-0.000 000 100 003 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 100 003(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
-0.000 000 100 003(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:

|-0.000 000 100 003| = 0.000 000 100 003


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: 0.000 000 100 003.

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) 0.000 000 100 003 × 2 = 0 + 0.000 000 200 006;
  • 2) 0.000 000 200 006 × 2 = 0 + 0.000 000 400 012;
  • 3) 0.000 000 400 012 × 2 = 0 + 0.000 000 800 024;
  • 4) 0.000 000 800 024 × 2 = 0 + 0.000 001 600 048;
  • 5) 0.000 001 600 048 × 2 = 0 + 0.000 003 200 096;
  • 6) 0.000 003 200 096 × 2 = 0 + 0.000 006 400 192;
  • 7) 0.000 006 400 192 × 2 = 0 + 0.000 012 800 384;
  • 8) 0.000 012 800 384 × 2 = 0 + 0.000 025 600 768;
  • 9) 0.000 025 600 768 × 2 = 0 + 0.000 051 201 536;
  • 10) 0.000 051 201 536 × 2 = 0 + 0.000 102 403 072;
  • 11) 0.000 102 403 072 × 2 = 0 + 0.000 204 806 144;
  • 12) 0.000 204 806 144 × 2 = 0 + 0.000 409 612 288;
  • 13) 0.000 409 612 288 × 2 = 0 + 0.000 819 224 576;
  • 14) 0.000 819 224 576 × 2 = 0 + 0.001 638 449 152;
  • 15) 0.001 638 449 152 × 2 = 0 + 0.003 276 898 304;
  • 16) 0.003 276 898 304 × 2 = 0 + 0.006 553 796 608;
  • 17) 0.006 553 796 608 × 2 = 0 + 0.013 107 593 216;
  • 18) 0.013 107 593 216 × 2 = 0 + 0.026 215 186 432;
  • 19) 0.026 215 186 432 × 2 = 0 + 0.052 430 372 864;
  • 20) 0.052 430 372 864 × 2 = 0 + 0.104 860 745 728;
  • 21) 0.104 860 745 728 × 2 = 0 + 0.209 721 491 456;
  • 22) 0.209 721 491 456 × 2 = 0 + 0.419 442 982 912;
  • 23) 0.419 442 982 912 × 2 = 0 + 0.838 885 965 824;
  • 24) 0.838 885 965 824 × 2 = 1 + 0.677 771 931 648;
  • 25) 0.677 771 931 648 × 2 = 1 + 0.355 543 863 296;
  • 26) 0.355 543 863 296 × 2 = 0 + 0.711 087 726 592;
  • 27) 0.711 087 726 592 × 2 = 1 + 0.422 175 453 184;
  • 28) 0.422 175 453 184 × 2 = 0 + 0.844 350 906 368;
  • 29) 0.844 350 906 368 × 2 = 1 + 0.688 701 812 736;
  • 30) 0.688 701 812 736 × 2 = 1 + 0.377 403 625 472;
  • 31) 0.377 403 625 472 × 2 = 0 + 0.754 807 250 944;
  • 32) 0.754 807 250 944 × 2 = 1 + 0.509 614 501 888;
  • 33) 0.509 614 501 888 × 2 = 1 + 0.019 229 003 776;
  • 34) 0.019 229 003 776 × 2 = 0 + 0.038 458 007 552;
  • 35) 0.038 458 007 552 × 2 = 0 + 0.076 916 015 104;
  • 36) 0.076 916 015 104 × 2 = 0 + 0.153 832 030 208;
  • 37) 0.153 832 030 208 × 2 = 0 + 0.307 664 060 416;
  • 38) 0.307 664 060 416 × 2 = 0 + 0.615 328 120 832;
  • 39) 0.615 328 120 832 × 2 = 1 + 0.230 656 241 664;
  • 40) 0.230 656 241 664 × 2 = 0 + 0.461 312 483 328;
  • 41) 0.461 312 483 328 × 2 = 0 + 0.922 624 966 656;
  • 42) 0.922 624 966 656 × 2 = 1 + 0.845 249 933 312;
  • 43) 0.845 249 933 312 × 2 = 1 + 0.690 499 866 624;
  • 44) 0.690 499 866 624 × 2 = 1 + 0.380 999 733 248;
  • 45) 0.380 999 733 248 × 2 = 0 + 0.761 999 466 496;
  • 46) 0.761 999 466 496 × 2 = 1 + 0.523 998 932 992;
  • 47) 0.523 998 932 992 × 2 = 1 + 0.047 997 865 984;

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:


0.000 000 100 003(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1000 0010 0111 011(2)

6. Positive number before normalization:

0.000 000 100 003(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1000 0010 0111 011(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:


0.000 000 100 003(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1000 0010 0111 011(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1000 0010 0111 011(2) × 20 =

1.1010 1101 1000 0010 0111 011(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 1000 0010 0111 011


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 1100 0001 0011 1011 =

101 0110 1100 0001 0011 1011


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 1100 0001 0011 1011


Decimal number -0.000 000 100 003 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1100 0001 0011 1011

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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