-0.000 030 518 533 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 030 518 533 1(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
-0.000 030 518 533 1(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:

|-0.000 030 518 533 1| = 0.000 030 518 533 1


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: 0.000 030 518 533 1.

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) 0.000 030 518 533 1 × 2 = 0 + 0.000 061 037 066 2;
  • 2) 0.000 061 037 066 2 × 2 = 0 + 0.000 122 074 132 4;
  • 3) 0.000 122 074 132 4 × 2 = 0 + 0.000 244 148 264 8;
  • 4) 0.000 244 148 264 8 × 2 = 0 + 0.000 488 296 529 6;
  • 5) 0.000 488 296 529 6 × 2 = 0 + 0.000 976 593 059 2;
  • 6) 0.000 976 593 059 2 × 2 = 0 + 0.001 953 186 118 4;
  • 7) 0.001 953 186 118 4 × 2 = 0 + 0.003 906 372 236 8;
  • 8) 0.003 906 372 236 8 × 2 = 0 + 0.007 812 744 473 6;
  • 9) 0.007 812 744 473 6 × 2 = 0 + 0.015 625 488 947 2;
  • 10) 0.015 625 488 947 2 × 2 = 0 + 0.031 250 977 894 4;
  • 11) 0.031 250 977 894 4 × 2 = 0 + 0.062 501 955 788 8;
  • 12) 0.062 501 955 788 8 × 2 = 0 + 0.125 003 911 577 6;
  • 13) 0.125 003 911 577 6 × 2 = 0 + 0.250 007 823 155 2;
  • 14) 0.250 007 823 155 2 × 2 = 0 + 0.500 015 646 310 4;
  • 15) 0.500 015 646 310 4 × 2 = 1 + 0.000 031 292 620 8;
  • 16) 0.000 031 292 620 8 × 2 = 0 + 0.000 062 585 241 6;
  • 17) 0.000 062 585 241 6 × 2 = 0 + 0.000 125 170 483 2;
  • 18) 0.000 125 170 483 2 × 2 = 0 + 0.000 250 340 966 4;
  • 19) 0.000 250 340 966 4 × 2 = 0 + 0.000 500 681 932 8;
  • 20) 0.000 500 681 932 8 × 2 = 0 + 0.001 001 363 865 6;
  • 21) 0.001 001 363 865 6 × 2 = 0 + 0.002 002 727 731 2;
  • 22) 0.002 002 727 731 2 × 2 = 0 + 0.004 005 455 462 4;
  • 23) 0.004 005 455 462 4 × 2 = 0 + 0.008 010 910 924 8;
  • 24) 0.008 010 910 924 8 × 2 = 0 + 0.016 021 821 849 6;
  • 25) 0.016 021 821 849 6 × 2 = 0 + 0.032 043 643 699 2;
  • 26) 0.032 043 643 699 2 × 2 = 0 + 0.064 087 287 398 4;
  • 27) 0.064 087 287 398 4 × 2 = 0 + 0.128 174 574 796 8;
  • 28) 0.128 174 574 796 8 × 2 = 0 + 0.256 349 149 593 6;
  • 29) 0.256 349 149 593 6 × 2 = 0 + 0.512 698 299 187 2;
  • 30) 0.512 698 299 187 2 × 2 = 1 + 0.025 396 598 374 4;
  • 31) 0.025 396 598 374 4 × 2 = 0 + 0.050 793 196 748 8;
  • 32) 0.050 793 196 748 8 × 2 = 0 + 0.101 586 393 497 6;
  • 33) 0.101 586 393 497 6 × 2 = 0 + 0.203 172 786 995 2;
  • 34) 0.203 172 786 995 2 × 2 = 0 + 0.406 345 573 990 4;
  • 35) 0.406 345 573 990 4 × 2 = 0 + 0.812 691 147 980 8;
  • 36) 0.812 691 147 980 8 × 2 = 1 + 0.625 382 295 961 6;
  • 37) 0.625 382 295 961 6 × 2 = 1 + 0.250 764 591 923 2;
  • 38) 0.250 764 591 923 2 × 2 = 0 + 0.501 529 183 846 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:


0.000 030 518 533 1(10) =

0.0000 0000 0000 0010 0000 0000 0000 0100 0001 10(2)

6. Positive number before normalization:

0.000 030 518 533 1(10) =

0.0000 0000 0000 0010 0000 0000 0000 0100 0001 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 030 518 533 1(10) =

0.0000 0000 0000 0010 0000 0000 0000 0100 0001 10(2) =

0.0000 0000 0000 0010 0000 0000 0000 0100 0001 10(2) × 20 =

1.0000 0000 0000 0010 0000 110(2) × 2-15


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): -15

Mantissa (not normalized):
1.0000 0000 0000 0010 0000 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(8-1) - 1 =

(-15 + 127)(10) =

112(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;
  • 112 ÷ 2 = 56 + 0;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

112(10) =

0111 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. 000 0000 0000 0001 0000 0110 =

000 0000 0000 0001 0000 0110


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) =
0111 0000

Mantissa (23 bits) =
000 0000 0000 0001 0000 0110


Decimal number -0.000 030 518 533 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0000 - 000 0000 0000 0001 0000 0110

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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