-0.000 000 032 427 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 032 427 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
-0.000 000 032 427 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:

|-0.000 000 032 427 4| = 0.000 000 032 427 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: 0.000 000 032 427 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) 0.000 000 032 427 4 × 2 = 0 + 0.000 000 064 854 8;
  • 2) 0.000 000 064 854 8 × 2 = 0 + 0.000 000 129 709 6;
  • 3) 0.000 000 129 709 6 × 2 = 0 + 0.000 000 259 419 2;
  • 4) 0.000 000 259 419 2 × 2 = 0 + 0.000 000 518 838 4;
  • 5) 0.000 000 518 838 4 × 2 = 0 + 0.000 001 037 676 8;
  • 6) 0.000 001 037 676 8 × 2 = 0 + 0.000 002 075 353 6;
  • 7) 0.000 002 075 353 6 × 2 = 0 + 0.000 004 150 707 2;
  • 8) 0.000 004 150 707 2 × 2 = 0 + 0.000 008 301 414 4;
  • 9) 0.000 008 301 414 4 × 2 = 0 + 0.000 016 602 828 8;
  • 10) 0.000 016 602 828 8 × 2 = 0 + 0.000 033 205 657 6;
  • 11) 0.000 033 205 657 6 × 2 = 0 + 0.000 066 411 315 2;
  • 12) 0.000 066 411 315 2 × 2 = 0 + 0.000 132 822 630 4;
  • 13) 0.000 132 822 630 4 × 2 = 0 + 0.000 265 645 260 8;
  • 14) 0.000 265 645 260 8 × 2 = 0 + 0.000 531 290 521 6;
  • 15) 0.000 531 290 521 6 × 2 = 0 + 0.001 062 581 043 2;
  • 16) 0.001 062 581 043 2 × 2 = 0 + 0.002 125 162 086 4;
  • 17) 0.002 125 162 086 4 × 2 = 0 + 0.004 250 324 172 8;
  • 18) 0.004 250 324 172 8 × 2 = 0 + 0.008 500 648 345 6;
  • 19) 0.008 500 648 345 6 × 2 = 0 + 0.017 001 296 691 2;
  • 20) 0.017 001 296 691 2 × 2 = 0 + 0.034 002 593 382 4;
  • 21) 0.034 002 593 382 4 × 2 = 0 + 0.068 005 186 764 8;
  • 22) 0.068 005 186 764 8 × 2 = 0 + 0.136 010 373 529 6;
  • 23) 0.136 010 373 529 6 × 2 = 0 + 0.272 020 747 059 2;
  • 24) 0.272 020 747 059 2 × 2 = 0 + 0.544 041 494 118 4;
  • 25) 0.544 041 494 118 4 × 2 = 1 + 0.088 082 988 236 8;
  • 26) 0.088 082 988 236 8 × 2 = 0 + 0.176 165 976 473 6;
  • 27) 0.176 165 976 473 6 × 2 = 0 + 0.352 331 952 947 2;
  • 28) 0.352 331 952 947 2 × 2 = 0 + 0.704 663 905 894 4;
  • 29) 0.704 663 905 894 4 × 2 = 1 + 0.409 327 811 788 8;
  • 30) 0.409 327 811 788 8 × 2 = 0 + 0.818 655 623 577 6;
  • 31) 0.818 655 623 577 6 × 2 = 1 + 0.637 311 247 155 2;
  • 32) 0.637 311 247 155 2 × 2 = 1 + 0.274 622 494 310 4;
  • 33) 0.274 622 494 310 4 × 2 = 0 + 0.549 244 988 620 8;
  • 34) 0.549 244 988 620 8 × 2 = 1 + 0.098 489 977 241 6;
  • 35) 0.098 489 977 241 6 × 2 = 0 + 0.196 979 954 483 2;
  • 36) 0.196 979 954 483 2 × 2 = 0 + 0.393 959 908 966 4;
  • 37) 0.393 959 908 966 4 × 2 = 0 + 0.787 919 817 932 8;
  • 38) 0.787 919 817 932 8 × 2 = 1 + 0.575 839 635 865 6;
  • 39) 0.575 839 635 865 6 × 2 = 1 + 0.151 679 271 731 2;
  • 40) 0.151 679 271 731 2 × 2 = 0 + 0.303 358 543 462 4;
  • 41) 0.303 358 543 462 4 × 2 = 0 + 0.606 717 086 924 8;
  • 42) 0.606 717 086 924 8 × 2 = 1 + 0.213 434 173 849 6;
  • 43) 0.213 434 173 849 6 × 2 = 0 + 0.426 868 347 699 2;
  • 44) 0.426 868 347 699 2 × 2 = 0 + 0.853 736 695 398 4;
  • 45) 0.853 736 695 398 4 × 2 = 1 + 0.707 473 390 796 8;
  • 46) 0.707 473 390 796 8 × 2 = 1 + 0.414 946 781 593 6;
  • 47) 0.414 946 781 593 6 × 2 = 0 + 0.829 893 563 187 2;
  • 48) 0.829 893 563 187 2 × 2 = 1 + 0.659 787 126 374 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:


0.000 000 032 427 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0110 0100 1101(2)

6. Positive number before normalization:

0.000 000 032 427 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0110 0100 1101(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:


0.000 000 032 427 4(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0110 0100 1101(2) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0110 0100 1101(2) × 20 =

1.0001 0110 1000 1100 1001 101(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 1100 1001 101


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 0110 0100 1101 =

000 1011 0100 0110 0100 1101


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 0110 0100 1101


Decimal number -0.000 000 032 427 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 1011 0100 0110 0100 1101

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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