-0.000 000 000 75 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 75(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 75(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 75| = 0.000 000 000 75


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 75.

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 75 × 2 = 0 + 0.000 000 001 5;
  • 2) 0.000 000 001 5 × 2 = 0 + 0.000 000 003;
  • 3) 0.000 000 003 × 2 = 0 + 0.000 000 006;
  • 4) 0.000 000 006 × 2 = 0 + 0.000 000 012;
  • 5) 0.000 000 012 × 2 = 0 + 0.000 000 024;
  • 6) 0.000 000 024 × 2 = 0 + 0.000 000 048;
  • 7) 0.000 000 048 × 2 = 0 + 0.000 000 096;
  • 8) 0.000 000 096 × 2 = 0 + 0.000 000 192;
  • 9) 0.000 000 192 × 2 = 0 + 0.000 000 384;
  • 10) 0.000 000 384 × 2 = 0 + 0.000 000 768;
  • 11) 0.000 000 768 × 2 = 0 + 0.000 001 536;
  • 12) 0.000 001 536 × 2 = 0 + 0.000 003 072;
  • 13) 0.000 003 072 × 2 = 0 + 0.000 006 144;
  • 14) 0.000 006 144 × 2 = 0 + 0.000 012 288;
  • 15) 0.000 012 288 × 2 = 0 + 0.000 024 576;
  • 16) 0.000 024 576 × 2 = 0 + 0.000 049 152;
  • 17) 0.000 049 152 × 2 = 0 + 0.000 098 304;
  • 18) 0.000 098 304 × 2 = 0 + 0.000 196 608;
  • 19) 0.000 196 608 × 2 = 0 + 0.000 393 216;
  • 20) 0.000 393 216 × 2 = 0 + 0.000 786 432;
  • 21) 0.000 786 432 × 2 = 0 + 0.001 572 864;
  • 22) 0.001 572 864 × 2 = 0 + 0.003 145 728;
  • 23) 0.003 145 728 × 2 = 0 + 0.006 291 456;
  • 24) 0.006 291 456 × 2 = 0 + 0.012 582 912;
  • 25) 0.012 582 912 × 2 = 0 + 0.025 165 824;
  • 26) 0.025 165 824 × 2 = 0 + 0.050 331 648;
  • 27) 0.050 331 648 × 2 = 0 + 0.100 663 296;
  • 28) 0.100 663 296 × 2 = 0 + 0.201 326 592;
  • 29) 0.201 326 592 × 2 = 0 + 0.402 653 184;
  • 30) 0.402 653 184 × 2 = 0 + 0.805 306 368;
  • 31) 0.805 306 368 × 2 = 1 + 0.610 612 736;
  • 32) 0.610 612 736 × 2 = 1 + 0.221 225 472;
  • 33) 0.221 225 472 × 2 = 0 + 0.442 450 944;
  • 34) 0.442 450 944 × 2 = 0 + 0.884 901 888;
  • 35) 0.884 901 888 × 2 = 1 + 0.769 803 776;
  • 36) 0.769 803 776 × 2 = 1 + 0.539 607 552;
  • 37) 0.539 607 552 × 2 = 1 + 0.079 215 104;
  • 38) 0.079 215 104 × 2 = 0 + 0.158 430 208;
  • 39) 0.158 430 208 × 2 = 0 + 0.316 860 416;
  • 40) 0.316 860 416 × 2 = 0 + 0.633 720 832;
  • 41) 0.633 720 832 × 2 = 1 + 0.267 441 664;
  • 42) 0.267 441 664 × 2 = 0 + 0.534 883 328;
  • 43) 0.534 883 328 × 2 = 1 + 0.069 766 656;
  • 44) 0.069 766 656 × 2 = 0 + 0.139 533 312;
  • 45) 0.139 533 312 × 2 = 0 + 0.279 066 624;
  • 46) 0.279 066 624 × 2 = 0 + 0.558 133 248;
  • 47) 0.558 133 248 × 2 = 1 + 0.116 266 496;
  • 48) 0.116 266 496 × 2 = 0 + 0.232 532 992;
  • 49) 0.232 532 992 × 2 = 0 + 0.465 065 984;
  • 50) 0.465 065 984 × 2 = 0 + 0.930 131 968;
  • 51) 0.930 131 968 × 2 = 1 + 0.860 263 936;
  • 52) 0.860 263 936 × 2 = 1 + 0.720 527 872;
  • 53) 0.720 527 872 × 2 = 1 + 0.441 055 744;
  • 54) 0.441 055 744 × 2 = 0 + 0.882 111 488;

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 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1000 1010 0010 0011 10(2)

6. Positive number before normalization:

0.000 000 000 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1000 1010 0010 0011 10(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 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1000 1010 0010 0011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1000 1010 0010 0011 10(2) × 20 =

1.1001 1100 0101 0001 0001 110(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):
1.1001 1100 0101 0001 0001 110


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. 100 1110 0010 1000 1000 1110 =

100 1110 0010 1000 1000 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) =
0110 0000

Mantissa (23 bits) =
100 1110 0010 1000 1000 1110


Decimal number -0.000 000 000 75 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1110 0010 1000 1000 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