 + 0.126 848;
  • 46) 0.126 848 × 2 = 0 + 0.253 696;
  • 47) 0.253 696 × 2 = 0 + 0.507 392;
  • 48) 0.507 392 × 2 = 1 + 0.014 784;
  • 49) 0.014 784 × 2 = 0 + 0.029 568;
  • 50) 0.029 568 × 2 = 0 + 0.059 136;
  • 51) 0.059 136 × 2 = 0 + 0.118 272;
  • 52) 0.118 272 × 2 = 0 + 0.236 544;
  • 53) 0.236 544 × 2 = 0 + 0.473 088;

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:

0.000 114(10) =


0.0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0(2)


5. Positive number before normalization:

111 011 010 001.000 114(10) =


1 1001 1101 1000 1100 0101 1001 0101 1101 0001.0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 36 positions to the left so that only one non zero digit remains to the left of it:

111 011 010 001.000 114(10) =


1 1001 1101 1000 1100 0101 1001 0101 1101 0001.0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0(2) =


1 1001 1101 1000 1100 0101 1001 0101 1101 0001.0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0(2) × 20 =


1.1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0(2) × 236


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): 36


Mantissa (not normalized):
1.1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111 0111 1000 1001 1010 0100 0101 1001 0001 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


36 + 2(11-1) - 1 =


(36 + 1 023)(10) =


1 059(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;
  • 1 059 ÷ 2 = 529 + 1;
  • 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;

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) =


1059(10) =


100 0010 0011(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, 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. 1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111 0 1111 0001 0011 0100 1000 1011 0010 0010 0000 =


1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111


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) =
100 0010 0011


Mantissa (52 bits) =
1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111


Number 111 011 010 001.000 114 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0010 0011 - 1001 1101 1000 1100 0101 1001 0101 1101 0001 0000 0000 0000 0111

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 1

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 1

      23
    • 1

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 1

      0

More operations of this kind:

111 011 010 001.000 113 = ? ... 111 011 010 001.000 115 = ?


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

111 011 010 001.000 114 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
0.402 5 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
0.402 4 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
15.131 36 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
726 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
98 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:03 UTC (GMT)
0.734 33 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:02 UTC (GMT)
-73.016 3 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:02 UTC (GMT)
10 697 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:02 UTC (GMT)
700 000 000 000 000 000 001 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:02 UTC (GMT)
123.234 234 234 234 23 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:01 UTC (GMT)
-9 222 517 684 032 011 906 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:01 UTC (GMT)
5 334 to 64 bit double precision IEEE 754 binary floating point = ? Jun 13 23:00 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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