-3.402 725 075 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.402 725 075 64(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
-3.402 725 075 64(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:

|-3.402 725 075 64| = 3.402 725 075 64


2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.402 725 075 64.

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) 0.402 725 075 64 × 2 = 0 + 0.805 450 151 28;
  • 2) 0.805 450 151 28 × 2 = 1 + 0.610 900 302 56;
  • 3) 0.610 900 302 56 × 2 = 1 + 0.221 800 605 12;
  • 4) 0.221 800 605 12 × 2 = 0 + 0.443 601 210 24;
  • 5) 0.443 601 210 24 × 2 = 0 + 0.887 202 420 48;
  • 6) 0.887 202 420 48 × 2 = 1 + 0.774 404 840 96;
  • 7) 0.774 404 840 96 × 2 = 1 + 0.548 809 681 92;
  • 8) 0.548 809 681 92 × 2 = 1 + 0.097 619 363 84;
  • 9) 0.097 619 363 84 × 2 = 0 + 0.195 238 727 68;
  • 10) 0.195 238 727 68 × 2 = 0 + 0.390 477 455 36;
  • 11) 0.390 477 455 36 × 2 = 0 + 0.780 954 910 72;
  • 12) 0.780 954 910 72 × 2 = 1 + 0.561 909 821 44;
  • 13) 0.561 909 821 44 × 2 = 1 + 0.123 819 642 88;
  • 14) 0.123 819 642 88 × 2 = 0 + 0.247 639 285 76;
  • 15) 0.247 639 285 76 × 2 = 0 + 0.495 278 571 52;
  • 16) 0.495 278 571 52 × 2 = 0 + 0.990 557 143 04;
  • 17) 0.990 557 143 04 × 2 = 1 + 0.981 114 286 08;
  • 18) 0.981 114 286 08 × 2 = 1 + 0.962 228 572 16;
  • 19) 0.962 228 572 16 × 2 = 1 + 0.924 457 144 32;
  • 20) 0.924 457 144 32 × 2 = 1 + 0.848 914 288 64;
  • 21) 0.848 914 288 64 × 2 = 1 + 0.697 828 577 28;
  • 22) 0.697 828 577 28 × 2 = 1 + 0.395 657 154 56;
  • 23) 0.395 657 154 56 × 2 = 0 + 0.791 314 309 12;
  • 24) 0.791 314 309 12 × 2 = 1 + 0.582 628 618 24;
  • 25) 0.582 628 618 24 × 2 = 1 + 0.165 257 236 48;
  • 26) 0.165 257 236 48 × 2 = 0 + 0.330 514 472 96;
  • 27) 0.330 514 472 96 × 2 = 0 + 0.661 028 945 92;
  • 28) 0.661 028 945 92 × 2 = 1 + 0.322 057 891 84;
  • 29) 0.322 057 891 84 × 2 = 0 + 0.644 115 783 68;
  • 30) 0.644 115 783 68 × 2 = 1 + 0.288 231 567 36;
  • 31) 0.288 231 567 36 × 2 = 0 + 0.576 463 134 72;
  • 32) 0.576 463 134 72 × 2 = 1 + 0.152 926 269 44;
  • 33) 0.152 926 269 44 × 2 = 0 + 0.305 852 538 88;
  • 34) 0.305 852 538 88 × 2 = 0 + 0.611 705 077 76;
  • 35) 0.611 705 077 76 × 2 = 1 + 0.223 410 155 52;
  • 36) 0.223 410 155 52 × 2 = 0 + 0.446 820 311 04;
  • 37) 0.446 820 311 04 × 2 = 0 + 0.893 640 622 08;
  • 38) 0.893 640 622 08 × 2 = 1 + 0.787 281 244 16;
  • 39) 0.787 281 244 16 × 2 = 1 + 0.574 562 488 32;
  • 40) 0.574 562 488 32 × 2 = 1 + 0.149 124 976 64;
  • 41) 0.149 124 976 64 × 2 = 0 + 0.298 249 953 28;
  • 42) 0.298 249 953 28 × 2 = 0 + 0.596 499 906 56;
  • 43) 0.596 499 906 56 × 2 = 1 + 0.192 999 813 12;
  • 44) 0.192 999 813 12 × 2 = 0 + 0.385 999 626 24;
  • 45) 0.385 999 626 24 × 2 = 0 + 0.771 999 252 48;
  • 46) 0.771 999 252 48 × 2 = 1 + 0.543 998 504 96;
  • 47) 0.543 998 504 96 × 2 = 1 + 0.087 997 009 92;
  • 48) 0.087 997 009 92 × 2 = 0 + 0.175 994 019 84;
  • 49) 0.175 994 019 84 × 2 = 0 + 0.351 988 039 68;
  • 50) 0.351 988 039 68 × 2 = 0 + 0.703 976 079 36;
  • 51) 0.703 976 079 36 × 2 = 1 + 0.407 952 158 72;
  • 52) 0.407 952 158 72 × 2 = 0 + 0.815 904 317 44;
  • 53) 0.815 904 317 44 × 2 = 1 + 0.631 808 634 88;

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:


0.402 725 075 64(10) =

0.0110 0111 0001 1000 1111 1101 1001 0101 0010 0111 0010 0110 0010 1(2)

6. Positive number before normalization:

3.402 725 075 64(10) =

11.0110 0111 0001 1000 1111 1101 1001 0101 0010 0111 0010 0110 0010 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.402 725 075 64(10) =

11.0110 0111 0001 1000 1111 1101 1001 0101 0010 0111 0010 0110 0010 1(2) =

11.0110 0111 0001 1000 1111 1101 1001 0101 0010 0111 0010 0110 0010 1(2) × 20 =

1.1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001 01(2) × 21


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): 1

Mantissa (not normalized):
1.1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001 01


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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;

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) =

1024(10) =

100 0000 0000(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, 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. 1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001 01 =

1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001


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) =
100 0000 0000

Mantissa (52 bits) =
1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001


Decimal number -3.402 725 075 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1011 0011 1000 1100 0111 1110 1100 1010 1001 0011 1001 0011 0001

» Convert decimal -3.402 725 075 12 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