-0.009 826 502 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.009 826 502 4(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
-0.009 826 502 4(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:

|-0.009 826 502 4| = 0.009 826 502 4


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: 0.009 826 502 4.

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) 0.009 826 502 4 × 2 = 0 + 0.019 653 004 8;
  • 2) 0.019 653 004 8 × 2 = 0 + 0.039 306 009 6;
  • 3) 0.039 306 009 6 × 2 = 0 + 0.078 612 019 2;
  • 4) 0.078 612 019 2 × 2 = 0 + 0.157 224 038 4;
  • 5) 0.157 224 038 4 × 2 = 0 + 0.314 448 076 8;
  • 6) 0.314 448 076 8 × 2 = 0 + 0.628 896 153 6;
  • 7) 0.628 896 153 6 × 2 = 1 + 0.257 792 307 2;
  • 8) 0.257 792 307 2 × 2 = 0 + 0.515 584 614 4;
  • 9) 0.515 584 614 4 × 2 = 1 + 0.031 169 228 8;
  • 10) 0.031 169 228 8 × 2 = 0 + 0.062 338 457 6;
  • 11) 0.062 338 457 6 × 2 = 0 + 0.124 676 915 2;
  • 12) 0.124 676 915 2 × 2 = 0 + 0.249 353 830 4;
  • 13) 0.249 353 830 4 × 2 = 0 + 0.498 707 660 8;
  • 14) 0.498 707 660 8 × 2 = 0 + 0.997 415 321 6;
  • 15) 0.997 415 321 6 × 2 = 1 + 0.994 830 643 2;
  • 16) 0.994 830 643 2 × 2 = 1 + 0.989 661 286 4;
  • 17) 0.989 661 286 4 × 2 = 1 + 0.979 322 572 8;
  • 18) 0.979 322 572 8 × 2 = 1 + 0.958 645 145 6;
  • 19) 0.958 645 145 6 × 2 = 1 + 0.917 290 291 2;
  • 20) 0.917 290 291 2 × 2 = 1 + 0.834 580 582 4;
  • 21) 0.834 580 582 4 × 2 = 1 + 0.669 161 164 8;
  • 22) 0.669 161 164 8 × 2 = 1 + 0.338 322 329 6;
  • 23) 0.338 322 329 6 × 2 = 0 + 0.676 644 659 2;
  • 24) 0.676 644 659 2 × 2 = 1 + 0.353 289 318 4;
  • 25) 0.353 289 318 4 × 2 = 0 + 0.706 578 636 8;
  • 26) 0.706 578 636 8 × 2 = 1 + 0.413 157 273 6;
  • 27) 0.413 157 273 6 × 2 = 0 + 0.826 314 547 2;
  • 28) 0.826 314 547 2 × 2 = 1 + 0.652 629 094 4;
  • 29) 0.652 629 094 4 × 2 = 1 + 0.305 258 188 8;
  • 30) 0.305 258 188 8 × 2 = 0 + 0.610 516 377 6;
  • 31) 0.610 516 377 6 × 2 = 1 + 0.221 032 755 2;
  • 32) 0.221 032 755 2 × 2 = 0 + 0.442 065 510 4;
  • 33) 0.442 065 510 4 × 2 = 0 + 0.884 131 020 8;
  • 34) 0.884 131 020 8 × 2 = 1 + 0.768 262 041 6;
  • 35) 0.768 262 041 6 × 2 = 1 + 0.536 524 083 2;
  • 36) 0.536 524 083 2 × 2 = 1 + 0.073 048 166 4;
  • 37) 0.073 048 166 4 × 2 = 0 + 0.146 096 332 8;
  • 38) 0.146 096 332 8 × 2 = 0 + 0.292 192 665 6;
  • 39) 0.292 192 665 6 × 2 = 0 + 0.584 385 331 2;
  • 40) 0.584 385 331 2 × 2 = 1 + 0.168 770 662 4;
  • 41) 0.168 770 662 4 × 2 = 0 + 0.337 541 324 8;
  • 42) 0.337 541 324 8 × 2 = 0 + 0.675 082 649 6;
  • 43) 0.675 082 649 6 × 2 = 1 + 0.350 165 299 2;
  • 44) 0.350 165 299 2 × 2 = 0 + 0.700 330 598 4;
  • 45) 0.700 330 598 4 × 2 = 1 + 0.400 661 196 8;
  • 46) 0.400 661 196 8 × 2 = 0 + 0.801 322 393 6;
  • 47) 0.801 322 393 6 × 2 = 1 + 0.602 644 787 2;
  • 48) 0.602 644 787 2 × 2 = 1 + 0.205 289 574 4;
  • 49) 0.205 289 574 4 × 2 = 0 + 0.410 579 148 8;
  • 50) 0.410 579 148 8 × 2 = 0 + 0.821 158 297 6;
  • 51) 0.821 158 297 6 × 2 = 1 + 0.642 316 595 2;
  • 52) 0.642 316 595 2 × 2 = 1 + 0.284 633 190 4;
  • 53) 0.284 633 190 4 × 2 = 0 + 0.569 266 380 8;
  • 54) 0.569 266 380 8 × 2 = 1 + 0.138 532 761 6;
  • 55) 0.138 532 761 6 × 2 = 0 + 0.277 065 523 2;
  • 56) 0.277 065 523 2 × 2 = 0 + 0.554 131 046 4;
  • 57) 0.554 131 046 4 × 2 = 1 + 0.108 262 092 8;
  • 58) 0.108 262 092 8 × 2 = 0 + 0.216 524 185 6;
  • 59) 0.216 524 185 6 × 2 = 0 + 0.433 048 371 2;

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:


0.009 826 502 4(10) =

0.0000 0010 1000 0011 1111 1101 0101 1010 0111 0001 0010 1011 0011 0100 100(2)

6. Positive number before normalization:

0.009 826 502 4(10) =

0.0000 0010 1000 0011 1111 1101 0101 1010 0111 0001 0010 1011 0011 0100 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.009 826 502 4(10) =

0.0000 0010 1000 0011 1111 1101 0101 1010 0111 0001 0010 1011 0011 0100 100(2) =

0.0000 0010 1000 0011 1111 1101 0101 1010 0111 0001 0010 1011 0011 0100 100(2) × 20 =

1.0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100(2) × 2-7


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): -7

Mantissa (not normalized):
1.0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100 =

0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100


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 1000

Mantissa (52 bits) =
0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100


Decimal number -0.009 826 502 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0100 0001 1111 1110 1010 1101 0011 1000 1001 0101 1001 1010 0100

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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