-0.105 000 024 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 024(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
-0.105 000 024(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:

|-0.105 000 024| = 0.105 000 024


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: 0.105 000 024.

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) 0.105 000 024 × 2 = 0 + 0.210 000 048;
  • 2) 0.210 000 048 × 2 = 0 + 0.420 000 096;
  • 3) 0.420 000 096 × 2 = 0 + 0.840 000 192;
  • 4) 0.840 000 192 × 2 = 1 + 0.680 000 384;
  • 5) 0.680 000 384 × 2 = 1 + 0.360 000 768;
  • 6) 0.360 000 768 × 2 = 0 + 0.720 001 536;
  • 7) 0.720 001 536 × 2 = 1 + 0.440 003 072;
  • 8) 0.440 003 072 × 2 = 0 + 0.880 006 144;
  • 9) 0.880 006 144 × 2 = 1 + 0.760 012 288;
  • 10) 0.760 012 288 × 2 = 1 + 0.520 024 576;
  • 11) 0.520 024 576 × 2 = 1 + 0.040 049 152;
  • 12) 0.040 049 152 × 2 = 0 + 0.080 098 304;
  • 13) 0.080 098 304 × 2 = 0 + 0.160 196 608;
  • 14) 0.160 196 608 × 2 = 0 + 0.320 393 216;
  • 15) 0.320 393 216 × 2 = 0 + 0.640 786 432;
  • 16) 0.640 786 432 × 2 = 1 + 0.281 572 864;
  • 17) 0.281 572 864 × 2 = 0 + 0.563 145 728;
  • 18) 0.563 145 728 × 2 = 1 + 0.126 291 456;
  • 19) 0.126 291 456 × 2 = 0 + 0.252 582 912;
  • 20) 0.252 582 912 × 2 = 0 + 0.505 165 824;
  • 21) 0.505 165 824 × 2 = 1 + 0.010 331 648;
  • 22) 0.010 331 648 × 2 = 0 + 0.020 663 296;
  • 23) 0.020 663 296 × 2 = 0 + 0.041 326 592;
  • 24) 0.041 326 592 × 2 = 0 + 0.082 653 184;
  • 25) 0.082 653 184 × 2 = 0 + 0.165 306 368;
  • 26) 0.165 306 368 × 2 = 0 + 0.330 612 736;
  • 27) 0.330 612 736 × 2 = 0 + 0.661 225 472;
  • 28) 0.661 225 472 × 2 = 1 + 0.322 450 944;
  • 29) 0.322 450 944 × 2 = 0 + 0.644 901 888;
  • 30) 0.644 901 888 × 2 = 1 + 0.289 803 776;
  • 31) 0.289 803 776 × 2 = 0 + 0.579 607 552;
  • 32) 0.579 607 552 × 2 = 1 + 0.159 215 104;
  • 33) 0.159 215 104 × 2 = 0 + 0.318 430 208;
  • 34) 0.318 430 208 × 2 = 0 + 0.636 860 416;
  • 35) 0.636 860 416 × 2 = 1 + 0.273 720 832;
  • 36) 0.273 720 832 × 2 = 0 + 0.547 441 664;
  • 37) 0.547 441 664 × 2 = 1 + 0.094 883 328;
  • 38) 0.094 883 328 × 2 = 0 + 0.189 766 656;
  • 39) 0.189 766 656 × 2 = 0 + 0.379 533 312;
  • 40) 0.379 533 312 × 2 = 0 + 0.759 066 624;
  • 41) 0.759 066 624 × 2 = 1 + 0.518 133 248;
  • 42) 0.518 133 248 × 2 = 1 + 0.036 266 496;
  • 43) 0.036 266 496 × 2 = 0 + 0.072 532 992;
  • 44) 0.072 532 992 × 2 = 0 + 0.145 065 984;
  • 45) 0.145 065 984 × 2 = 0 + 0.290 131 968;
  • 46) 0.290 131 968 × 2 = 0 + 0.580 263 936;
  • 47) 0.580 263 936 × 2 = 1 + 0.160 527 872;
  • 48) 0.160 527 872 × 2 = 0 + 0.321 055 744;
  • 49) 0.321 055 744 × 2 = 0 + 0.642 111 488;
  • 50) 0.642 111 488 × 2 = 1 + 0.284 222 976;
  • 51) 0.284 222 976 × 2 = 0 + 0.568 445 952;
  • 52) 0.568 445 952 × 2 = 1 + 0.136 891 904;
  • 53) 0.136 891 904 × 2 = 0 + 0.273 783 808;
  • 54) 0.273 783 808 × 2 = 0 + 0.547 567 616;
  • 55) 0.547 567 616 × 2 = 1 + 0.095 135 232;
  • 56) 0.095 135 232 × 2 = 0 + 0.190 270 464;

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:


0.105 000 024(10) =

0.0001 1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010(2)

6. Positive number before normalization:

0.105 000 024(10) =

0.0001 1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010(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:


0.105 000 024(10) =

0.0001 1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010(2) =

0.0001 1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010(2) × 20 =

1.1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010(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 1000 0001 0101 0010 1000 1100 0010 0101 0010


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 1000 0001 0101 0010 1000 1100 0010 0101 0010 =

1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010


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 1000 0001 0101 0010 1000 1100 0010 0101 0010


Decimal number -0.105 000 024 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 1000 0001 0101 0010 1000 1100 0010 0101 0010

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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