-0.365 481 236 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.365 481 236 16(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
-0.365 481 236 16(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:

|-0.365 481 236 16| = 0.365 481 236 16


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: 0.365 481 236 16.

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) 0.365 481 236 16 × 2 = 0 + 0.730 962 472 32;
  • 2) 0.730 962 472 32 × 2 = 1 + 0.461 924 944 64;
  • 3) 0.461 924 944 64 × 2 = 0 + 0.923 849 889 28;
  • 4) 0.923 849 889 28 × 2 = 1 + 0.847 699 778 56;
  • 5) 0.847 699 778 56 × 2 = 1 + 0.695 399 557 12;
  • 6) 0.695 399 557 12 × 2 = 1 + 0.390 799 114 24;
  • 7) 0.390 799 114 24 × 2 = 0 + 0.781 598 228 48;
  • 8) 0.781 598 228 48 × 2 = 1 + 0.563 196 456 96;
  • 9) 0.563 196 456 96 × 2 = 1 + 0.126 392 913 92;
  • 10) 0.126 392 913 92 × 2 = 0 + 0.252 785 827 84;
  • 11) 0.252 785 827 84 × 2 = 0 + 0.505 571 655 68;
  • 12) 0.505 571 655 68 × 2 = 1 + 0.011 143 311 36;
  • 13) 0.011 143 311 36 × 2 = 0 + 0.022 286 622 72;
  • 14) 0.022 286 622 72 × 2 = 0 + 0.044 573 245 44;
  • 15) 0.044 573 245 44 × 2 = 0 + 0.089 146 490 88;
  • 16) 0.089 146 490 88 × 2 = 0 + 0.178 292 981 76;
  • 17) 0.178 292 981 76 × 2 = 0 + 0.356 585 963 52;
  • 18) 0.356 585 963 52 × 2 = 0 + 0.713 171 927 04;
  • 19) 0.713 171 927 04 × 2 = 1 + 0.426 343 854 08;
  • 20) 0.426 343 854 08 × 2 = 0 + 0.852 687 708 16;
  • 21) 0.852 687 708 16 × 2 = 1 + 0.705 375 416 32;
  • 22) 0.705 375 416 32 × 2 = 1 + 0.410 750 832 64;
  • 23) 0.410 750 832 64 × 2 = 0 + 0.821 501 665 28;
  • 24) 0.821 501 665 28 × 2 = 1 + 0.643 003 330 56;
  • 25) 0.643 003 330 56 × 2 = 1 + 0.286 006 661 12;
  • 26) 0.286 006 661 12 × 2 = 0 + 0.572 013 322 24;
  • 27) 0.572 013 322 24 × 2 = 1 + 0.144 026 644 48;
  • 28) 0.144 026 644 48 × 2 = 0 + 0.288 053 288 96;
  • 29) 0.288 053 288 96 × 2 = 0 + 0.576 106 577 92;
  • 30) 0.576 106 577 92 × 2 = 1 + 0.152 213 155 84;
  • 31) 0.152 213 155 84 × 2 = 0 + 0.304 426 311 68;
  • 32) 0.304 426 311 68 × 2 = 0 + 0.608 852 623 36;
  • 33) 0.608 852 623 36 × 2 = 1 + 0.217 705 246 72;
  • 34) 0.217 705 246 72 × 2 = 0 + 0.435 410 493 44;
  • 35) 0.435 410 493 44 × 2 = 0 + 0.870 820 986 88;
  • 36) 0.870 820 986 88 × 2 = 1 + 0.741 641 973 76;
  • 37) 0.741 641 973 76 × 2 = 1 + 0.483 283 947 52;
  • 38) 0.483 283 947 52 × 2 = 0 + 0.966 567 895 04;
  • 39) 0.966 567 895 04 × 2 = 1 + 0.933 135 790 08;
  • 40) 0.933 135 790 08 × 2 = 1 + 0.866 271 580 16;
  • 41) 0.866 271 580 16 × 2 = 1 + 0.732 543 160 32;
  • 42) 0.732 543 160 32 × 2 = 1 + 0.465 086 320 64;
  • 43) 0.465 086 320 64 × 2 = 0 + 0.930 172 641 28;
  • 44) 0.930 172 641 28 × 2 = 1 + 0.860 345 282 56;
  • 45) 0.860 345 282 56 × 2 = 1 + 0.720 690 565 12;
  • 46) 0.720 690 565 12 × 2 = 1 + 0.441 381 130 24;
  • 47) 0.441 381 130 24 × 2 = 0 + 0.882 762 260 48;
  • 48) 0.882 762 260 48 × 2 = 1 + 0.765 524 520 96;
  • 49) 0.765 524 520 96 × 2 = 1 + 0.531 049 041 92;
  • 50) 0.531 049 041 92 × 2 = 1 + 0.062 098 083 84;
  • 51) 0.062 098 083 84 × 2 = 0 + 0.124 196 167 68;
  • 52) 0.124 196 167 68 × 2 = 0 + 0.248 392 335 36;
  • 53) 0.248 392 335 36 × 2 = 0 + 0.496 784 670 72;
  • 54) 0.496 784 670 72 × 2 = 0 + 0.993 569 341 44;

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:


0.365 481 236 16(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 1001 1011 1101 1101 1100 00(2)

6. Positive number before normalization:

0.365 481 236 16(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 1001 1011 1101 1101 1100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.365 481 236 16(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 1001 1011 1101 1101 1100 00(2) =

0.0101 1101 1001 0000 0010 1101 1010 0100 1001 1011 1101 1101 1100 00(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000(2) × 2-2


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): -2

Mantissa (not normalized):
1.0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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;

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) =

1021(10) =

011 1111 1101(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. 0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000 =

0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000


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 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000


Decimal number -0.365 481 236 16 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 0010 0110 1111 0111 0111 0000

» Convert decimal -0.365 481 236 19 to 64 bit double precision IEEE 754 binary floating point representation standard

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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