-0.000 000 000 939 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 939(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
-0.000 000 000 939(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:

|-0.000 000 000 939| = 0.000 000 000 939


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: 0.000 000 000 939.

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) 0.000 000 000 939 × 2 = 0 + 0.000 000 001 878;
  • 2) 0.000 000 001 878 × 2 = 0 + 0.000 000 003 756;
  • 3) 0.000 000 003 756 × 2 = 0 + 0.000 000 007 512;
  • 4) 0.000 000 007 512 × 2 = 0 + 0.000 000 015 024;
  • 5) 0.000 000 015 024 × 2 = 0 + 0.000 000 030 048;
  • 6) 0.000 000 030 048 × 2 = 0 + 0.000 000 060 096;
  • 7) 0.000 000 060 096 × 2 = 0 + 0.000 000 120 192;
  • 8) 0.000 000 120 192 × 2 = 0 + 0.000 000 240 384;
  • 9) 0.000 000 240 384 × 2 = 0 + 0.000 000 480 768;
  • 10) 0.000 000 480 768 × 2 = 0 + 0.000 000 961 536;
  • 11) 0.000 000 961 536 × 2 = 0 + 0.000 001 923 072;
  • 12) 0.000 001 923 072 × 2 = 0 + 0.000 003 846 144;
  • 13) 0.000 003 846 144 × 2 = 0 + 0.000 007 692 288;
  • 14) 0.000 007 692 288 × 2 = 0 + 0.000 015 384 576;
  • 15) 0.000 015 384 576 × 2 = 0 + 0.000 030 769 152;
  • 16) 0.000 030 769 152 × 2 = 0 + 0.000 061 538 304;
  • 17) 0.000 061 538 304 × 2 = 0 + 0.000 123 076 608;
  • 18) 0.000 123 076 608 × 2 = 0 + 0.000 246 153 216;
  • 19) 0.000 246 153 216 × 2 = 0 + 0.000 492 306 432;
  • 20) 0.000 492 306 432 × 2 = 0 + 0.000 984 612 864;
  • 21) 0.000 984 612 864 × 2 = 0 + 0.001 969 225 728;
  • 22) 0.001 969 225 728 × 2 = 0 + 0.003 938 451 456;
  • 23) 0.003 938 451 456 × 2 = 0 + 0.007 876 902 912;
  • 24) 0.007 876 902 912 × 2 = 0 + 0.015 753 805 824;
  • 25) 0.015 753 805 824 × 2 = 0 + 0.031 507 611 648;
  • 26) 0.031 507 611 648 × 2 = 0 + 0.063 015 223 296;
  • 27) 0.063 015 223 296 × 2 = 0 + 0.126 030 446 592;
  • 28) 0.126 030 446 592 × 2 = 0 + 0.252 060 893 184;
  • 29) 0.252 060 893 184 × 2 = 0 + 0.504 121 786 368;
  • 30) 0.504 121 786 368 × 2 = 1 + 0.008 243 572 736;
  • 31) 0.008 243 572 736 × 2 = 0 + 0.016 487 145 472;
  • 32) 0.016 487 145 472 × 2 = 0 + 0.032 974 290 944;
  • 33) 0.032 974 290 944 × 2 = 0 + 0.065 948 581 888;
  • 34) 0.065 948 581 888 × 2 = 0 + 0.131 897 163 776;
  • 35) 0.131 897 163 776 × 2 = 0 + 0.263 794 327 552;
  • 36) 0.263 794 327 552 × 2 = 0 + 0.527 588 655 104;
  • 37) 0.527 588 655 104 × 2 = 1 + 0.055 177 310 208;
  • 38) 0.055 177 310 208 × 2 = 0 + 0.110 354 620 416;
  • 39) 0.110 354 620 416 × 2 = 0 + 0.220 709 240 832;
  • 40) 0.220 709 240 832 × 2 = 0 + 0.441 418 481 664;
  • 41) 0.441 418 481 664 × 2 = 0 + 0.882 836 963 328;
  • 42) 0.882 836 963 328 × 2 = 1 + 0.765 673 926 656;
  • 43) 0.765 673 926 656 × 2 = 1 + 0.531 347 853 312;
  • 44) 0.531 347 853 312 × 2 = 1 + 0.062 695 706 624;
  • 45) 0.062 695 706 624 × 2 = 0 + 0.125 391 413 248;
  • 46) 0.125 391 413 248 × 2 = 0 + 0.250 782 826 496;
  • 47) 0.250 782 826 496 × 2 = 0 + 0.501 565 652 992;
  • 48) 0.501 565 652 992 × 2 = 1 + 0.003 131 305 984;
  • 49) 0.003 131 305 984 × 2 = 0 + 0.006 262 611 968;
  • 50) 0.006 262 611 968 × 2 = 0 + 0.012 525 223 936;
  • 51) 0.012 525 223 936 × 2 = 0 + 0.025 050 447 872;
  • 52) 0.025 050 447 872 × 2 = 0 + 0.050 100 895 744;
  • 53) 0.050 100 895 744 × 2 = 0 + 0.100 201 791 488;

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:


0.000 000 000 939(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1000 0111 0001 0000 0(2)

6. Positive number before normalization:

0.000 000 000 939(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1000 0111 0001 0000 0(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:


0.000 000 000 939(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1000 0111 0001 0000 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0000 1000 0111 0001 0000 0(2) × 20 =

1.0000 0010 0001 1100 0100 000(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 0001 1100 0100 000


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 0000 1110 0010 0000 =

000 0001 0000 1110 0010 0000


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 0000 1110 0010 0000


Decimal number -0.000 000 000 939 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 000 0001 0000 1110 0010 0000

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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