-0.000 000 000 942 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 942(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
-0.000 000 000 942(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:

|-0.000 000 000 942| = 0.000 000 000 942


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: 0.000 000 000 942.

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) 0.000 000 000 942 × 2 = 0 + 0.000 000 001 884;
  • 2) 0.000 000 001 884 × 2 = 0 + 0.000 000 003 768;
  • 3) 0.000 000 003 768 × 2 = 0 + 0.000 000 007 536;
  • 4) 0.000 000 007 536 × 2 = 0 + 0.000 000 015 072;
  • 5) 0.000 000 015 072 × 2 = 0 + 0.000 000 030 144;
  • 6) 0.000 000 030 144 × 2 = 0 + 0.000 000 060 288;
  • 7) 0.000 000 060 288 × 2 = 0 + 0.000 000 120 576;
  • 8) 0.000 000 120 576 × 2 = 0 + 0.000 000 241 152;
  • 9) 0.000 000 241 152 × 2 = 0 + 0.000 000 482 304;
  • 10) 0.000 000 482 304 × 2 = 0 + 0.000 000 964 608;
  • 11) 0.000 000 964 608 × 2 = 0 + 0.000 001 929 216;
  • 12) 0.000 001 929 216 × 2 = 0 + 0.000 003 858 432;
  • 13) 0.000 003 858 432 × 2 = 0 + 0.000 007 716 864;
  • 14) 0.000 007 716 864 × 2 = 0 + 0.000 015 433 728;
  • 15) 0.000 015 433 728 × 2 = 0 + 0.000 030 867 456;
  • 16) 0.000 030 867 456 × 2 = 0 + 0.000 061 734 912;
  • 17) 0.000 061 734 912 × 2 = 0 + 0.000 123 469 824;
  • 18) 0.000 123 469 824 × 2 = 0 + 0.000 246 939 648;
  • 19) 0.000 246 939 648 × 2 = 0 + 0.000 493 879 296;
  • 20) 0.000 493 879 296 × 2 = 0 + 0.000 987 758 592;
  • 21) 0.000 987 758 592 × 2 = 0 + 0.001 975 517 184;
  • 22) 0.001 975 517 184 × 2 = 0 + 0.003 951 034 368;
  • 23) 0.003 951 034 368 × 2 = 0 + 0.007 902 068 736;
  • 24) 0.007 902 068 736 × 2 = 0 + 0.015 804 137 472;
  • 25) 0.015 804 137 472 × 2 = 0 + 0.031 608 274 944;
  • 26) 0.031 608 274 944 × 2 = 0 + 0.063 216 549 888;
  • 27) 0.063 216 549 888 × 2 = 0 + 0.126 433 099 776;
  • 28) 0.126 433 099 776 × 2 = 0 + 0.252 866 199 552;
  • 29) 0.252 866 199 552 × 2 = 0 + 0.505 732 399 104;
  • 30) 0.505 732 399 104 × 2 = 1 + 0.011 464 798 208;
  • 31) 0.011 464 798 208 × 2 = 0 + 0.022 929 596 416;
  • 32) 0.022 929 596 416 × 2 = 0 + 0.045 859 192 832;
  • 33) 0.045 859 192 832 × 2 = 0 + 0.091 718 385 664;
  • 34) 0.091 718 385 664 × 2 = 0 + 0.183 436 771 328;
  • 35) 0.183 436 771 328 × 2 = 0 + 0.366 873 542 656;
  • 36) 0.366 873 542 656 × 2 = 0 + 0.733 747 085 312;
  • 37) 0.733 747 085 312 × 2 = 1 + 0.467 494 170 624;
  • 38) 0.467 494 170 624 × 2 = 0 + 0.934 988 341 248;
  • 39) 0.934 988 341 248 × 2 = 1 + 0.869 976 682 496;
  • 40) 0.869 976 682 496 × 2 = 1 + 0.739 953 364 992;
  • 41) 0.739 953 364 992 × 2 = 1 + 0.479 906 729 984;
  • 42) 0.479 906 729 984 × 2 = 0 + 0.959 813 459 968;
  • 43) 0.959 813 459 968 × 2 = 1 + 0.919 626 919 936;
  • 44) 0.919 626 919 936 × 2 = 1 + 0.839 253 839 872;
  • 45) 0.839 253 839 872 × 2 = 1 + 0.678 507 679 744;
  • 46) 0.678 507 679 744 × 2 = 1 + 0.357 015 359 488;
  • 47) 0.357 015 359 488 × 2 = 0 + 0.714 030 718 976;
  • 48) 0.714 030 718 976 × 2 = 1 + 0.428 061 437 952;
  • 49) 0.428 061 437 952 × 2 = 0 + 0.856 122 875 904;
  • 50) 0.856 122 875 904 × 2 = 1 + 0.712 245 751 808;
  • 51) 0.712 245 751 808 × 2 = 1 + 0.424 491 503 616;
  • 52) 0.424 491 503 616 × 2 = 0 + 0.848 983 007 232;
  • 53) 0.848 983 007 232 × 2 = 1 + 0.697 966 014 464;

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:


0.000 000 000 942(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 0110 1(2)

6. Positive number before normalization:

0.000 000 000 942(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 0110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 942(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 0110 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 0110 1(2) × 20 =

1.0000 0010 1110 1111 0101 101(2) × 2-30


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): -30

Mantissa (not normalized):
1.0000 0010 1110 1111 0101 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 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) =

97(10) =

0110 0001(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 0001 0111 0111 1010 1101 =

000 0001 0111 0111 1010 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 0001

Mantissa (23 bits) =
000 0001 0111 0111 1010 1101


Decimal number -0.000 000 000 942 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 000 0001 0111 0111 1010 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