8.535 179 606 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 8.535 179 606(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
8.535 179 606(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. First, convert to binary (in base 2) the integer part: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

8(10) =

1000(2)


3. Convert to binary (base 2) the fractional part: 0.535 179 606.

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) 0.535 179 606 × 2 = 1 + 0.070 359 212;
  • 2) 0.070 359 212 × 2 = 0 + 0.140 718 424;
  • 3) 0.140 718 424 × 2 = 0 + 0.281 436 848;
  • 4) 0.281 436 848 × 2 = 0 + 0.562 873 696;
  • 5) 0.562 873 696 × 2 = 1 + 0.125 747 392;
  • 6) 0.125 747 392 × 2 = 0 + 0.251 494 784;
  • 7) 0.251 494 784 × 2 = 0 + 0.502 989 568;
  • 8) 0.502 989 568 × 2 = 1 + 0.005 979 136;
  • 9) 0.005 979 136 × 2 = 0 + 0.011 958 272;
  • 10) 0.011 958 272 × 2 = 0 + 0.023 916 544;
  • 11) 0.023 916 544 × 2 = 0 + 0.047 833 088;
  • 12) 0.047 833 088 × 2 = 0 + 0.095 666 176;
  • 13) 0.095 666 176 × 2 = 0 + 0.191 332 352;
  • 14) 0.191 332 352 × 2 = 0 + 0.382 664 704;
  • 15) 0.382 664 704 × 2 = 0 + 0.765 329 408;
  • 16) 0.765 329 408 × 2 = 1 + 0.530 658 816;
  • 17) 0.530 658 816 × 2 = 1 + 0.061 317 632;
  • 18) 0.061 317 632 × 2 = 0 + 0.122 635 264;
  • 19) 0.122 635 264 × 2 = 0 + 0.245 270 528;
  • 20) 0.245 270 528 × 2 = 0 + 0.490 541 056;
  • 21) 0.490 541 056 × 2 = 0 + 0.981 082 112;
  • 22) 0.981 082 112 × 2 = 1 + 0.962 164 224;
  • 23) 0.962 164 224 × 2 = 1 + 0.924 328 448;
  • 24) 0.924 328 448 × 2 = 1 + 0.848 656 896;
  • 25) 0.848 656 896 × 2 = 1 + 0.697 313 792;
  • 26) 0.697 313 792 × 2 = 1 + 0.394 627 584;
  • 27) 0.394 627 584 × 2 = 0 + 0.789 255 168;
  • 28) 0.789 255 168 × 2 = 1 + 0.578 510 336;
  • 29) 0.578 510 336 × 2 = 1 + 0.157 020 672;
  • 30) 0.157 020 672 × 2 = 0 + 0.314 041 344;
  • 31) 0.314 041 344 × 2 = 0 + 0.628 082 688;
  • 32) 0.628 082 688 × 2 = 1 + 0.256 165 376;
  • 33) 0.256 165 376 × 2 = 0 + 0.512 330 752;
  • 34) 0.512 330 752 × 2 = 1 + 0.024 661 504;
  • 35) 0.024 661 504 × 2 = 0 + 0.049 323 008;
  • 36) 0.049 323 008 × 2 = 0 + 0.098 646 016;
  • 37) 0.098 646 016 × 2 = 0 + 0.197 292 032;
  • 38) 0.197 292 032 × 2 = 0 + 0.394 584 064;
  • 39) 0.394 584 064 × 2 = 0 + 0.789 168 128;
  • 40) 0.789 168 128 × 2 = 1 + 0.578 336 256;
  • 41) 0.578 336 256 × 2 = 1 + 0.156 672 512;
  • 42) 0.156 672 512 × 2 = 0 + 0.313 345 024;
  • 43) 0.313 345 024 × 2 = 0 + 0.626 690 048;
  • 44) 0.626 690 048 × 2 = 1 + 0.253 380 096;
  • 45) 0.253 380 096 × 2 = 0 + 0.506 760 192;
  • 46) 0.506 760 192 × 2 = 1 + 0.013 520 384;
  • 47) 0.013 520 384 × 2 = 0 + 0.027 040 768;
  • 48) 0.027 040 768 × 2 = 0 + 0.054 081 536;
  • 49) 0.054 081 536 × 2 = 0 + 0.108 163 072;
  • 50) 0.108 163 072 × 2 = 0 + 0.216 326 144;
  • 51) 0.216 326 144 × 2 = 0 + 0.432 652 288;
  • 52) 0.432 652 288 × 2 = 0 + 0.865 304 576;
  • 53) 0.865 304 576 × 2 = 1 + 0.730 609 152;

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).


4. 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:


0.535 179 606(10) =

0.1000 1001 0000 0001 1000 0111 1101 1001 0100 0001 1001 0100 0000 1(2)

5. Positive number before normalization:

8.535 179 606(10) =

1000.1000 1001 0000 0001 1000 0111 1101 1001 0100 0001 1001 0100 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.535 179 606(10) =

1000.1000 1001 0000 0001 1000 0111 1101 1001 0100 0001 1001 0100 0000 1(2) =

1000.1000 1001 0000 0001 1000 0111 1101 1001 0100 0001 1001 0100 0000 1(2) × 20 =

1.0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000 0001(2) × 23


7. 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 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(10)


9. 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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

1026(10) =

100 0000 0010(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000 0001 =

0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0010

Mantissa (52 bits) =
0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000


Decimal number 8.535 179 606 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0010 - 0001 0001 0010 0000 0011 0000 1111 1011 0010 1000 0011 0010 1000

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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