-0.000 000 000 46 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 46(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 46(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 46| = 0.000 000 000 46


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 46.

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 46 × 2 = 0 + 0.000 000 000 92;
  • 2) 0.000 000 000 92 × 2 = 0 + 0.000 000 001 84;
  • 3) 0.000 000 001 84 × 2 = 0 + 0.000 000 003 68;
  • 4) 0.000 000 003 68 × 2 = 0 + 0.000 000 007 36;
  • 5) 0.000 000 007 36 × 2 = 0 + 0.000 000 014 72;
  • 6) 0.000 000 014 72 × 2 = 0 + 0.000 000 029 44;
  • 7) 0.000 000 029 44 × 2 = 0 + 0.000 000 058 88;
  • 8) 0.000 000 058 88 × 2 = 0 + 0.000 000 117 76;
  • 9) 0.000 000 117 76 × 2 = 0 + 0.000 000 235 52;
  • 10) 0.000 000 235 52 × 2 = 0 + 0.000 000 471 04;
  • 11) 0.000 000 471 04 × 2 = 0 + 0.000 000 942 08;
  • 12) 0.000 000 942 08 × 2 = 0 + 0.000 001 884 16;
  • 13) 0.000 001 884 16 × 2 = 0 + 0.000 003 768 32;
  • 14) 0.000 003 768 32 × 2 = 0 + 0.000 007 536 64;
  • 15) 0.000 007 536 64 × 2 = 0 + 0.000 015 073 28;
  • 16) 0.000 015 073 28 × 2 = 0 + 0.000 030 146 56;
  • 17) 0.000 030 146 56 × 2 = 0 + 0.000 060 293 12;
  • 18) 0.000 060 293 12 × 2 = 0 + 0.000 120 586 24;
  • 19) 0.000 120 586 24 × 2 = 0 + 0.000 241 172 48;
  • 20) 0.000 241 172 48 × 2 = 0 + 0.000 482 344 96;
  • 21) 0.000 482 344 96 × 2 = 0 + 0.000 964 689 92;
  • 22) 0.000 964 689 92 × 2 = 0 + 0.001 929 379 84;
  • 23) 0.001 929 379 84 × 2 = 0 + 0.003 858 759 68;
  • 24) 0.003 858 759 68 × 2 = 0 + 0.007 717 519 36;
  • 25) 0.007 717 519 36 × 2 = 0 + 0.015 435 038 72;
  • 26) 0.015 435 038 72 × 2 = 0 + 0.030 870 077 44;
  • 27) 0.030 870 077 44 × 2 = 0 + 0.061 740 154 88;
  • 28) 0.061 740 154 88 × 2 = 0 + 0.123 480 309 76;
  • 29) 0.123 480 309 76 × 2 = 0 + 0.246 960 619 52;
  • 30) 0.246 960 619 52 × 2 = 0 + 0.493 921 239 04;
  • 31) 0.493 921 239 04 × 2 = 0 + 0.987 842 478 08;
  • 32) 0.987 842 478 08 × 2 = 1 + 0.975 684 956 16;
  • 33) 0.975 684 956 16 × 2 = 1 + 0.951 369 912 32;
  • 34) 0.951 369 912 32 × 2 = 1 + 0.902 739 824 64;
  • 35) 0.902 739 824 64 × 2 = 1 + 0.805 479 649 28;
  • 36) 0.805 479 649 28 × 2 = 1 + 0.610 959 298 56;
  • 37) 0.610 959 298 56 × 2 = 1 + 0.221 918 597 12;
  • 38) 0.221 918 597 12 × 2 = 0 + 0.443 837 194 24;
  • 39) 0.443 837 194 24 × 2 = 0 + 0.887 674 388 48;
  • 40) 0.887 674 388 48 × 2 = 1 + 0.775 348 776 96;
  • 41) 0.775 348 776 96 × 2 = 1 + 0.550 697 553 92;
  • 42) 0.550 697 553 92 × 2 = 1 + 0.101 395 107 84;
  • 43) 0.101 395 107 84 × 2 = 0 + 0.202 790 215 68;
  • 44) 0.202 790 215 68 × 2 = 0 + 0.405 580 431 36;
  • 45) 0.405 580 431 36 × 2 = 0 + 0.811 160 862 72;
  • 46) 0.811 160 862 72 × 2 = 1 + 0.622 321 725 44;
  • 47) 0.622 321 725 44 × 2 = 1 + 0.244 643 450 88;
  • 48) 0.244 643 450 88 × 2 = 0 + 0.489 286 901 76;
  • 49) 0.489 286 901 76 × 2 = 0 + 0.978 573 803 52;
  • 50) 0.978 573 803 52 × 2 = 1 + 0.957 147 607 04;
  • 51) 0.957 147 607 04 × 2 = 1 + 0.914 295 214 08;
  • 52) 0.914 295 214 08 × 2 = 1 + 0.828 590 428 16;
  • 53) 0.828 590 428 16 × 2 = 1 + 0.657 180 856 32;
  • 54) 0.657 180 856 32 × 2 = 1 + 0.314 361 712 64;
  • 55) 0.314 361 712 64 × 2 = 0 + 0.628 723 425 28;

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 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1111 1001 1100 0110 0111 110(2)

6. Positive number before normalization:

0.000 000 000 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1111 1001 1100 0110 0111 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1111 1001 1100 0110 0111 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1111 1001 1100 0110 0111 110(2) × 20 =

1.1111 1001 1100 0110 0111 110(2) × 2-32


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): -32

Mantissa (not normalized):
1.1111 1001 1100 0110 0111 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

95(10) =

0101 1111(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. 111 1100 1110 0011 0011 1110 =

111 1100 1110 0011 0011 1110


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) =
0101 1111

Mantissa (23 bits) =
111 1100 1110 0011 0011 1110


Decimal number -0.000 000 000 46 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 111 1100 1110 0011 0011 1110

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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