-0.000 000 000 842 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 842(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
-0.000 000 000 842(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:

|-0.000 000 000 842| = 0.000 000 000 842


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: 0.000 000 000 842.

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) 0.000 000 000 842 × 2 = 0 + 0.000 000 001 684;
  • 2) 0.000 000 001 684 × 2 = 0 + 0.000 000 003 368;
  • 3) 0.000 000 003 368 × 2 = 0 + 0.000 000 006 736;
  • 4) 0.000 000 006 736 × 2 = 0 + 0.000 000 013 472;
  • 5) 0.000 000 013 472 × 2 = 0 + 0.000 000 026 944;
  • 6) 0.000 000 026 944 × 2 = 0 + 0.000 000 053 888;
  • 7) 0.000 000 053 888 × 2 = 0 + 0.000 000 107 776;
  • 8) 0.000 000 107 776 × 2 = 0 + 0.000 000 215 552;
  • 9) 0.000 000 215 552 × 2 = 0 + 0.000 000 431 104;
  • 10) 0.000 000 431 104 × 2 = 0 + 0.000 000 862 208;
  • 11) 0.000 000 862 208 × 2 = 0 + 0.000 001 724 416;
  • 12) 0.000 001 724 416 × 2 = 0 + 0.000 003 448 832;
  • 13) 0.000 003 448 832 × 2 = 0 + 0.000 006 897 664;
  • 14) 0.000 006 897 664 × 2 = 0 + 0.000 013 795 328;
  • 15) 0.000 013 795 328 × 2 = 0 + 0.000 027 590 656;
  • 16) 0.000 027 590 656 × 2 = 0 + 0.000 055 181 312;
  • 17) 0.000 055 181 312 × 2 = 0 + 0.000 110 362 624;
  • 18) 0.000 110 362 624 × 2 = 0 + 0.000 220 725 248;
  • 19) 0.000 220 725 248 × 2 = 0 + 0.000 441 450 496;
  • 20) 0.000 441 450 496 × 2 = 0 + 0.000 882 900 992;
  • 21) 0.000 882 900 992 × 2 = 0 + 0.001 765 801 984;
  • 22) 0.001 765 801 984 × 2 = 0 + 0.003 531 603 968;
  • 23) 0.003 531 603 968 × 2 = 0 + 0.007 063 207 936;
  • 24) 0.007 063 207 936 × 2 = 0 + 0.014 126 415 872;
  • 25) 0.014 126 415 872 × 2 = 0 + 0.028 252 831 744;
  • 26) 0.028 252 831 744 × 2 = 0 + 0.056 505 663 488;
  • 27) 0.056 505 663 488 × 2 = 0 + 0.113 011 326 976;
  • 28) 0.113 011 326 976 × 2 = 0 + 0.226 022 653 952;
  • 29) 0.226 022 653 952 × 2 = 0 + 0.452 045 307 904;
  • 30) 0.452 045 307 904 × 2 = 0 + 0.904 090 615 808;
  • 31) 0.904 090 615 808 × 2 = 1 + 0.808 181 231 616;
  • 32) 0.808 181 231 616 × 2 = 1 + 0.616 362 463 232;
  • 33) 0.616 362 463 232 × 2 = 1 + 0.232 724 926 464;
  • 34) 0.232 724 926 464 × 2 = 0 + 0.465 449 852 928;
  • 35) 0.465 449 852 928 × 2 = 0 + 0.930 899 705 856;
  • 36) 0.930 899 705 856 × 2 = 1 + 0.861 799 411 712;
  • 37) 0.861 799 411 712 × 2 = 1 + 0.723 598 823 424;
  • 38) 0.723 598 823 424 × 2 = 1 + 0.447 197 646 848;
  • 39) 0.447 197 646 848 × 2 = 0 + 0.894 395 293 696;
  • 40) 0.894 395 293 696 × 2 = 1 + 0.788 790 587 392;
  • 41) 0.788 790 587 392 × 2 = 1 + 0.577 581 174 784;
  • 42) 0.577 581 174 784 × 2 = 1 + 0.155 162 349 568;
  • 43) 0.155 162 349 568 × 2 = 0 + 0.310 324 699 136;
  • 44) 0.310 324 699 136 × 2 = 0 + 0.620 649 398 272;
  • 45) 0.620 649 398 272 × 2 = 1 + 0.241 298 796 544;
  • 46) 0.241 298 796 544 × 2 = 0 + 0.482 597 593 088;
  • 47) 0.482 597 593 088 × 2 = 0 + 0.965 195 186 176;
  • 48) 0.965 195 186 176 × 2 = 1 + 0.930 390 372 352;
  • 49) 0.930 390 372 352 × 2 = 1 + 0.860 780 744 704;
  • 50) 0.860 780 744 704 × 2 = 1 + 0.721 561 489 408;
  • 51) 0.721 561 489 408 × 2 = 1 + 0.443 122 978 816;
  • 52) 0.443 122 978 816 × 2 = 0 + 0.886 245 957 632;
  • 53) 0.886 245 957 632 × 2 = 1 + 0.772 491 915 264;
  • 54) 0.772 491 915 264 × 2 = 1 + 0.544 983 830 528;

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:


0.000 000 000 842(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1101 1100 1001 1110 11(2)

6. Positive number before normalization:

0.000 000 000 842(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1101 1100 1001 1110 11(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:


0.000 000 000 842(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1101 1100 1001 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1101 1100 1001 1110 11(2) × 20 =

1.1100 1110 1110 0100 1111 011(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):
1.1100 1110 1110 0100 1111 011


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. 110 0111 0111 0010 0111 1011 =

110 0111 0111 0010 0111 1011


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) =
110 0111 0111 0010 0111 1011


Decimal number -0.000 000 000 842 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0111 0111 0010 0111 1011

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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