-0.105 000 000 049 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 000 049(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
-0.105 000 000 049(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:

|-0.105 000 000 049| = 0.105 000 000 049


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: 0.105 000 000 049.

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) 0.105 000 000 049 × 2 = 0 + 0.210 000 000 098;
  • 2) 0.210 000 000 098 × 2 = 0 + 0.420 000 000 196;
  • 3) 0.420 000 000 196 × 2 = 0 + 0.840 000 000 392;
  • 4) 0.840 000 000 392 × 2 = 1 + 0.680 000 000 784;
  • 5) 0.680 000 000 784 × 2 = 1 + 0.360 000 001 568;
  • 6) 0.360 000 001 568 × 2 = 0 + 0.720 000 003 136;
  • 7) 0.720 000 003 136 × 2 = 1 + 0.440 000 006 272;
  • 8) 0.440 000 006 272 × 2 = 0 + 0.880 000 012 544;
  • 9) 0.880 000 012 544 × 2 = 1 + 0.760 000 025 088;
  • 10) 0.760 000 025 088 × 2 = 1 + 0.520 000 050 176;
  • 11) 0.520 000 050 176 × 2 = 1 + 0.040 000 100 352;
  • 12) 0.040 000 100 352 × 2 = 0 + 0.080 000 200 704;
  • 13) 0.080 000 200 704 × 2 = 0 + 0.160 000 401 408;
  • 14) 0.160 000 401 408 × 2 = 0 + 0.320 000 802 816;
  • 15) 0.320 000 802 816 × 2 = 0 + 0.640 001 605 632;
  • 16) 0.640 001 605 632 × 2 = 1 + 0.280 003 211 264;
  • 17) 0.280 003 211 264 × 2 = 0 + 0.560 006 422 528;
  • 18) 0.560 006 422 528 × 2 = 1 + 0.120 012 845 056;
  • 19) 0.120 012 845 056 × 2 = 0 + 0.240 025 690 112;
  • 20) 0.240 025 690 112 × 2 = 0 + 0.480 051 380 224;
  • 21) 0.480 051 380 224 × 2 = 0 + 0.960 102 760 448;
  • 22) 0.960 102 760 448 × 2 = 1 + 0.920 205 520 896;
  • 23) 0.920 205 520 896 × 2 = 1 + 0.840 411 041 792;
  • 24) 0.840 411 041 792 × 2 = 1 + 0.680 822 083 584;
  • 25) 0.680 822 083 584 × 2 = 1 + 0.361 644 167 168;
  • 26) 0.361 644 167 168 × 2 = 0 + 0.723 288 334 336;
  • 27) 0.723 288 334 336 × 2 = 1 + 0.446 576 668 672;
  • 28) 0.446 576 668 672 × 2 = 0 + 0.893 153 337 344;
  • 29) 0.893 153 337 344 × 2 = 1 + 0.786 306 674 688;
  • 30) 0.786 306 674 688 × 2 = 1 + 0.572 613 349 376;
  • 31) 0.572 613 349 376 × 2 = 1 + 0.145 226 698 752;
  • 32) 0.145 226 698 752 × 2 = 0 + 0.290 453 397 504;
  • 33) 0.290 453 397 504 × 2 = 0 + 0.580 906 795 008;
  • 34) 0.580 906 795 008 × 2 = 1 + 0.161 813 590 016;
  • 35) 0.161 813 590 016 × 2 = 0 + 0.323 627 180 032;
  • 36) 0.323 627 180 032 × 2 = 0 + 0.647 254 360 064;
  • 37) 0.647 254 360 064 × 2 = 1 + 0.294 508 720 128;
  • 38) 0.294 508 720 128 × 2 = 0 + 0.589 017 440 256;
  • 39) 0.589 017 440 256 × 2 = 1 + 0.178 034 880 512;
  • 40) 0.178 034 880 512 × 2 = 0 + 0.356 069 761 024;
  • 41) 0.356 069 761 024 × 2 = 0 + 0.712 139 522 048;
  • 42) 0.712 139 522 048 × 2 = 1 + 0.424 279 044 096;
  • 43) 0.424 279 044 096 × 2 = 0 + 0.848 558 088 192;
  • 44) 0.848 558 088 192 × 2 = 1 + 0.697 116 176 384;
  • 45) 0.697 116 176 384 × 2 = 1 + 0.394 232 352 768;
  • 46) 0.394 232 352 768 × 2 = 0 + 0.788 464 705 536;
  • 47) 0.788 464 705 536 × 2 = 1 + 0.576 929 411 072;
  • 48) 0.576 929 411 072 × 2 = 1 + 0.153 858 822 144;
  • 49) 0.153 858 822 144 × 2 = 0 + 0.307 717 644 288;
  • 50) 0.307 717 644 288 × 2 = 0 + 0.615 435 288 576;
  • 51) 0.615 435 288 576 × 2 = 1 + 0.230 870 577 152;
  • 52) 0.230 870 577 152 × 2 = 0 + 0.461 741 154 304;
  • 53) 0.461 741 154 304 × 2 = 0 + 0.923 482 308 608;
  • 54) 0.923 482 308 608 × 2 = 1 + 0.846 964 617 216;
  • 55) 0.846 964 617 216 × 2 = 1 + 0.693 929 234 432;
  • 56) 0.693 929 234 432 × 2 = 1 + 0.387 858 468 864;

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:


0.105 000 000 049(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 0111(2)

6. Positive number before normalization:

0.105 000 000 049(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 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:


0.105 000 000 049(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 0111(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 0111(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 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 0100 1010 0101 1011 0010 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 0100 1010 0101 1011 0010 0111 =

1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 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 0100 1010 0101 1011 0010 0111


Decimal number -0.105 000 000 049 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 0100 1010 0101 1011 0010 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