-0.000 000 000 732 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 732 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
-0.000 000 000 732 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:

|-0.000 000 000 732 1| = 0.000 000 000 732 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: 0.000 000 000 732 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) 0.000 000 000 732 1 × 2 = 0 + 0.000 000 001 464 2;
  • 2) 0.000 000 001 464 2 × 2 = 0 + 0.000 000 002 928 4;
  • 3) 0.000 000 002 928 4 × 2 = 0 + 0.000 000 005 856 8;
  • 4) 0.000 000 005 856 8 × 2 = 0 + 0.000 000 011 713 6;
  • 5) 0.000 000 011 713 6 × 2 = 0 + 0.000 000 023 427 2;
  • 6) 0.000 000 023 427 2 × 2 = 0 + 0.000 000 046 854 4;
  • 7) 0.000 000 046 854 4 × 2 = 0 + 0.000 000 093 708 8;
  • 8) 0.000 000 093 708 8 × 2 = 0 + 0.000 000 187 417 6;
  • 9) 0.000 000 187 417 6 × 2 = 0 + 0.000 000 374 835 2;
  • 10) 0.000 000 374 835 2 × 2 = 0 + 0.000 000 749 670 4;
  • 11) 0.000 000 749 670 4 × 2 = 0 + 0.000 001 499 340 8;
  • 12) 0.000 001 499 340 8 × 2 = 0 + 0.000 002 998 681 6;
  • 13) 0.000 002 998 681 6 × 2 = 0 + 0.000 005 997 363 2;
  • 14) 0.000 005 997 363 2 × 2 = 0 + 0.000 011 994 726 4;
  • 15) 0.000 011 994 726 4 × 2 = 0 + 0.000 023 989 452 8;
  • 16) 0.000 023 989 452 8 × 2 = 0 + 0.000 047 978 905 6;
  • 17) 0.000 047 978 905 6 × 2 = 0 + 0.000 095 957 811 2;
  • 18) 0.000 095 957 811 2 × 2 = 0 + 0.000 191 915 622 4;
  • 19) 0.000 191 915 622 4 × 2 = 0 + 0.000 383 831 244 8;
  • 20) 0.000 383 831 244 8 × 2 = 0 + 0.000 767 662 489 6;
  • 21) 0.000 767 662 489 6 × 2 = 0 + 0.001 535 324 979 2;
  • 22) 0.001 535 324 979 2 × 2 = 0 + 0.003 070 649 958 4;
  • 23) 0.003 070 649 958 4 × 2 = 0 + 0.006 141 299 916 8;
  • 24) 0.006 141 299 916 8 × 2 = 0 + 0.012 282 599 833 6;
  • 25) 0.012 282 599 833 6 × 2 = 0 + 0.024 565 199 667 2;
  • 26) 0.024 565 199 667 2 × 2 = 0 + 0.049 130 399 334 4;
  • 27) 0.049 130 399 334 4 × 2 = 0 + 0.098 260 798 668 8;
  • 28) 0.098 260 798 668 8 × 2 = 0 + 0.196 521 597 337 6;
  • 29) 0.196 521 597 337 6 × 2 = 0 + 0.393 043 194 675 2;
  • 30) 0.393 043 194 675 2 × 2 = 0 + 0.786 086 389 350 4;
  • 31) 0.786 086 389 350 4 × 2 = 1 + 0.572 172 778 700 8;
  • 32) 0.572 172 778 700 8 × 2 = 1 + 0.144 345 557 401 6;
  • 33) 0.144 345 557 401 6 × 2 = 0 + 0.288 691 114 803 2;
  • 34) 0.288 691 114 803 2 × 2 = 0 + 0.577 382 229 606 4;
  • 35) 0.577 382 229 606 4 × 2 = 1 + 0.154 764 459 212 8;
  • 36) 0.154 764 459 212 8 × 2 = 0 + 0.309 528 918 425 6;
  • 37) 0.309 528 918 425 6 × 2 = 0 + 0.619 057 836 851 2;
  • 38) 0.619 057 836 851 2 × 2 = 1 + 0.238 115 673 702 4;
  • 39) 0.238 115 673 702 4 × 2 = 0 + 0.476 231 347 404 8;
  • 40) 0.476 231 347 404 8 × 2 = 0 + 0.952 462 694 809 6;
  • 41) 0.952 462 694 809 6 × 2 = 1 + 0.904 925 389 619 2;
  • 42) 0.904 925 389 619 2 × 2 = 1 + 0.809 850 779 238 4;
  • 43) 0.809 850 779 238 4 × 2 = 1 + 0.619 701 558 476 8;
  • 44) 0.619 701 558 476 8 × 2 = 1 + 0.239 403 116 953 6;
  • 45) 0.239 403 116 953 6 × 2 = 0 + 0.478 806 233 907 2;
  • 46) 0.478 806 233 907 2 × 2 = 0 + 0.957 612 467 814 4;
  • 47) 0.957 612 467 814 4 × 2 = 1 + 0.915 224 935 628 8;
  • 48) 0.915 224 935 628 8 × 2 = 1 + 0.830 449 871 257 6;
  • 49) 0.830 449 871 257 6 × 2 = 1 + 0.660 899 742 515 2;
  • 50) 0.660 899 742 515 2 × 2 = 1 + 0.321 799 485 030 4;
  • 51) 0.321 799 485 030 4 × 2 = 0 + 0.643 598 970 060 8;
  • 52) 0.643 598 970 060 8 × 2 = 1 + 0.287 197 940 121 6;
  • 53) 0.287 197 940 121 6 × 2 = 0 + 0.574 395 880 243 2;
  • 54) 0.574 395 880 243 2 × 2 = 1 + 0.148 791 760 486 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:


0.000 000 000 732 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0100 1111 0011 1101 01(2)

6. Positive number before normalization:

0.000 000 000 732 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0100 1111 0011 1101 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:


0.000 000 000 732 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0100 1111 0011 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0100 1111 0011 1101 01(2) × 20 =

1.1001 0010 0111 1001 1110 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 0010 0111 1001 1110 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 1001 0011 1100 1111 0101 =

100 1001 0011 1100 1111 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 1001 0011 1100 1111 0101


Decimal number -0.000 000 000 732 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1001 0011 1100 1111 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