-0.000 000 000 684 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 684(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
-0.000 000 000 684(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:

|-0.000 000 000 684| = 0.000 000 000 684


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: 0.000 000 000 684.

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) 0.000 000 000 684 × 2 = 0 + 0.000 000 001 368;
  • 2) 0.000 000 001 368 × 2 = 0 + 0.000 000 002 736;
  • 3) 0.000 000 002 736 × 2 = 0 + 0.000 000 005 472;
  • 4) 0.000 000 005 472 × 2 = 0 + 0.000 000 010 944;
  • 5) 0.000 000 010 944 × 2 = 0 + 0.000 000 021 888;
  • 6) 0.000 000 021 888 × 2 = 0 + 0.000 000 043 776;
  • 7) 0.000 000 043 776 × 2 = 0 + 0.000 000 087 552;
  • 8) 0.000 000 087 552 × 2 = 0 + 0.000 000 175 104;
  • 9) 0.000 000 175 104 × 2 = 0 + 0.000 000 350 208;
  • 10) 0.000 000 350 208 × 2 = 0 + 0.000 000 700 416;
  • 11) 0.000 000 700 416 × 2 = 0 + 0.000 001 400 832;
  • 12) 0.000 001 400 832 × 2 = 0 + 0.000 002 801 664;
  • 13) 0.000 002 801 664 × 2 = 0 + 0.000 005 603 328;
  • 14) 0.000 005 603 328 × 2 = 0 + 0.000 011 206 656;
  • 15) 0.000 011 206 656 × 2 = 0 + 0.000 022 413 312;
  • 16) 0.000 022 413 312 × 2 = 0 + 0.000 044 826 624;
  • 17) 0.000 044 826 624 × 2 = 0 + 0.000 089 653 248;
  • 18) 0.000 089 653 248 × 2 = 0 + 0.000 179 306 496;
  • 19) 0.000 179 306 496 × 2 = 0 + 0.000 358 612 992;
  • 20) 0.000 358 612 992 × 2 = 0 + 0.000 717 225 984;
  • 21) 0.000 717 225 984 × 2 = 0 + 0.001 434 451 968;
  • 22) 0.001 434 451 968 × 2 = 0 + 0.002 868 903 936;
  • 23) 0.002 868 903 936 × 2 = 0 + 0.005 737 807 872;
  • 24) 0.005 737 807 872 × 2 = 0 + 0.011 475 615 744;
  • 25) 0.011 475 615 744 × 2 = 0 + 0.022 951 231 488;
  • 26) 0.022 951 231 488 × 2 = 0 + 0.045 902 462 976;
  • 27) 0.045 902 462 976 × 2 = 0 + 0.091 804 925 952;
  • 28) 0.091 804 925 952 × 2 = 0 + 0.183 609 851 904;
  • 29) 0.183 609 851 904 × 2 = 0 + 0.367 219 703 808;
  • 30) 0.367 219 703 808 × 2 = 0 + 0.734 439 407 616;
  • 31) 0.734 439 407 616 × 2 = 1 + 0.468 878 815 232;
  • 32) 0.468 878 815 232 × 2 = 0 + 0.937 757 630 464;
  • 33) 0.937 757 630 464 × 2 = 1 + 0.875 515 260 928;
  • 34) 0.875 515 260 928 × 2 = 1 + 0.751 030 521 856;
  • 35) 0.751 030 521 856 × 2 = 1 + 0.502 061 043 712;
  • 36) 0.502 061 043 712 × 2 = 1 + 0.004 122 087 424;
  • 37) 0.004 122 087 424 × 2 = 0 + 0.008 244 174 848;
  • 38) 0.008 244 174 848 × 2 = 0 + 0.016 488 349 696;
  • 39) 0.016 488 349 696 × 2 = 0 + 0.032 976 699 392;
  • 40) 0.032 976 699 392 × 2 = 0 + 0.065 953 398 784;
  • 41) 0.065 953 398 784 × 2 = 0 + 0.131 906 797 568;
  • 42) 0.131 906 797 568 × 2 = 0 + 0.263 813 595 136;
  • 43) 0.263 813 595 136 × 2 = 0 + 0.527 627 190 272;
  • 44) 0.527 627 190 272 × 2 = 1 + 0.055 254 380 544;
  • 45) 0.055 254 380 544 × 2 = 0 + 0.110 508 761 088;
  • 46) 0.110 508 761 088 × 2 = 0 + 0.221 017 522 176;
  • 47) 0.221 017 522 176 × 2 = 0 + 0.442 035 044 352;
  • 48) 0.442 035 044 352 × 2 = 0 + 0.884 070 088 704;
  • 49) 0.884 070 088 704 × 2 = 1 + 0.768 140 177 408;
  • 50) 0.768 140 177 408 × 2 = 1 + 0.536 280 354 816;
  • 51) 0.536 280 354 816 × 2 = 1 + 0.072 560 709 632;
  • 52) 0.072 560 709 632 × 2 = 0 + 0.145 121 419 264;
  • 53) 0.145 121 419 264 × 2 = 0 + 0.290 242 838 528;
  • 54) 0.290 242 838 528 × 2 = 0 + 0.580 485 677 056;

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:


0.000 000 000 684(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0000 0001 0000 1110 00(2)

6. Positive number before normalization:

0.000 000 000 684(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0000 0001 0000 1110 00(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:


0.000 000 000 684(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0000 0001 0000 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 0000 0001 0000 1110 00(2) × 20 =

1.0111 1000 0000 1000 0111 000(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):
1.0111 1000 0000 1000 0111 000


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. 011 1100 0000 0100 0011 1000 =

011 1100 0000 0100 0011 1000


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) =
011 1100 0000 0100 0011 1000


Decimal number -0.000 000 000 684 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 1100 0000 0100 0011 1000

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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