-0.000 000 032 428 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 032 428 4(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
-0.000 000 032 428 4(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:

|-0.000 000 032 428 4| = 0.000 000 032 428 4


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: 0.000 000 032 428 4.

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) 0.000 000 032 428 4 × 2 = 0 + 0.000 000 064 856 8;
  • 2) 0.000 000 064 856 8 × 2 = 0 + 0.000 000 129 713 6;
  • 3) 0.000 000 129 713 6 × 2 = 0 + 0.000 000 259 427 2;
  • 4) 0.000 000 259 427 2 × 2 = 0 + 0.000 000 518 854 4;
  • 5) 0.000 000 518 854 4 × 2 = 0 + 0.000 001 037 708 8;
  • 6) 0.000 001 037 708 8 × 2 = 0 + 0.000 002 075 417 6;
  • 7) 0.000 002 075 417 6 × 2 = 0 + 0.000 004 150 835 2;
  • 8) 0.000 004 150 835 2 × 2 = 0 + 0.000 008 301 670 4;
  • 9) 0.000 008 301 670 4 × 2 = 0 + 0.000 016 603 340 8;
  • 10) 0.000 016 603 340 8 × 2 = 0 + 0.000 033 206 681 6;
  • 11) 0.000 033 206 681 6 × 2 = 0 + 0.000 066 413 363 2;
  • 12) 0.000 066 413 363 2 × 2 = 0 + 0.000 132 826 726 4;
  • 13) 0.000 132 826 726 4 × 2 = 0 + 0.000 265 653 452 8;
  • 14) 0.000 265 653 452 8 × 2 = 0 + 0.000 531 306 905 6;
  • 15) 0.000 531 306 905 6 × 2 = 0 + 0.001 062 613 811 2;
  • 16) 0.001 062 613 811 2 × 2 = 0 + 0.002 125 227 622 4;
  • 17) 0.002 125 227 622 4 × 2 = 0 + 0.004 250 455 244 8;
  • 18) 0.004 250 455 244 8 × 2 = 0 + 0.008 500 910 489 6;
  • 19) 0.008 500 910 489 6 × 2 = 0 + 0.017 001 820 979 2;
  • 20) 0.017 001 820 979 2 × 2 = 0 + 0.034 003 641 958 4;
  • 21) 0.034 003 641 958 4 × 2 = 0 + 0.068 007 283 916 8;
  • 22) 0.068 007 283 916 8 × 2 = 0 + 0.136 014 567 833 6;
  • 23) 0.136 014 567 833 6 × 2 = 0 + 0.272 029 135 667 2;
  • 24) 0.272 029 135 667 2 × 2 = 0 + 0.544 058 271 334 4;
  • 25) 0.544 058 271 334 4 × 2 = 1 + 0.088 116 542 668 8;
  • 26) 0.088 116 542 668 8 × 2 = 0 + 0.176 233 085 337 6;
  • 27) 0.176 233 085 337 6 × 2 = 0 + 0.352 466 170 675 2;
  • 28) 0.352 466 170 675 2 × 2 = 0 + 0.704 932 341 350 4;
  • 29) 0.704 932 341 350 4 × 2 = 1 + 0.409 864 682 700 8;
  • 30) 0.409 864 682 700 8 × 2 = 0 + 0.819 729 365 401 6;
  • 31) 0.819 729 365 401 6 × 2 = 1 + 0.639 458 730 803 2;
  • 32) 0.639 458 730 803 2 × 2 = 1 + 0.278 917 461 606 4;
  • 33) 0.278 917 461 606 4 × 2 = 0 + 0.557 834 923 212 8;
  • 34) 0.557 834 923 212 8 × 2 = 1 + 0.115 669 846 425 6;
  • 35) 0.115 669 846 425 6 × 2 = 0 + 0.231 339 692 851 2;
  • 36) 0.231 339 692 851 2 × 2 = 0 + 0.462 679 385 702 4;
  • 37) 0.462 679 385 702 4 × 2 = 0 + 0.925 358 771 404 8;
  • 38) 0.925 358 771 404 8 × 2 = 1 + 0.850 717 542 809 6;
  • 39) 0.850 717 542 809 6 × 2 = 1 + 0.701 435 085 619 2;
  • 40) 0.701 435 085 619 2 × 2 = 1 + 0.402 870 171 238 4;
  • 41) 0.402 870 171 238 4 × 2 = 0 + 0.805 740 342 476 8;
  • 42) 0.805 740 342 476 8 × 2 = 1 + 0.611 480 684 953 6;
  • 43) 0.611 480 684 953 6 × 2 = 1 + 0.222 961 369 907 2;
  • 44) 0.222 961 369 907 2 × 2 = 0 + 0.445 922 739 814 4;
  • 45) 0.445 922 739 814 4 × 2 = 0 + 0.891 845 479 628 8;
  • 46) 0.891 845 479 628 8 × 2 = 1 + 0.783 690 959 257 6;
  • 47) 0.783 690 959 257 6 × 2 = 1 + 0.567 381 918 515 2;
  • 48) 0.567 381 918 515 2 × 2 = 1 + 0.134 763 837 030 4;

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:


0.000 000 032 428 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0111 0110 0111(2)

6. Positive number before normalization:

0.000 000 032 428 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0111 0110 0111(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:


0.000 000 032 428 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0111 0110 0111(2) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0111 0110 0111(2) × 20 =

1.0001 0110 1000 1110 1100 111(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 1000 1110 1100 111


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 0100 0111 0110 0111 =

000 1011 0100 0111 0110 0111


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 0100 0111 0110 0111


Decimal number -0.000 000 032 428 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 1011 0100 0111 0110 0111

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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