-0.000 000 000 737 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 737 7(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
-0.000 000 000 737 7(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:

|-0.000 000 000 737 7| = 0.000 000 000 737 7


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: 0.000 000 000 737 7.

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) 0.000 000 000 737 7 × 2 = 0 + 0.000 000 001 475 4;
  • 2) 0.000 000 001 475 4 × 2 = 0 + 0.000 000 002 950 8;
  • 3) 0.000 000 002 950 8 × 2 = 0 + 0.000 000 005 901 6;
  • 4) 0.000 000 005 901 6 × 2 = 0 + 0.000 000 011 803 2;
  • 5) 0.000 000 011 803 2 × 2 = 0 + 0.000 000 023 606 4;
  • 6) 0.000 000 023 606 4 × 2 = 0 + 0.000 000 047 212 8;
  • 7) 0.000 000 047 212 8 × 2 = 0 + 0.000 000 094 425 6;
  • 8) 0.000 000 094 425 6 × 2 = 0 + 0.000 000 188 851 2;
  • 9) 0.000 000 188 851 2 × 2 = 0 + 0.000 000 377 702 4;
  • 10) 0.000 000 377 702 4 × 2 = 0 + 0.000 000 755 404 8;
  • 11) 0.000 000 755 404 8 × 2 = 0 + 0.000 001 510 809 6;
  • 12) 0.000 001 510 809 6 × 2 = 0 + 0.000 003 021 619 2;
  • 13) 0.000 003 021 619 2 × 2 = 0 + 0.000 006 043 238 4;
  • 14) 0.000 006 043 238 4 × 2 = 0 + 0.000 012 086 476 8;
  • 15) 0.000 012 086 476 8 × 2 = 0 + 0.000 024 172 953 6;
  • 16) 0.000 024 172 953 6 × 2 = 0 + 0.000 048 345 907 2;
  • 17) 0.000 048 345 907 2 × 2 = 0 + 0.000 096 691 814 4;
  • 18) 0.000 096 691 814 4 × 2 = 0 + 0.000 193 383 628 8;
  • 19) 0.000 193 383 628 8 × 2 = 0 + 0.000 386 767 257 6;
  • 20) 0.000 386 767 257 6 × 2 = 0 + 0.000 773 534 515 2;
  • 21) 0.000 773 534 515 2 × 2 = 0 + 0.001 547 069 030 4;
  • 22) 0.001 547 069 030 4 × 2 = 0 + 0.003 094 138 060 8;
  • 23) 0.003 094 138 060 8 × 2 = 0 + 0.006 188 276 121 6;
  • 24) 0.006 188 276 121 6 × 2 = 0 + 0.012 376 552 243 2;
  • 25) 0.012 376 552 243 2 × 2 = 0 + 0.024 753 104 486 4;
  • 26) 0.024 753 104 486 4 × 2 = 0 + 0.049 506 208 972 8;
  • 27) 0.049 506 208 972 8 × 2 = 0 + 0.099 012 417 945 6;
  • 28) 0.099 012 417 945 6 × 2 = 0 + 0.198 024 835 891 2;
  • 29) 0.198 024 835 891 2 × 2 = 0 + 0.396 049 671 782 4;
  • 30) 0.396 049 671 782 4 × 2 = 0 + 0.792 099 343 564 8;
  • 31) 0.792 099 343 564 8 × 2 = 1 + 0.584 198 687 129 6;
  • 32) 0.584 198 687 129 6 × 2 = 1 + 0.168 397 374 259 2;
  • 33) 0.168 397 374 259 2 × 2 = 0 + 0.336 794 748 518 4;
  • 34) 0.336 794 748 518 4 × 2 = 0 + 0.673 589 497 036 8;
  • 35) 0.673 589 497 036 8 × 2 = 1 + 0.347 178 994 073 6;
  • 36) 0.347 178 994 073 6 × 2 = 0 + 0.694 357 988 147 2;
  • 37) 0.694 357 988 147 2 × 2 = 1 + 0.388 715 976 294 4;
  • 38) 0.388 715 976 294 4 × 2 = 0 + 0.777 431 952 588 8;
  • 39) 0.777 431 952 588 8 × 2 = 1 + 0.554 863 905 177 6;
  • 40) 0.554 863 905 177 6 × 2 = 1 + 0.109 727 810 355 2;
  • 41) 0.109 727 810 355 2 × 2 = 0 + 0.219 455 620 710 4;
  • 42) 0.219 455 620 710 4 × 2 = 0 + 0.438 911 241 420 8;
  • 43) 0.438 911 241 420 8 × 2 = 0 + 0.877 822 482 841 6;
  • 44) 0.877 822 482 841 6 × 2 = 1 + 0.755 644 965 683 2;
  • 45) 0.755 644 965 683 2 × 2 = 1 + 0.511 289 931 366 4;
  • 46) 0.511 289 931 366 4 × 2 = 1 + 0.022 579 862 732 8;
  • 47) 0.022 579 862 732 8 × 2 = 0 + 0.045 159 725 465 6;
  • 48) 0.045 159 725 465 6 × 2 = 0 + 0.090 319 450 931 2;
  • 49) 0.090 319 450 931 2 × 2 = 0 + 0.180 638 901 862 4;
  • 50) 0.180 638 901 862 4 × 2 = 0 + 0.361 277 803 724 8;
  • 51) 0.361 277 803 724 8 × 2 = 0 + 0.722 555 607 449 6;
  • 52) 0.722 555 607 449 6 × 2 = 1 + 0.445 111 214 899 2;
  • 53) 0.445 111 214 899 2 × 2 = 0 + 0.890 222 429 798 4;
  • 54) 0.890 222 429 798 4 × 2 = 1 + 0.780 444 859 596 8;

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:


0.000 000 000 737 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1011 0001 1100 0001 01(2)

6. Positive number before normalization:

0.000 000 000 737 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1011 0001 1100 0001 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 737 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1011 0001 1100 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1011 0001 1100 0001 01(2) × 20 =

1.1001 0101 1000 1110 0000 101(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0101 1000 1110 0000 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

96(10) =

0110 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. 100 1010 1100 0111 0000 0101 =

100 1010 1100 0111 0000 0101


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 0000

Mantissa (23 bits) =
100 1010 1100 0111 0000 0101


Decimal number -0.000 000 000 737 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 1100 0111 0000 0101

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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