-0.000 000 000 740 45 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 45(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
-0.000 000 000 740 45(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:

|-0.000 000 000 740 45| = 0.000 000 000 740 45


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: 0.000 000 000 740 45.

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) 0.000 000 000 740 45 × 2 = 0 + 0.000 000 001 480 9;
  • 2) 0.000 000 001 480 9 × 2 = 0 + 0.000 000 002 961 8;
  • 3) 0.000 000 002 961 8 × 2 = 0 + 0.000 000 005 923 6;
  • 4) 0.000 000 005 923 6 × 2 = 0 + 0.000 000 011 847 2;
  • 5) 0.000 000 011 847 2 × 2 = 0 + 0.000 000 023 694 4;
  • 6) 0.000 000 023 694 4 × 2 = 0 + 0.000 000 047 388 8;
  • 7) 0.000 000 047 388 8 × 2 = 0 + 0.000 000 094 777 6;
  • 8) 0.000 000 094 777 6 × 2 = 0 + 0.000 000 189 555 2;
  • 9) 0.000 000 189 555 2 × 2 = 0 + 0.000 000 379 110 4;
  • 10) 0.000 000 379 110 4 × 2 = 0 + 0.000 000 758 220 8;
  • 11) 0.000 000 758 220 8 × 2 = 0 + 0.000 001 516 441 6;
  • 12) 0.000 001 516 441 6 × 2 = 0 + 0.000 003 032 883 2;
  • 13) 0.000 003 032 883 2 × 2 = 0 + 0.000 006 065 766 4;
  • 14) 0.000 006 065 766 4 × 2 = 0 + 0.000 012 131 532 8;
  • 15) 0.000 012 131 532 8 × 2 = 0 + 0.000 024 263 065 6;
  • 16) 0.000 024 263 065 6 × 2 = 0 + 0.000 048 526 131 2;
  • 17) 0.000 048 526 131 2 × 2 = 0 + 0.000 097 052 262 4;
  • 18) 0.000 097 052 262 4 × 2 = 0 + 0.000 194 104 524 8;
  • 19) 0.000 194 104 524 8 × 2 = 0 + 0.000 388 209 049 6;
  • 20) 0.000 388 209 049 6 × 2 = 0 + 0.000 776 418 099 2;
  • 21) 0.000 776 418 099 2 × 2 = 0 + 0.001 552 836 198 4;
  • 22) 0.001 552 836 198 4 × 2 = 0 + 0.003 105 672 396 8;
  • 23) 0.003 105 672 396 8 × 2 = 0 + 0.006 211 344 793 6;
  • 24) 0.006 211 344 793 6 × 2 = 0 + 0.012 422 689 587 2;
  • 25) 0.012 422 689 587 2 × 2 = 0 + 0.024 845 379 174 4;
  • 26) 0.024 845 379 174 4 × 2 = 0 + 0.049 690 758 348 8;
  • 27) 0.049 690 758 348 8 × 2 = 0 + 0.099 381 516 697 6;
  • 28) 0.099 381 516 697 6 × 2 = 0 + 0.198 763 033 395 2;
  • 29) 0.198 763 033 395 2 × 2 = 0 + 0.397 526 066 790 4;
  • 30) 0.397 526 066 790 4 × 2 = 0 + 0.795 052 133 580 8;
  • 31) 0.795 052 133 580 8 × 2 = 1 + 0.590 104 267 161 6;
  • 32) 0.590 104 267 161 6 × 2 = 1 + 0.180 208 534 323 2;
  • 33) 0.180 208 534 323 2 × 2 = 0 + 0.360 417 068 646 4;
  • 34) 0.360 417 068 646 4 × 2 = 0 + 0.720 834 137 292 8;
  • 35) 0.720 834 137 292 8 × 2 = 1 + 0.441 668 274 585 6;
  • 36) 0.441 668 274 585 6 × 2 = 0 + 0.883 336 549 171 2;
  • 37) 0.883 336 549 171 2 × 2 = 1 + 0.766 673 098 342 4;
  • 38) 0.766 673 098 342 4 × 2 = 1 + 0.533 346 196 684 8;
  • 39) 0.533 346 196 684 8 × 2 = 1 + 0.066 692 393 369 6;
  • 40) 0.066 692 393 369 6 × 2 = 0 + 0.133 384 786 739 2;
  • 41) 0.133 384 786 739 2 × 2 = 0 + 0.266 769 573 478 4;
  • 42) 0.266 769 573 478 4 × 2 = 0 + 0.533 539 146 956 8;
  • 43) 0.533 539 146 956 8 × 2 = 1 + 0.067 078 293 913 6;
  • 44) 0.067 078 293 913 6 × 2 = 0 + 0.134 156 587 827 2;
  • 45) 0.134 156 587 827 2 × 2 = 0 + 0.268 313 175 654 4;
  • 46) 0.268 313 175 654 4 × 2 = 0 + 0.536 626 351 308 8;
  • 47) 0.536 626 351 308 8 × 2 = 1 + 0.073 252 702 617 6;
  • 48) 0.073 252 702 617 6 × 2 = 0 + 0.146 505 405 235 2;
  • 49) 0.146 505 405 235 2 × 2 = 0 + 0.293 010 810 470 4;
  • 50) 0.293 010 810 470 4 × 2 = 0 + 0.586 021 620 940 8;
  • 51) 0.586 021 620 940 8 × 2 = 1 + 0.172 043 241 881 6;
  • 52) 0.172 043 241 881 6 × 2 = 0 + 0.344 086 483 763 2;
  • 53) 0.344 086 483 763 2 × 2 = 0 + 0.688 172 967 526 4;
  • 54) 0.688 172 967 526 4 × 2 = 1 + 0.376 345 935 052 8;

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:


0.000 000 000 740 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0010 0010 01(2)

6. Positive number before normalization:

0.000 000 000 740 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0010 0010 01(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:


0.000 000 000 740 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0010 0010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0010 0010 01(2) × 20 =

1.1001 0111 0001 0001 0001 001(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 0111 0001 0001 0001 001


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 1011 1000 1000 1000 1001 =

100 1011 1000 1000 1000 1001


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 1011 1000 1000 1000 1001


Decimal number -0.000 000 000 740 45 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1000 1000 1000 1001

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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