1.750 176 989 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.750 176 989 8(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
1.750 176 989 8(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: 0.750 176 989 8.

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) 0.750 176 989 8 × 2 = 1 + 0.500 353 979 6;
  • 2) 0.500 353 979 6 × 2 = 1 + 0.000 707 959 2;
  • 3) 0.000 707 959 2 × 2 = 0 + 0.001 415 918 4;
  • 4) 0.001 415 918 4 × 2 = 0 + 0.002 831 836 8;
  • 5) 0.002 831 836 8 × 2 = 0 + 0.005 663 673 6;
  • 6) 0.005 663 673 6 × 2 = 0 + 0.011 327 347 2;
  • 7) 0.011 327 347 2 × 2 = 0 + 0.022 654 694 4;
  • 8) 0.022 654 694 4 × 2 = 0 + 0.045 309 388 8;
  • 9) 0.045 309 388 8 × 2 = 0 + 0.090 618 777 6;
  • 10) 0.090 618 777 6 × 2 = 0 + 0.181 237 555 2;
  • 11) 0.181 237 555 2 × 2 = 0 + 0.362 475 110 4;
  • 12) 0.362 475 110 4 × 2 = 0 + 0.724 950 220 8;
  • 13) 0.724 950 220 8 × 2 = 1 + 0.449 900 441 6;
  • 14) 0.449 900 441 6 × 2 = 0 + 0.899 800 883 2;
  • 15) 0.899 800 883 2 × 2 = 1 + 0.799 601 766 4;
  • 16) 0.799 601 766 4 × 2 = 1 + 0.599 203 532 8;
  • 17) 0.599 203 532 8 × 2 = 1 + 0.198 407 065 6;
  • 18) 0.198 407 065 6 × 2 = 0 + 0.396 814 131 2;
  • 19) 0.396 814 131 2 × 2 = 0 + 0.793 628 262 4;
  • 20) 0.793 628 262 4 × 2 = 1 + 0.587 256 524 8;
  • 21) 0.587 256 524 8 × 2 = 1 + 0.174 513 049 6;
  • 22) 0.174 513 049 6 × 2 = 0 + 0.349 026 099 2;
  • 23) 0.349 026 099 2 × 2 = 0 + 0.698 052 198 4;
  • 24) 0.698 052 198 4 × 2 = 1 + 0.396 104 396 8;
  • 25) 0.396 104 396 8 × 2 = 0 + 0.792 208 793 6;
  • 26) 0.792 208 793 6 × 2 = 1 + 0.584 417 587 2;
  • 27) 0.584 417 587 2 × 2 = 1 + 0.168 835 174 4;
  • 28) 0.168 835 174 4 × 2 = 0 + 0.337 670 348 8;
  • 29) 0.337 670 348 8 × 2 = 0 + 0.675 340 697 6;
  • 30) 0.675 340 697 6 × 2 = 1 + 0.350 681 395 2;
  • 31) 0.350 681 395 2 × 2 = 0 + 0.701 362 790 4;
  • 32) 0.701 362 790 4 × 2 = 1 + 0.402 725 580 8;
  • 33) 0.402 725 580 8 × 2 = 0 + 0.805 451 161 6;
  • 34) 0.805 451 161 6 × 2 = 1 + 0.610 902 323 2;
  • 35) 0.610 902 323 2 × 2 = 1 + 0.221 804 646 4;
  • 36) 0.221 804 646 4 × 2 = 0 + 0.443 609 292 8;
  • 37) 0.443 609 292 8 × 2 = 0 + 0.887 218 585 6;
  • 38) 0.887 218 585 6 × 2 = 1 + 0.774 437 171 2;
  • 39) 0.774 437 171 2 × 2 = 1 + 0.548 874 342 4;
  • 40) 0.548 874 342 4 × 2 = 1 + 0.097 748 684 8;
  • 41) 0.097 748 684 8 × 2 = 0 + 0.195 497 369 6;
  • 42) 0.195 497 369 6 × 2 = 0 + 0.390 994 739 2;
  • 43) 0.390 994 739 2 × 2 = 0 + 0.781 989 478 4;
  • 44) 0.781 989 478 4 × 2 = 1 + 0.563 978 956 8;
  • 45) 0.563 978 956 8 × 2 = 1 + 0.127 957 913 6;
  • 46) 0.127 957 913 6 × 2 = 0 + 0.255 915 827 2;
  • 47) 0.255 915 827 2 × 2 = 0 + 0.511 831 654 4;
  • 48) 0.511 831 654 4 × 2 = 1 + 0.023 663 308 8;
  • 49) 0.023 663 308 8 × 2 = 0 + 0.047 326 617 6;
  • 50) 0.047 326 617 6 × 2 = 0 + 0.094 653 235 2;
  • 51) 0.094 653 235 2 × 2 = 0 + 0.189 306 470 4;
  • 52) 0.189 306 470 4 × 2 = 0 + 0.378 612 940 8;
  • 53) 0.378 612 940 8 × 2 = 0 + 0.757 225 881 6;

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:


0.750 176 989 8(10) =

0.1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 0(2)

5. Positive number before normalization:

1.750 176 989 8(10) =

1.1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 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:


1.750 176 989 8(10) =

1.1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 0(2) =

1.1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 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):
1.1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 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. 1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000 0 =

1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000


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) =
1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000


Decimal number 1.750 176 989 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 0000 0000 1011 1001 1001 0110 0101 0110 0111 0001 1001 0000

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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