-0.000 000 000 804 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 804(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 804(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 804| = 0.000 000 000 804


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: 0.000 000 000 804.

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) 0.000 000 000 804 × 2 = 0 + 0.000 000 001 608;
  • 2) 0.000 000 001 608 × 2 = 0 + 0.000 000 003 216;
  • 3) 0.000 000 003 216 × 2 = 0 + 0.000 000 006 432;
  • 4) 0.000 000 006 432 × 2 = 0 + 0.000 000 012 864;
  • 5) 0.000 000 012 864 × 2 = 0 + 0.000 000 025 728;
  • 6) 0.000 000 025 728 × 2 = 0 + 0.000 000 051 456;
  • 7) 0.000 000 051 456 × 2 = 0 + 0.000 000 102 912;
  • 8) 0.000 000 102 912 × 2 = 0 + 0.000 000 205 824;
  • 9) 0.000 000 205 824 × 2 = 0 + 0.000 000 411 648;
  • 10) 0.000 000 411 648 × 2 = 0 + 0.000 000 823 296;
  • 11) 0.000 000 823 296 × 2 = 0 + 0.000 001 646 592;
  • 12) 0.000 001 646 592 × 2 = 0 + 0.000 003 293 184;
  • 13) 0.000 003 293 184 × 2 = 0 + 0.000 006 586 368;
  • 14) 0.000 006 586 368 × 2 = 0 + 0.000 013 172 736;
  • 15) 0.000 013 172 736 × 2 = 0 + 0.000 026 345 472;
  • 16) 0.000 026 345 472 × 2 = 0 + 0.000 052 690 944;
  • 17) 0.000 052 690 944 × 2 = 0 + 0.000 105 381 888;
  • 18) 0.000 105 381 888 × 2 = 0 + 0.000 210 763 776;
  • 19) 0.000 210 763 776 × 2 = 0 + 0.000 421 527 552;
  • 20) 0.000 421 527 552 × 2 = 0 + 0.000 843 055 104;
  • 21) 0.000 843 055 104 × 2 = 0 + 0.001 686 110 208;
  • 22) 0.001 686 110 208 × 2 = 0 + 0.003 372 220 416;
  • 23) 0.003 372 220 416 × 2 = 0 + 0.006 744 440 832;
  • 24) 0.006 744 440 832 × 2 = 0 + 0.013 488 881 664;
  • 25) 0.013 488 881 664 × 2 = 0 + 0.026 977 763 328;
  • 26) 0.026 977 763 328 × 2 = 0 + 0.053 955 526 656;
  • 27) 0.053 955 526 656 × 2 = 0 + 0.107 911 053 312;
  • 28) 0.107 911 053 312 × 2 = 0 + 0.215 822 106 624;
  • 29) 0.215 822 106 624 × 2 = 0 + 0.431 644 213 248;
  • 30) 0.431 644 213 248 × 2 = 0 + 0.863 288 426 496;
  • 31) 0.863 288 426 496 × 2 = 1 + 0.726 576 852 992;
  • 32) 0.726 576 852 992 × 2 = 1 + 0.453 153 705 984;
  • 33) 0.453 153 705 984 × 2 = 0 + 0.906 307 411 968;
  • 34) 0.906 307 411 968 × 2 = 1 + 0.812 614 823 936;
  • 35) 0.812 614 823 936 × 2 = 1 + 0.625 229 647 872;
  • 36) 0.625 229 647 872 × 2 = 1 + 0.250 459 295 744;
  • 37) 0.250 459 295 744 × 2 = 0 + 0.500 918 591 488;
  • 38) 0.500 918 591 488 × 2 = 1 + 0.001 837 182 976;
  • 39) 0.001 837 182 976 × 2 = 0 + 0.003 674 365 952;
  • 40) 0.003 674 365 952 × 2 = 0 + 0.007 348 731 904;
  • 41) 0.007 348 731 904 × 2 = 0 + 0.014 697 463 808;
  • 42) 0.014 697 463 808 × 2 = 0 + 0.029 394 927 616;
  • 43) 0.029 394 927 616 × 2 = 0 + 0.058 789 855 232;
  • 44) 0.058 789 855 232 × 2 = 0 + 0.117 579 710 464;
  • 45) 0.117 579 710 464 × 2 = 0 + 0.235 159 420 928;
  • 46) 0.235 159 420 928 × 2 = 0 + 0.470 318 841 856;
  • 47) 0.470 318 841 856 × 2 = 0 + 0.940 637 683 712;
  • 48) 0.940 637 683 712 × 2 = 1 + 0.881 275 367 424;
  • 49) 0.881 275 367 424 × 2 = 1 + 0.762 550 734 848;
  • 50) 0.762 550 734 848 × 2 = 1 + 0.525 101 469 696;
  • 51) 0.525 101 469 696 × 2 = 1 + 0.050 202 939 392;
  • 52) 0.050 202 939 392 × 2 = 0 + 0.100 405 878 784;
  • 53) 0.100 405 878 784 × 2 = 0 + 0.200 811 757 568;
  • 54) 0.200 811 757 568 × 2 = 0 + 0.401 623 515 136;

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:


0.000 000 000 804(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0100 0000 0001 1110 00(2)

6. Positive number before normalization:

0.000 000 000 804(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0100 0000 0001 1110 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 804(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0100 0000 0001 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0100 0000 0001 1110 00(2) × 20 =

1.1011 1010 0000 0000 1111 000(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1011 1010 0000 0000 1111 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 101 1101 0000 0000 0111 1000 =

101 1101 0000 0000 0111 1000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
101 1101 0000 0000 0111 1000


Decimal number -0.000 000 000 804 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1101 0000 0000 0111 1000

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111