-0.105 000 000 001 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 000 001 5(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.105 000 000 001 5(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.105 000 000 001 5| = 0.105 000 000 001 5


2. First, convert to binary (in 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 binary (base 2) the fractional part: 0.105 000 000 001 5.

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) 0.105 000 000 001 5 × 2 = 0 + 0.210 000 000 003;
  • 2) 0.210 000 000 003 × 2 = 0 + 0.420 000 000 006;
  • 3) 0.420 000 000 006 × 2 = 0 + 0.840 000 000 012;
  • 4) 0.840 000 000 012 × 2 = 1 + 0.680 000 000 024;
  • 5) 0.680 000 000 024 × 2 = 1 + 0.360 000 000 048;
  • 6) 0.360 000 000 048 × 2 = 0 + 0.720 000 000 096;
  • 7) 0.720 000 000 096 × 2 = 1 + 0.440 000 000 192;
  • 8) 0.440 000 000 192 × 2 = 0 + 0.880 000 000 384;
  • 9) 0.880 000 000 384 × 2 = 1 + 0.760 000 000 768;
  • 10) 0.760 000 000 768 × 2 = 1 + 0.520 000 001 536;
  • 11) 0.520 000 001 536 × 2 = 1 + 0.040 000 003 072;
  • 12) 0.040 000 003 072 × 2 = 0 + 0.080 000 006 144;
  • 13) 0.080 000 006 144 × 2 = 0 + 0.160 000 012 288;
  • 14) 0.160 000 012 288 × 2 = 0 + 0.320 000 024 576;
  • 15) 0.320 000 024 576 × 2 = 0 + 0.640 000 049 152;
  • 16) 0.640 000 049 152 × 2 = 1 + 0.280 000 098 304;
  • 17) 0.280 000 098 304 × 2 = 0 + 0.560 000 196 608;
  • 18) 0.560 000 196 608 × 2 = 1 + 0.120 000 393 216;
  • 19) 0.120 000 393 216 × 2 = 0 + 0.240 000 786 432;
  • 20) 0.240 000 786 432 × 2 = 0 + 0.480 001 572 864;
  • 21) 0.480 001 572 864 × 2 = 0 + 0.960 003 145 728;
  • 22) 0.960 003 145 728 × 2 = 1 + 0.920 006 291 456;
  • 23) 0.920 006 291 456 × 2 = 1 + 0.840 012 582 912;
  • 24) 0.840 012 582 912 × 2 = 1 + 0.680 025 165 824;
  • 25) 0.680 025 165 824 × 2 = 1 + 0.360 050 331 648;
  • 26) 0.360 050 331 648 × 2 = 0 + 0.720 100 663 296;
  • 27) 0.720 100 663 296 × 2 = 1 + 0.440 201 326 592;
  • 28) 0.440 201 326 592 × 2 = 0 + 0.880 402 653 184;
  • 29) 0.880 402 653 184 × 2 = 1 + 0.760 805 306 368;
  • 30) 0.760 805 306 368 × 2 = 1 + 0.521 610 612 736;
  • 31) 0.521 610 612 736 × 2 = 1 + 0.043 221 225 472;
  • 32) 0.043 221 225 472 × 2 = 0 + 0.086 442 450 944;
  • 33) 0.086 442 450 944 × 2 = 0 + 0.172 884 901 888;
  • 34) 0.172 884 901 888 × 2 = 0 + 0.345 769 803 776;
  • 35) 0.345 769 803 776 × 2 = 0 + 0.691 539 607 552;
  • 36) 0.691 539 607 552 × 2 = 1 + 0.383 079 215 104;
  • 37) 0.383 079 215 104 × 2 = 0 + 0.766 158 430 208;
  • 38) 0.766 158 430 208 × 2 = 1 + 0.532 316 860 416;
  • 39) 0.532 316 860 416 × 2 = 1 + 0.064 633 720 832;
  • 40) 0.064 633 720 832 × 2 = 0 + 0.129 267 441 664;
  • 41) 0.129 267 441 664 × 2 = 0 + 0.258 534 883 328;
  • 42) 0.258 534 883 328 × 2 = 0 + 0.517 069 766 656;
  • 43) 0.517 069 766 656 × 2 = 1 + 0.034 139 533 312;
  • 44) 0.034 139 533 312 × 2 = 0 + 0.068 279 066 624;
  • 45) 0.068 279 066 624 × 2 = 0 + 0.136 558 133 248;
  • 46) 0.136 558 133 248 × 2 = 0 + 0.273 116 266 496;
  • 47) 0.273 116 266 496 × 2 = 0 + 0.546 232 532 992;
  • 48) 0.546 232 532 992 × 2 = 1 + 0.092 465 065 984;
  • 49) 0.092 465 065 984 × 2 = 0 + 0.184 930 131 968;
  • 50) 0.184 930 131 968 × 2 = 0 + 0.369 860 263 936;
  • 51) 0.369 860 263 936 × 2 = 0 + 0.739 720 527 872;
  • 52) 0.739 720 527 872 × 2 = 1 + 0.479 441 055 744;
  • 53) 0.479 441 055 744 × 2 = 0 + 0.958 882 111 488;
  • 54) 0.958 882 111 488 × 2 = 1 + 0.917 764 222 976;
  • 55) 0.917 764 222 976 × 2 = 1 + 0.835 528 445 952;
  • 56) 0.835 528 445 952 × 2 = 1 + 0.671 056 891 904;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.105 000 000 001 5(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111(2)

6. Positive number before normalization:

0.105 000 000 001 5(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111(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:


0.105 000 000 001 5(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111(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):
1.1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111


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. 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111 =

1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111


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) =
1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111


Decimal number -0.105 000 000 001 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 0001 0110 0010 0001 0001 0111

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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