Convert 53.232 860 565 188 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

53.232 860 565 188(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: 53.
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;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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.

53(10) =


11 0101(2)


3. Convert to the binary (base 2) the fractional part: 0.232 860 565 188.

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) 0.232 860 565 188 × 2 = 0 + 0.465 721 130 376;
  • 2) 0.465 721 130 376 × 2 = 0 + 0.931 442 260 752;
  • 3) 0.931 442 260 752 × 2 = 1 + 0.862 884 521 504;
  • 4) 0.862 884 521 504 × 2 = 1 + 0.725 769 043 008;
  • 5) 0.725 769 043 008 × 2 = 1 + 0.451 538 086 016;
  • 6) 0.451 538 086 016 × 2 = 0 + 0.903 076 172 032;
  • 7) 0.903 076 172 032 × 2 = 1 + 0.806 152 344 064;
  • 8) 0.806 152 344 064 × 2 = 1 + 0.612 304 688 128;
  • 9) 0.612 304 688 128 × 2 = 1 + 0.224 609 376 256;
  • 10) 0.224 609 376 256 × 2 = 0 + 0.449 218 752 512;
  • 11) 0.449 218 752 512 × 2 = 0 + 0.898 437 505 024;
  • 12) 0.898 437 505 024 × 2 = 1 + 0.796 875 010 048;
  • 13) 0.796 875 010 048 × 2 = 1 + 0.593 750 020 096;
  • 14) 0.593 750 020 096 × 2 = 1 + 0.187 500 040 192;
  • 15) 0.187 500 040 192 × 2 = 0 + 0.375 000 080 384;
  • 16) 0.375 000 080 384 × 2 = 0 + 0.750 000 160 768;
  • 17) 0.750 000 160 768 × 2 = 1 + 0.500 000 321 536;
  • 18) 0.500 000 321 536 × 2 = 1 + 0.000 000 643 072;
  • 19) 0.000 000 643 072 × 2 = 0 + 0.000 001 286 144;
  • 20) 0.000 001 286 144 × 2 = 0 + 0.000 002 572 288;
  • 21) 0.000 002 572 288 × 2 = 0 + 0.000 005 144 576;
  • 22) 0.000 005 144 576 × 2 = 0 + 0.000 010 289 152;
  • 23) 0.000 010 289 152 × 2 = 0 + 0.000 020 578 304;
  • 24) 0.000 020 578 304 × 2 = 0 + 0.000 041 156 608;
  • 25) 0.000 041 156 608 × 2 = 0 + 0.000 082 313 216;
  • 26) 0.000 082 313 216 × 2 = 0 + 0.000 164 626 432;
  • 27) 0.000 164 626 432 × 2 = 0 + 0.000 329 252 864;
  • 28) 0.000 329 252 864 × 2 = 0 + 0.000 658 505 728;
  • 29) 0.000 658 505 728 × 2 = 0 + 0.001 317 011 456;
  • 30) 0.001 317 011 456 × 2 = 0 + 0.002 634 022 912;
  • 31) 0.002 634 022 912 × 2 = 0 + 0.005 268 045 824;
  • 32) 0.005 268 045 824 × 2 = 0 + 0.010 536 091 648;
  • 33) 0.010 536 091 648 × 2 = 0 + 0.021 072 183 296;
  • 34) 0.021 072 183 296 × 2 = 0 + 0.042 144 366 592;
  • 35) 0.042 144 366 592 × 2 = 0 + 0.084 288 733 184;
  • 36) 0.084 288 733 184 × 2 = 0 + 0.168 577 466 368;
  • 37) 0.168 577 466 368 × 2 = 0 + 0.337 154 932 736;
  • 38) 0.337 154 932 736 × 2 = 0 + 0.674 309 865 472;
  • 39) 0.674 309 865 472 × 2 = 1 + 0.348 619 730 944;
  • 40) 0.348 619 730 944 × 2 = 0 + 0.697 239 461 888;
  • 41) 0.697 239 461 888 × 2 = 1 + 0.394 478 923 776;
  • 42) 0.394 478 923 776 × 2 = 0 + 0.788 957 847 552;
  • 43) 0.788 957 847 552 × 2 = 1 + 0.577 915 695 104;
  • 44) 0.577 915 695 104 × 2 = 1 + 0.155 831 390 208;
  • 45) 0.155 831 390 208 × 2 = 0 + 0.311 662 780 416;
  • 46) 0.311 662 780 416 × 2 = 0 + 0.623 325 560 832;
  • 47) 0.623 325 560 832 × 2 = 1 + 0.246 651 121 664;
  • 48) 0.246 651 121 664 × 2 = 0 + 0.493 302 243 328;
  • 49) 0.493 302 243 328 × 2 = 0 + 0.986 604 486 656;
  • 50) 0.986 604 486 656 × 2 = 1 + 0.973 208 973 312;
  • 51) 0.973 208 973 312 × 2 = 1 + 0.946 417 946 624;
  • 52) 0.946 417 946 624 × 2 = 1 + 0.892 835 893 248;
  • 53) 0.892 835 893 248 × 2 = 1 + 0.785 671 786 496;

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:

0.232 860 565 188(10) =


0.0011 1011 1001 1100 1100 0000 0000 0000 0000 0010 1011 0010 0111 1(2)


5. Positive number before normalization:

53.232 860 565 188(10) =


11 0101.0011 1011 1001 1100 1100 0000 0000 0000 0000 0010 1011 0010 0111 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left so that only one non zero digit remains to the left of it:

53.232 860 565 188(10) =


11 0101.0011 1011 1001 1100 1100 0000 0000 0000 0000 0010 1011 0010 0111 1(2) =


11 0101.0011 1011 1001 1100 1100 0000 0000 0000 0000 0010 1011 0010 0111 1(2) × 20 =


1.1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 1001 0011 11(2) × 25


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): 5


Mantissa (not normalized):
1.1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 1001 0011 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


5 + 2(11-1) - 1 =


(5 + 1 023)(10) =


1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 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) =


1028(10) =


100 0000 0100(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. 1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 1001 00 1111 =


1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 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) =
100 0000 0100


Mantissa (52 bits) =
1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 1001


Number 53.232 860 565 188 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0100 - 1010 1001 1101 1100 1110 0110 0000 0000 0000 0000 0001 0101 1001

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 1

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 1

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 1

      34
    • 1

      33
    • 0

      32
    • 0

      31
    • 1

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

53.232 860 565 187 = ? ... 53.232 860 565 189 = ?


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

53.232 860 565 188 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 03:58 UTC (GMT)
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0.555 555 555 555 555 580 227 178 325 003 478 676 080 703 735 351 562 4 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 03:58 UTC (GMT)
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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