-0.000 000 000 898 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 898(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
-0.000 000 000 898(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:

|-0.000 000 000 898| = 0.000 000 000 898


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: 0.000 000 000 898.

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) 0.000 000 000 898 × 2 = 0 + 0.000 000 001 796;
  • 2) 0.000 000 001 796 × 2 = 0 + 0.000 000 003 592;
  • 3) 0.000 000 003 592 × 2 = 0 + 0.000 000 007 184;
  • 4) 0.000 000 007 184 × 2 = 0 + 0.000 000 014 368;
  • 5) 0.000 000 014 368 × 2 = 0 + 0.000 000 028 736;
  • 6) 0.000 000 028 736 × 2 = 0 + 0.000 000 057 472;
  • 7) 0.000 000 057 472 × 2 = 0 + 0.000 000 114 944;
  • 8) 0.000 000 114 944 × 2 = 0 + 0.000 000 229 888;
  • 9) 0.000 000 229 888 × 2 = 0 + 0.000 000 459 776;
  • 10) 0.000 000 459 776 × 2 = 0 + 0.000 000 919 552;
  • 11) 0.000 000 919 552 × 2 = 0 + 0.000 001 839 104;
  • 12) 0.000 001 839 104 × 2 = 0 + 0.000 003 678 208;
  • 13) 0.000 003 678 208 × 2 = 0 + 0.000 007 356 416;
  • 14) 0.000 007 356 416 × 2 = 0 + 0.000 014 712 832;
  • 15) 0.000 014 712 832 × 2 = 0 + 0.000 029 425 664;
  • 16) 0.000 029 425 664 × 2 = 0 + 0.000 058 851 328;
  • 17) 0.000 058 851 328 × 2 = 0 + 0.000 117 702 656;
  • 18) 0.000 117 702 656 × 2 = 0 + 0.000 235 405 312;
  • 19) 0.000 235 405 312 × 2 = 0 + 0.000 470 810 624;
  • 20) 0.000 470 810 624 × 2 = 0 + 0.000 941 621 248;
  • 21) 0.000 941 621 248 × 2 = 0 + 0.001 883 242 496;
  • 22) 0.001 883 242 496 × 2 = 0 + 0.003 766 484 992;
  • 23) 0.003 766 484 992 × 2 = 0 + 0.007 532 969 984;
  • 24) 0.007 532 969 984 × 2 = 0 + 0.015 065 939 968;
  • 25) 0.015 065 939 968 × 2 = 0 + 0.030 131 879 936;
  • 26) 0.030 131 879 936 × 2 = 0 + 0.060 263 759 872;
  • 27) 0.060 263 759 872 × 2 = 0 + 0.120 527 519 744;
  • 28) 0.120 527 519 744 × 2 = 0 + 0.241 055 039 488;
  • 29) 0.241 055 039 488 × 2 = 0 + 0.482 110 078 976;
  • 30) 0.482 110 078 976 × 2 = 0 + 0.964 220 157 952;
  • 31) 0.964 220 157 952 × 2 = 1 + 0.928 440 315 904;
  • 32) 0.928 440 315 904 × 2 = 1 + 0.856 880 631 808;
  • 33) 0.856 880 631 808 × 2 = 1 + 0.713 761 263 616;
  • 34) 0.713 761 263 616 × 2 = 1 + 0.427 522 527 232;
  • 35) 0.427 522 527 232 × 2 = 0 + 0.855 045 054 464;
  • 36) 0.855 045 054 464 × 2 = 1 + 0.710 090 108 928;
  • 37) 0.710 090 108 928 × 2 = 1 + 0.420 180 217 856;
  • 38) 0.420 180 217 856 × 2 = 0 + 0.840 360 435 712;
  • 39) 0.840 360 435 712 × 2 = 1 + 0.680 720 871 424;
  • 40) 0.680 720 871 424 × 2 = 1 + 0.361 441 742 848;
  • 41) 0.361 441 742 848 × 2 = 0 + 0.722 883 485 696;
  • 42) 0.722 883 485 696 × 2 = 1 + 0.445 766 971 392;
  • 43) 0.445 766 971 392 × 2 = 0 + 0.891 533 942 784;
  • 44) 0.891 533 942 784 × 2 = 1 + 0.783 067 885 568;
  • 45) 0.783 067 885 568 × 2 = 1 + 0.566 135 771 136;
  • 46) 0.566 135 771 136 × 2 = 1 + 0.132 271 542 272;
  • 47) 0.132 271 542 272 × 2 = 0 + 0.264 543 084 544;
  • 48) 0.264 543 084 544 × 2 = 0 + 0.529 086 169 088;
  • 49) 0.529 086 169 088 × 2 = 1 + 0.058 172 338 176;
  • 50) 0.058 172 338 176 × 2 = 0 + 0.116 344 676 352;
  • 51) 0.116 344 676 352 × 2 = 0 + 0.232 689 352 704;
  • 52) 0.232 689 352 704 × 2 = 0 + 0.465 378 705 408;
  • 53) 0.465 378 705 408 × 2 = 0 + 0.930 757 410 816;
  • 54) 0.930 757 410 816 × 2 = 1 + 0.861 514 821 632;

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:


0.000 000 000 898(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 1011 0101 1100 1000 01(2)

6. Positive number before normalization:

0.000 000 000 898(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 1011 0101 1100 1000 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:


0.000 000 000 898(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 1011 0101 1100 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 1011 0101 1100 1000 01(2) × 20 =

1.1110 1101 1010 1110 0100 001(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):
1.1110 1101 1010 1110 0100 001


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. 111 0110 1101 0111 0010 0001 =

111 0110 1101 0111 0010 0001


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) =
111 0110 1101 0111 0010 0001


Decimal number -0.000 000 000 898 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0110 1101 0111 0010 0001

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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