-0.014 657 786 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.014 657 786 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
-0.014 657 786 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:

|-0.014 657 786 5| = 0.014 657 786 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: 0.014 657 786 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) 0.014 657 786 5 × 2 = 0 + 0.029 315 573;
  • 2) 0.029 315 573 × 2 = 0 + 0.058 631 146;
  • 3) 0.058 631 146 × 2 = 0 + 0.117 262 292;
  • 4) 0.117 262 292 × 2 = 0 + 0.234 524 584;
  • 5) 0.234 524 584 × 2 = 0 + 0.469 049 168;
  • 6) 0.469 049 168 × 2 = 0 + 0.938 098 336;
  • 7) 0.938 098 336 × 2 = 1 + 0.876 196 672;
  • 8) 0.876 196 672 × 2 = 1 + 0.752 393 344;
  • 9) 0.752 393 344 × 2 = 1 + 0.504 786 688;
  • 10) 0.504 786 688 × 2 = 1 + 0.009 573 376;
  • 11) 0.009 573 376 × 2 = 0 + 0.019 146 752;
  • 12) 0.019 146 752 × 2 = 0 + 0.038 293 504;
  • 13) 0.038 293 504 × 2 = 0 + 0.076 587 008;
  • 14) 0.076 587 008 × 2 = 0 + 0.153 174 016;
  • 15) 0.153 174 016 × 2 = 0 + 0.306 348 032;
  • 16) 0.306 348 032 × 2 = 0 + 0.612 696 064;
  • 17) 0.612 696 064 × 2 = 1 + 0.225 392 128;
  • 18) 0.225 392 128 × 2 = 0 + 0.450 784 256;
  • 19) 0.450 784 256 × 2 = 0 + 0.901 568 512;
  • 20) 0.901 568 512 × 2 = 1 + 0.803 137 024;
  • 21) 0.803 137 024 × 2 = 1 + 0.606 274 048;
  • 22) 0.606 274 048 × 2 = 1 + 0.212 548 096;
  • 23) 0.212 548 096 × 2 = 0 + 0.425 096 192;
  • 24) 0.425 096 192 × 2 = 0 + 0.850 192 384;
  • 25) 0.850 192 384 × 2 = 1 + 0.700 384 768;
  • 26) 0.700 384 768 × 2 = 1 + 0.400 769 536;
  • 27) 0.400 769 536 × 2 = 0 + 0.801 539 072;
  • 28) 0.801 539 072 × 2 = 1 + 0.603 078 144;
  • 29) 0.603 078 144 × 2 = 1 + 0.206 156 288;
  • 30) 0.206 156 288 × 2 = 0 + 0.412 312 576;
  • 31) 0.412 312 576 × 2 = 0 + 0.824 625 152;
  • 32) 0.824 625 152 × 2 = 1 + 0.649 250 304;
  • 33) 0.649 250 304 × 2 = 1 + 0.298 500 608;
  • 34) 0.298 500 608 × 2 = 0 + 0.597 001 216;
  • 35) 0.597 001 216 × 2 = 1 + 0.194 002 432;
  • 36) 0.194 002 432 × 2 = 0 + 0.388 004 864;
  • 37) 0.388 004 864 × 2 = 0 + 0.776 009 728;
  • 38) 0.776 009 728 × 2 = 1 + 0.552 019 456;
  • 39) 0.552 019 456 × 2 = 1 + 0.104 038 912;
  • 40) 0.104 038 912 × 2 = 0 + 0.208 077 824;
  • 41) 0.208 077 824 × 2 = 0 + 0.416 155 648;
  • 42) 0.416 155 648 × 2 = 0 + 0.832 311 296;
  • 43) 0.832 311 296 × 2 = 1 + 0.664 622 592;
  • 44) 0.664 622 592 × 2 = 1 + 0.329 245 184;
  • 45) 0.329 245 184 × 2 = 0 + 0.658 490 368;
  • 46) 0.658 490 368 × 2 = 1 + 0.316 980 736;
  • 47) 0.316 980 736 × 2 = 0 + 0.633 961 472;
  • 48) 0.633 961 472 × 2 = 1 + 0.267 922 944;
  • 49) 0.267 922 944 × 2 = 0 + 0.535 845 888;
  • 50) 0.535 845 888 × 2 = 1 + 0.071 691 776;
  • 51) 0.071 691 776 × 2 = 0 + 0.143 383 552;
  • 52) 0.143 383 552 × 2 = 0 + 0.286 767 104;
  • 53) 0.286 767 104 × 2 = 0 + 0.573 534 208;
  • 54) 0.573 534 208 × 2 = 1 + 0.147 068 416;
  • 55) 0.147 068 416 × 2 = 0 + 0.294 136 832;
  • 56) 0.294 136 832 × 2 = 0 + 0.588 273 664;
  • 57) 0.588 273 664 × 2 = 1 + 0.176 547 328;
  • 58) 0.176 547 328 × 2 = 0 + 0.353 094 656;
  • 59) 0.353 094 656 × 2 = 0 + 0.706 189 312;

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:


0.014 657 786 5(10) =

0.0000 0011 1100 0000 1001 1100 1101 1001 1010 0110 0011 0101 0100 0100 100(2)

6. Positive number before normalization:

0.014 657 786 5(10) =

0.0000 0011 1100 0000 1001 1100 1101 1001 1010 0110 0011 0101 0100 0100 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.014 657 786 5(10) =

0.0000 0011 1100 0000 1001 1100 1101 1001 1010 0110 0011 0101 0100 0100 100(2) =

0.0000 0011 1100 0000 1001 1100 1101 1001 1010 0110 0011 0101 0100 0100 100(2) × 20 =

1.1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 0100(2) × 2-7


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): -7

Mantissa (not normalized):
1.1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 0100 =

1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 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 1000

Mantissa (52 bits) =
1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 0100


Decimal number -0.014 657 786 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1110 0000 0100 1110 0110 1100 1101 0011 0001 1010 1010 0010 0100

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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