1.400 000 024 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.400 000 024 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
1.400 000 024 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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.400 000 024 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) 0.400 000 024 5 × 2 = 0 + 0.800 000 049;
  • 2) 0.800 000 049 × 2 = 1 + 0.600 000 098;
  • 3) 0.600 000 098 × 2 = 1 + 0.200 000 196;
  • 4) 0.200 000 196 × 2 = 0 + 0.400 000 392;
  • 5) 0.400 000 392 × 2 = 0 + 0.800 000 784;
  • 6) 0.800 000 784 × 2 = 1 + 0.600 001 568;
  • 7) 0.600 001 568 × 2 = 1 + 0.200 003 136;
  • 8) 0.200 003 136 × 2 = 0 + 0.400 006 272;
  • 9) 0.400 006 272 × 2 = 0 + 0.800 012 544;
  • 10) 0.800 012 544 × 2 = 1 + 0.600 025 088;
  • 11) 0.600 025 088 × 2 = 1 + 0.200 050 176;
  • 12) 0.200 050 176 × 2 = 0 + 0.400 100 352;
  • 13) 0.400 100 352 × 2 = 0 + 0.800 200 704;
  • 14) 0.800 200 704 × 2 = 1 + 0.600 401 408;
  • 15) 0.600 401 408 × 2 = 1 + 0.200 802 816;
  • 16) 0.200 802 816 × 2 = 0 + 0.401 605 632;
  • 17) 0.401 605 632 × 2 = 0 + 0.803 211 264;
  • 18) 0.803 211 264 × 2 = 1 + 0.606 422 528;
  • 19) 0.606 422 528 × 2 = 1 + 0.212 845 056;
  • 20) 0.212 845 056 × 2 = 0 + 0.425 690 112;
  • 21) 0.425 690 112 × 2 = 0 + 0.851 380 224;
  • 22) 0.851 380 224 × 2 = 1 + 0.702 760 448;
  • 23) 0.702 760 448 × 2 = 1 + 0.405 520 896;
  • 24) 0.405 520 896 × 2 = 0 + 0.811 041 792;
  • 25) 0.811 041 792 × 2 = 1 + 0.622 083 584;
  • 26) 0.622 083 584 × 2 = 1 + 0.244 167 168;
  • 27) 0.244 167 168 × 2 = 0 + 0.488 334 336;
  • 28) 0.488 334 336 × 2 = 0 + 0.976 668 672;
  • 29) 0.976 668 672 × 2 = 1 + 0.953 337 344;
  • 30) 0.953 337 344 × 2 = 1 + 0.906 674 688;
  • 31) 0.906 674 688 × 2 = 1 + 0.813 349 376;
  • 32) 0.813 349 376 × 2 = 1 + 0.626 698 752;
  • 33) 0.626 698 752 × 2 = 1 + 0.253 397 504;
  • 34) 0.253 397 504 × 2 = 0 + 0.506 795 008;
  • 35) 0.506 795 008 × 2 = 1 + 0.013 590 016;
  • 36) 0.013 590 016 × 2 = 0 + 0.027 180 032;
  • 37) 0.027 180 032 × 2 = 0 + 0.054 360 064;
  • 38) 0.054 360 064 × 2 = 0 + 0.108 720 128;
  • 39) 0.108 720 128 × 2 = 0 + 0.217 440 256;
  • 40) 0.217 440 256 × 2 = 0 + 0.434 880 512;
  • 41) 0.434 880 512 × 2 = 0 + 0.869 761 024;
  • 42) 0.869 761 024 × 2 = 1 + 0.739 522 048;
  • 43) 0.739 522 048 × 2 = 1 + 0.479 044 096;
  • 44) 0.479 044 096 × 2 = 0 + 0.958 088 192;
  • 45) 0.958 088 192 × 2 = 1 + 0.916 176 384;
  • 46) 0.916 176 384 × 2 = 1 + 0.832 352 768;
  • 47) 0.832 352 768 × 2 = 1 + 0.664 705 536;
  • 48) 0.664 705 536 × 2 = 1 + 0.329 411 072;
  • 49) 0.329 411 072 × 2 = 0 + 0.658 822 144;
  • 50) 0.658 822 144 × 2 = 1 + 0.317 644 288;
  • 51) 0.317 644 288 × 2 = 0 + 0.635 288 576;
  • 52) 0.635 288 576 × 2 = 1 + 0.270 577 152;
  • 53) 0.270 577 152 × 2 = 0 + 0.541 154 304;

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:


0.400 000 024 5(10) =

0.0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0(2)

5. Positive number before normalization:

1.400 000 024 5(10) =

1.0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.400 000 024 5(10) =

1.0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0(2) =

1.0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0(2) × 20


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): 0

Mantissa (not normalized):
1.0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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. 0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101 0 =

0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101


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) =
011 1111 1111

Mantissa (52 bits) =
0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101


Decimal number 1.400 000 024 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0110 0110 0110 0110 0110 0110 1100 1111 1010 0000 0110 1111 0101

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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