-0.000 000 099 937 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 099 937(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
-0.000 000 099 937(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:

|-0.000 000 099 937| = 0.000 000 099 937


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: 0.000 000 099 937.

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) 0.000 000 099 937 × 2 = 0 + 0.000 000 199 874;
  • 2) 0.000 000 199 874 × 2 = 0 + 0.000 000 399 748;
  • 3) 0.000 000 399 748 × 2 = 0 + 0.000 000 799 496;
  • 4) 0.000 000 799 496 × 2 = 0 + 0.000 001 598 992;
  • 5) 0.000 001 598 992 × 2 = 0 + 0.000 003 197 984;
  • 6) 0.000 003 197 984 × 2 = 0 + 0.000 006 395 968;
  • 7) 0.000 006 395 968 × 2 = 0 + 0.000 012 791 936;
  • 8) 0.000 012 791 936 × 2 = 0 + 0.000 025 583 872;
  • 9) 0.000 025 583 872 × 2 = 0 + 0.000 051 167 744;
  • 10) 0.000 051 167 744 × 2 = 0 + 0.000 102 335 488;
  • 11) 0.000 102 335 488 × 2 = 0 + 0.000 204 670 976;
  • 12) 0.000 204 670 976 × 2 = 0 + 0.000 409 341 952;
  • 13) 0.000 409 341 952 × 2 = 0 + 0.000 818 683 904;
  • 14) 0.000 818 683 904 × 2 = 0 + 0.001 637 367 808;
  • 15) 0.001 637 367 808 × 2 = 0 + 0.003 274 735 616;
  • 16) 0.003 274 735 616 × 2 = 0 + 0.006 549 471 232;
  • 17) 0.006 549 471 232 × 2 = 0 + 0.013 098 942 464;
  • 18) 0.013 098 942 464 × 2 = 0 + 0.026 197 884 928;
  • 19) 0.026 197 884 928 × 2 = 0 + 0.052 395 769 856;
  • 20) 0.052 395 769 856 × 2 = 0 + 0.104 791 539 712;
  • 21) 0.104 791 539 712 × 2 = 0 + 0.209 583 079 424;
  • 22) 0.209 583 079 424 × 2 = 0 + 0.419 166 158 848;
  • 23) 0.419 166 158 848 × 2 = 0 + 0.838 332 317 696;
  • 24) 0.838 332 317 696 × 2 = 1 + 0.676 664 635 392;
  • 25) 0.676 664 635 392 × 2 = 1 + 0.353 329 270 784;
  • 26) 0.353 329 270 784 × 2 = 0 + 0.706 658 541 568;
  • 27) 0.706 658 541 568 × 2 = 1 + 0.413 317 083 136;
  • 28) 0.413 317 083 136 × 2 = 0 + 0.826 634 166 272;
  • 29) 0.826 634 166 272 × 2 = 1 + 0.653 268 332 544;
  • 30) 0.653 268 332 544 × 2 = 1 + 0.306 536 665 088;
  • 31) 0.306 536 665 088 × 2 = 0 + 0.613 073 330 176;
  • 32) 0.613 073 330 176 × 2 = 1 + 0.226 146 660 352;
  • 33) 0.226 146 660 352 × 2 = 0 + 0.452 293 320 704;
  • 34) 0.452 293 320 704 × 2 = 0 + 0.904 586 641 408;
  • 35) 0.904 586 641 408 × 2 = 1 + 0.809 173 282 816;
  • 36) 0.809 173 282 816 × 2 = 1 + 0.618 346 565 632;
  • 37) 0.618 346 565 632 × 2 = 1 + 0.236 693 131 264;
  • 38) 0.236 693 131 264 × 2 = 0 + 0.473 386 262 528;
  • 39) 0.473 386 262 528 × 2 = 0 + 0.946 772 525 056;
  • 40) 0.946 772 525 056 × 2 = 1 + 0.893 545 050 112;
  • 41) 0.893 545 050 112 × 2 = 1 + 0.787 090 100 224;
  • 42) 0.787 090 100 224 × 2 = 1 + 0.574 180 200 448;
  • 43) 0.574 180 200 448 × 2 = 1 + 0.148 360 400 896;
  • 44) 0.148 360 400 896 × 2 = 0 + 0.296 720 801 792;
  • 45) 0.296 720 801 792 × 2 = 0 + 0.593 441 603 584;
  • 46) 0.593 441 603 584 × 2 = 1 + 0.186 883 207 168;
  • 47) 0.186 883 207 168 × 2 = 0 + 0.373 766 414 336;

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:


0.000 000 099 937(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0011 1001 1110 010(2)

6. Positive number before normalization:

0.000 000 099 937(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0011 1001 1110 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 099 937(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0011 1001 1110 010(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0011 1001 1110 010(2) × 20 =

1.1010 1101 0011 1001 1110 010(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 0011 1001 1110 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1001 1100 1111 0010 =

101 0110 1001 1100 1111 0010


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 0111

Mantissa (23 bits) =
101 0110 1001 1100 1111 0010


Decimal number -0.000 000 099 937 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1001 1100 1111 0010

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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