-0.000 000 000 724 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 724 9(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
-0.000 000 000 724 9(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:

|-0.000 000 000 724 9| = 0.000 000 000 724 9


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: 0.000 000 000 724 9.

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) 0.000 000 000 724 9 × 2 = 0 + 0.000 000 001 449 8;
  • 2) 0.000 000 001 449 8 × 2 = 0 + 0.000 000 002 899 6;
  • 3) 0.000 000 002 899 6 × 2 = 0 + 0.000 000 005 799 2;
  • 4) 0.000 000 005 799 2 × 2 = 0 + 0.000 000 011 598 4;
  • 5) 0.000 000 011 598 4 × 2 = 0 + 0.000 000 023 196 8;
  • 6) 0.000 000 023 196 8 × 2 = 0 + 0.000 000 046 393 6;
  • 7) 0.000 000 046 393 6 × 2 = 0 + 0.000 000 092 787 2;
  • 8) 0.000 000 092 787 2 × 2 = 0 + 0.000 000 185 574 4;
  • 9) 0.000 000 185 574 4 × 2 = 0 + 0.000 000 371 148 8;
  • 10) 0.000 000 371 148 8 × 2 = 0 + 0.000 000 742 297 6;
  • 11) 0.000 000 742 297 6 × 2 = 0 + 0.000 001 484 595 2;
  • 12) 0.000 001 484 595 2 × 2 = 0 + 0.000 002 969 190 4;
  • 13) 0.000 002 969 190 4 × 2 = 0 + 0.000 005 938 380 8;
  • 14) 0.000 005 938 380 8 × 2 = 0 + 0.000 011 876 761 6;
  • 15) 0.000 011 876 761 6 × 2 = 0 + 0.000 023 753 523 2;
  • 16) 0.000 023 753 523 2 × 2 = 0 + 0.000 047 507 046 4;
  • 17) 0.000 047 507 046 4 × 2 = 0 + 0.000 095 014 092 8;
  • 18) 0.000 095 014 092 8 × 2 = 0 + 0.000 190 028 185 6;
  • 19) 0.000 190 028 185 6 × 2 = 0 + 0.000 380 056 371 2;
  • 20) 0.000 380 056 371 2 × 2 = 0 + 0.000 760 112 742 4;
  • 21) 0.000 760 112 742 4 × 2 = 0 + 0.001 520 225 484 8;
  • 22) 0.001 520 225 484 8 × 2 = 0 + 0.003 040 450 969 6;
  • 23) 0.003 040 450 969 6 × 2 = 0 + 0.006 080 901 939 2;
  • 24) 0.006 080 901 939 2 × 2 = 0 + 0.012 161 803 878 4;
  • 25) 0.012 161 803 878 4 × 2 = 0 + 0.024 323 607 756 8;
  • 26) 0.024 323 607 756 8 × 2 = 0 + 0.048 647 215 513 6;
  • 27) 0.048 647 215 513 6 × 2 = 0 + 0.097 294 431 027 2;
  • 28) 0.097 294 431 027 2 × 2 = 0 + 0.194 588 862 054 4;
  • 29) 0.194 588 862 054 4 × 2 = 0 + 0.389 177 724 108 8;
  • 30) 0.389 177 724 108 8 × 2 = 0 + 0.778 355 448 217 6;
  • 31) 0.778 355 448 217 6 × 2 = 1 + 0.556 710 896 435 2;
  • 32) 0.556 710 896 435 2 × 2 = 1 + 0.113 421 792 870 4;
  • 33) 0.113 421 792 870 4 × 2 = 0 + 0.226 843 585 740 8;
  • 34) 0.226 843 585 740 8 × 2 = 0 + 0.453 687 171 481 6;
  • 35) 0.453 687 171 481 6 × 2 = 0 + 0.907 374 342 963 2;
  • 36) 0.907 374 342 963 2 × 2 = 1 + 0.814 748 685 926 4;
  • 37) 0.814 748 685 926 4 × 2 = 1 + 0.629 497 371 852 8;
  • 38) 0.629 497 371 852 8 × 2 = 1 + 0.258 994 743 705 6;
  • 39) 0.258 994 743 705 6 × 2 = 0 + 0.517 989 487 411 2;
  • 40) 0.517 989 487 411 2 × 2 = 1 + 0.035 978 974 822 4;
  • 41) 0.035 978 974 822 4 × 2 = 0 + 0.071 957 949 644 8;
  • 42) 0.071 957 949 644 8 × 2 = 0 + 0.143 915 899 289 6;
  • 43) 0.143 915 899 289 6 × 2 = 0 + 0.287 831 798 579 2;
  • 44) 0.287 831 798 579 2 × 2 = 0 + 0.575 663 597 158 4;
  • 45) 0.575 663 597 158 4 × 2 = 1 + 0.151 327 194 316 8;
  • 46) 0.151 327 194 316 8 × 2 = 0 + 0.302 654 388 633 6;
  • 47) 0.302 654 388 633 6 × 2 = 0 + 0.605 308 777 267 2;
  • 48) 0.605 308 777 267 2 × 2 = 1 + 0.210 617 554 534 4;
  • 49) 0.210 617 554 534 4 × 2 = 0 + 0.421 235 109 068 8;
  • 50) 0.421 235 109 068 8 × 2 = 0 + 0.842 470 218 137 6;
  • 51) 0.842 470 218 137 6 × 2 = 1 + 0.684 940 436 275 2;
  • 52) 0.684 940 436 275 2 × 2 = 1 + 0.369 880 872 550 4;
  • 53) 0.369 880 872 550 4 × 2 = 0 + 0.739 761 745 100 8;
  • 54) 0.739 761 745 100 8 × 2 = 1 + 0.479 523 490 201 6;

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:


0.000 000 000 724 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0000 1001 0011 01(2)

6. Positive number before normalization:

0.000 000 000 724 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0000 1001 0011 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:


0.000 000 000 724 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0000 1001 0011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0000 1001 0011 01(2) × 20 =

1.1000 1110 1000 0100 1001 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):
1.1000 1110 1000 0100 1001 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 0111 0100 0010 0100 1101 =

100 0111 0100 0010 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 0000

Mantissa (23 bits) =
100 0111 0100 0010 0100 1101


Decimal number -0.000 000 000 724 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0111 0100 0010 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