Convert -0.095 000 000 000 000 03 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

-0.095 000 000 000 000 03(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. Start with the positive version of the number:

|-0.095 000 000 000 000 03| = 0.095 000 000 000 000 03

2. 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;

3. 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)


4. Convert to the binary (base 2) the fractional part: 0.095 000 000 000 000 03.

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) 0.095 000 000 000 000 03 × 2 = 0 + 0.190 000 000 000 000 06;
  • 2) 0.190 000 000 000 000 06 × 2 = 0 + 0.380 000 000 000 000 12;
  • 3) 0.380 000 000 000 000 12 × 2 = 0 + 0.760 000 000 000 000 24;
  • 4) 0.760 000 000 000 000 24 × 2 = 1 + 0.520 000 000 000 000 48;
  • 5) 0.520 000 000 000 000 48 × 2 = 1 + 0.040 000 000 000 000 96;
  • 6) 0.040 000 000 000 000 96 × 2 = 0 + 0.080 000 000 000 001 92;
  • 7) 0.080 000 000 000 001 92 × 2 = 0 + 0.160 000 000 000 003 84;
  • 8) 0.160 000 000 000 003 84 × 2 = 0 + 0.320 000 000 000 007 68;
  • 9) 0.320 000 000 000 007 68 × 2 = 0 + 0.640 000 000 000 015 36;
  • 10) 0.640 000 000 000 015 36 × 2 = 1 + 0.280 000 000 000 030 72;
  • 11) 0.280 000 000 000 030 72 × 2 = 0 + 0.560 000 000 000 061 44;
  • 12) 0.560 000 000 000 061 44 × 2 = 1 + 0.120 000 000 000 122 88;
  • 13) 0.120 000 000 000 122 88 × 2 = 0 + 0.240 000 000 000 245 76;
  • 14) 0.240 000 000 000 245 76 × 2 = 0 + 0.480 000 000 000 491 52;
  • 15) 0.480 000 000 000 491 52 × 2 = 0 + 0.960 000 000 000 983 04;
  • 16) 0.960 000 000 000 983 04 × 2 = 1 + 0.920 000 000 001 966 08;
  • 17) 0.920 000 000 001 966 08 × 2 = 1 + 0.840 000 000 003 932 16;
  • 18) 0.840 000 000 003 932 16 × 2 = 1 + 0.680 000 000 007 864 32;
  • 19) 0.680 000 000 007 864 32 × 2 = 1 + 0.360 000 000 015 728 64;
  • 20) 0.360 000 000 015 728 64 × 2 = 0 + 0.720 000 000 031 457 28;
  • 21) 0.720 000 000 031 457 28 × 2 = 1 + 0.440 000 000 062 914 56;
  • 22) 0.440 000 000 062 914 56 × 2 = 0 + 0.880 000 000 125 829 12;
  • 23) 0.880 000 000 125 829 12 × 2 = 1 + 0.760 000 000 251 658 24;
  • 24) 0.760 000 000 251 658 24 × 2 = 1 + 0.520 000 000 503 316 48;
  • 25) 0.520 000 000 503 316 48 × 2 = 1 + 0.040 000 001 006 632 96;
  • 26) 0.040 000 001 006 632 96 × 2 = 0 + 0.080 000 002 013 265 92;
  • 27) 0.080 000 002 013 265 92 × 2 = 0 + 0.160 000 004 026 531 84;
  • 28) 0.160 000 004 026 531 84 × 2 = 0 + 0.320 000 008 053 063 68;
  • 29) 0.320 000 008 053 063 68 × 2 = 0 + 0.640 000 016 106 127 36;
  • 30) 0.640 000 016 106 127 36 × 2 = 1 + 0.280 000 032 212 254 72;
  • 31) 0.280 000 032 212 254 72 × 2 = 0 + 0.560 000 064 424 509 44;
  • 32) 0.560 000 064 424 509 44 × 2 = 1 + 0.120 000 128 849 018 88;
  • 33) 0.120 000 128 849 018 88 × 2 = 0 + 0.240 000 257 698 037 76;
  • 34) 0.240 000 257 698 037 76 × 2 = 0 + 0.480 000 515 396 075 52;
  • 35) 0.480 000 515 396 075 52 × 2 = 0 + 0.960 001 030 792 151 04;
  • 36) 0.960 001 030 792 151 04 × 2 = 1 + 0.920 002 061 584 302 08;
  • 37) 0.920 002 061 584 302 08 × 2 = 1 + 0.840 004 123 168 604 16;
  • 38) 0.840 004 123 168 604 16 × 2 = 1 + 0.680 008 246 337 208 32;
  • 39) 0.680 008 246 337 208 32 × 2 = 1 + 0.360 016 492 674 416 64;
  • 40) 0.360 016 492 674 416 64 × 2 = 0 + 0.720 032 985 348 833 28;
  • 41) 0.720 032 985 348 833 28 × 2 = 1 + 0.440 065 970 697 666 56;
  • 42) 0.440 065 970 697 666 56 × 2 = 0 + 0.880 131 941 395 333 12;
  • 43) 0.880 131 941 395 333 12 × 2 = 1 + 0.760 263 882 790 666 24;
  • 44) 0.760 263 882 790 666 24 × 2 = 1 + 0.520 527 765 581 332 48;
  • 45) 0.520 527 765 581 332 48 × 2 = 1 + 0.041 055 531 162 664 96;
  • 46) 0.041 055 531 162 664 96 × 2 = 0 + 0.082 111 062 325 329 92;
  • 47) 0.082 111 062 325 329 92 × 2 = 0 + 0.164 222 124 650 659 84;
  • 48) 0.164 222 124 650 659 84 × 2 = 0 + 0.328 444 249 301 319 68;
  • 49) 0.328 444 249 301 319 68 × 2 = 0 + 0.656 888 498 602 639 36;
  • 50) 0.656 888 498 602 639 36 × 2 = 1 + 0.313 776 997 205 278 72;
  • 51) 0.313 776 997 205 278 72 × 2 = 0 + 0.627 553 994 410 557 44;
  • 52) 0.627 553 994 410 557 44 × 2 = 1 + 0.255 107 988 821 114 88;
  • 53) 0.255 107 988 821 114 88 × 2 = 0 + 0.510 215 977 642 229 76;
  • 54) 0.510 215 977 642 229 76 × 2 = 1 + 0.020 431 955 284 459 52;
  • 55) 0.020 431 955 284 459 52 × 2 = 0 + 0.040 863 910 568 919 04;
  • 56) 0.040 863 910 568 919 04 × 2 = 0 + 0.081 727 821 137 838 08;

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...)


5. 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:

0.095 000 000 000 000 03(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100(2)


6. Positive number before normalization:

0.095 000 000 000 000 03(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right so that only one non zero digit remains to the left of it:

0.095 000 000 000 000 03(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100(2) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100(2) × 20 =


1.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100(2) × 2-4


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):
1.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-4 + 2(11-1) - 1 =


(-4 + 1 023)(10) =


1 019(10)


10. 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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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

Exponent (adjusted) =


1019(10) =


011 1111 1011(2)


12. 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. 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100 =


1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100


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

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
011 1111 1011


Mantissa (52 bits) =
1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100


Number -0.095 000 000 000 000 03 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
1 - 011 1111 1011 - 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 0

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

-0.095 000 000 000 000 04 = ? ... -0.095 000 000 000 000 02 = ?


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

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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