-0.000 000 032 418 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 032 418 3(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
-0.000 000 032 418 3(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:

|-0.000 000 032 418 3| = 0.000 000 032 418 3


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: 0.000 000 032 418 3.

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) 0.000 000 032 418 3 × 2 = 0 + 0.000 000 064 836 6;
  • 2) 0.000 000 064 836 6 × 2 = 0 + 0.000 000 129 673 2;
  • 3) 0.000 000 129 673 2 × 2 = 0 + 0.000 000 259 346 4;
  • 4) 0.000 000 259 346 4 × 2 = 0 + 0.000 000 518 692 8;
  • 5) 0.000 000 518 692 8 × 2 = 0 + 0.000 001 037 385 6;
  • 6) 0.000 001 037 385 6 × 2 = 0 + 0.000 002 074 771 2;
  • 7) 0.000 002 074 771 2 × 2 = 0 + 0.000 004 149 542 4;
  • 8) 0.000 004 149 542 4 × 2 = 0 + 0.000 008 299 084 8;
  • 9) 0.000 008 299 084 8 × 2 = 0 + 0.000 016 598 169 6;
  • 10) 0.000 016 598 169 6 × 2 = 0 + 0.000 033 196 339 2;
  • 11) 0.000 033 196 339 2 × 2 = 0 + 0.000 066 392 678 4;
  • 12) 0.000 066 392 678 4 × 2 = 0 + 0.000 132 785 356 8;
  • 13) 0.000 132 785 356 8 × 2 = 0 + 0.000 265 570 713 6;
  • 14) 0.000 265 570 713 6 × 2 = 0 + 0.000 531 141 427 2;
  • 15) 0.000 531 141 427 2 × 2 = 0 + 0.001 062 282 854 4;
  • 16) 0.001 062 282 854 4 × 2 = 0 + 0.002 124 565 708 8;
  • 17) 0.002 124 565 708 8 × 2 = 0 + 0.004 249 131 417 6;
  • 18) 0.004 249 131 417 6 × 2 = 0 + 0.008 498 262 835 2;
  • 19) 0.008 498 262 835 2 × 2 = 0 + 0.016 996 525 670 4;
  • 20) 0.016 996 525 670 4 × 2 = 0 + 0.033 993 051 340 8;
  • 21) 0.033 993 051 340 8 × 2 = 0 + 0.067 986 102 681 6;
  • 22) 0.067 986 102 681 6 × 2 = 0 + 0.135 972 205 363 2;
  • 23) 0.135 972 205 363 2 × 2 = 0 + 0.271 944 410 726 4;
  • 24) 0.271 944 410 726 4 × 2 = 0 + 0.543 888 821 452 8;
  • 25) 0.543 888 821 452 8 × 2 = 1 + 0.087 777 642 905 6;
  • 26) 0.087 777 642 905 6 × 2 = 0 + 0.175 555 285 811 2;
  • 27) 0.175 555 285 811 2 × 2 = 0 + 0.351 110 571 622 4;
  • 28) 0.351 110 571 622 4 × 2 = 0 + 0.702 221 143 244 8;
  • 29) 0.702 221 143 244 8 × 2 = 1 + 0.404 442 286 489 6;
  • 30) 0.404 442 286 489 6 × 2 = 0 + 0.808 884 572 979 2;
  • 31) 0.808 884 572 979 2 × 2 = 1 + 0.617 769 145 958 4;
  • 32) 0.617 769 145 958 4 × 2 = 1 + 0.235 538 291 916 8;
  • 33) 0.235 538 291 916 8 × 2 = 0 + 0.471 076 583 833 6;
  • 34) 0.471 076 583 833 6 × 2 = 0 + 0.942 153 167 667 2;
  • 35) 0.942 153 167 667 2 × 2 = 1 + 0.884 306 335 334 4;
  • 36) 0.884 306 335 334 4 × 2 = 1 + 0.768 612 670 668 8;
  • 37) 0.768 612 670 668 8 × 2 = 1 + 0.537 225 341 337 6;
  • 38) 0.537 225 341 337 6 × 2 = 1 + 0.074 450 682 675 2;
  • 39) 0.074 450 682 675 2 × 2 = 0 + 0.148 901 365 350 4;
  • 40) 0.148 901 365 350 4 × 2 = 0 + 0.297 802 730 700 8;
  • 41) 0.297 802 730 700 8 × 2 = 0 + 0.595 605 461 401 6;
  • 42) 0.595 605 461 401 6 × 2 = 1 + 0.191 210 922 803 2;
  • 43) 0.191 210 922 803 2 × 2 = 0 + 0.382 421 845 606 4;
  • 44) 0.382 421 845 606 4 × 2 = 0 + 0.764 843 691 212 8;
  • 45) 0.764 843 691 212 8 × 2 = 1 + 0.529 687 382 425 6;
  • 46) 0.529 687 382 425 6 × 2 = 1 + 0.059 374 764 851 2;
  • 47) 0.059 374 764 851 2 × 2 = 0 + 0.118 749 529 702 4;
  • 48) 0.118 749 529 702 4 × 2 = 0 + 0.237 499 059 404 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:


0.000 000 032 418 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0011 1100 0100 1100(2)

6. Positive number before normalization:

0.000 000 032 418 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0011 1100 0100 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 032 418 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0011 1100 0100 1100(2) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0011 1100 0100 1100(2) × 20 =

1.0001 0110 0111 1000 1001 100(2) × 2-25


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): -25

Mantissa (not normalized):
1.0001 0110 0111 1000 1001 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

102(10) =

0110 0110(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. 000 1011 0011 1100 0100 1100 =

000 1011 0011 1100 0100 1100


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 0110

Mantissa (23 bits) =
000 1011 0011 1100 0100 1100


Decimal number -0.000 000 032 418 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 1011 0011 1100 0100 1100

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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