-0.000 000 000 802 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 802(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
-0.000 000 000 802(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:

|-0.000 000 000 802| = 0.000 000 000 802


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: 0.000 000 000 802.

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) 0.000 000 000 802 × 2 = 0 + 0.000 000 001 604;
  • 2) 0.000 000 001 604 × 2 = 0 + 0.000 000 003 208;
  • 3) 0.000 000 003 208 × 2 = 0 + 0.000 000 006 416;
  • 4) 0.000 000 006 416 × 2 = 0 + 0.000 000 012 832;
  • 5) 0.000 000 012 832 × 2 = 0 + 0.000 000 025 664;
  • 6) 0.000 000 025 664 × 2 = 0 + 0.000 000 051 328;
  • 7) 0.000 000 051 328 × 2 = 0 + 0.000 000 102 656;
  • 8) 0.000 000 102 656 × 2 = 0 + 0.000 000 205 312;
  • 9) 0.000 000 205 312 × 2 = 0 + 0.000 000 410 624;
  • 10) 0.000 000 410 624 × 2 = 0 + 0.000 000 821 248;
  • 11) 0.000 000 821 248 × 2 = 0 + 0.000 001 642 496;
  • 12) 0.000 001 642 496 × 2 = 0 + 0.000 003 284 992;
  • 13) 0.000 003 284 992 × 2 = 0 + 0.000 006 569 984;
  • 14) 0.000 006 569 984 × 2 = 0 + 0.000 013 139 968;
  • 15) 0.000 013 139 968 × 2 = 0 + 0.000 026 279 936;
  • 16) 0.000 026 279 936 × 2 = 0 + 0.000 052 559 872;
  • 17) 0.000 052 559 872 × 2 = 0 + 0.000 105 119 744;
  • 18) 0.000 105 119 744 × 2 = 0 + 0.000 210 239 488;
  • 19) 0.000 210 239 488 × 2 = 0 + 0.000 420 478 976;
  • 20) 0.000 420 478 976 × 2 = 0 + 0.000 840 957 952;
  • 21) 0.000 840 957 952 × 2 = 0 + 0.001 681 915 904;
  • 22) 0.001 681 915 904 × 2 = 0 + 0.003 363 831 808;
  • 23) 0.003 363 831 808 × 2 = 0 + 0.006 727 663 616;
  • 24) 0.006 727 663 616 × 2 = 0 + 0.013 455 327 232;
  • 25) 0.013 455 327 232 × 2 = 0 + 0.026 910 654 464;
  • 26) 0.026 910 654 464 × 2 = 0 + 0.053 821 308 928;
  • 27) 0.053 821 308 928 × 2 = 0 + 0.107 642 617 856;
  • 28) 0.107 642 617 856 × 2 = 0 + 0.215 285 235 712;
  • 29) 0.215 285 235 712 × 2 = 0 + 0.430 570 471 424;
  • 30) 0.430 570 471 424 × 2 = 0 + 0.861 140 942 848;
  • 31) 0.861 140 942 848 × 2 = 1 + 0.722 281 885 696;
  • 32) 0.722 281 885 696 × 2 = 1 + 0.444 563 771 392;
  • 33) 0.444 563 771 392 × 2 = 0 + 0.889 127 542 784;
  • 34) 0.889 127 542 784 × 2 = 1 + 0.778 255 085 568;
  • 35) 0.778 255 085 568 × 2 = 1 + 0.556 510 171 136;
  • 36) 0.556 510 171 136 × 2 = 1 + 0.113 020 342 272;
  • 37) 0.113 020 342 272 × 2 = 0 + 0.226 040 684 544;
  • 38) 0.226 040 684 544 × 2 = 0 + 0.452 081 369 088;
  • 39) 0.452 081 369 088 × 2 = 0 + 0.904 162 738 176;
  • 40) 0.904 162 738 176 × 2 = 1 + 0.808 325 476 352;
  • 41) 0.808 325 476 352 × 2 = 1 + 0.616 650 952 704;
  • 42) 0.616 650 952 704 × 2 = 1 + 0.233 301 905 408;
  • 43) 0.233 301 905 408 × 2 = 0 + 0.466 603 810 816;
  • 44) 0.466 603 810 816 × 2 = 0 + 0.933 207 621 632;
  • 45) 0.933 207 621 632 × 2 = 1 + 0.866 415 243 264;
  • 46) 0.866 415 243 264 × 2 = 1 + 0.732 830 486 528;
  • 47) 0.732 830 486 528 × 2 = 1 + 0.465 660 973 056;
  • 48) 0.465 660 973 056 × 2 = 0 + 0.931 321 946 112;
  • 49) 0.931 321 946 112 × 2 = 1 + 0.862 643 892 224;
  • 50) 0.862 643 892 224 × 2 = 1 + 0.725 287 784 448;
  • 51) 0.725 287 784 448 × 2 = 1 + 0.450 575 568 896;
  • 52) 0.450 575 568 896 × 2 = 0 + 0.901 151 137 792;
  • 53) 0.901 151 137 792 × 2 = 1 + 0.802 302 275 584;
  • 54) 0.802 302 275 584 × 2 = 1 + 0.604 604 551 168;

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:


0.000 000 000 802(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0001 1100 1110 1110 11(2)

6. Positive number before normalization:

0.000 000 000 802(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0001 1100 1110 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:


0.000 000 000 802(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0001 1100 1110 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 0001 1100 1110 1110 11(2) × 20 =

1.1011 1000 1110 0111 0111 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):
1.1011 1000 1110 0111 0111 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. 101 1100 0111 0011 1011 1011 =

101 1100 0111 0011 1011 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) =
101 1100 0111 0011 1011 1011


Decimal number -0.000 000 000 802 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1100 0111 0011 1011 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