-0.000 000 000 789 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 789(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
-0.000 000 000 789(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:

|-0.000 000 000 789| = 0.000 000 000 789


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: 0.000 000 000 789.

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) 0.000 000 000 789 × 2 = 0 + 0.000 000 001 578;
  • 2) 0.000 000 001 578 × 2 = 0 + 0.000 000 003 156;
  • 3) 0.000 000 003 156 × 2 = 0 + 0.000 000 006 312;
  • 4) 0.000 000 006 312 × 2 = 0 + 0.000 000 012 624;
  • 5) 0.000 000 012 624 × 2 = 0 + 0.000 000 025 248;
  • 6) 0.000 000 025 248 × 2 = 0 + 0.000 000 050 496;
  • 7) 0.000 000 050 496 × 2 = 0 + 0.000 000 100 992;
  • 8) 0.000 000 100 992 × 2 = 0 + 0.000 000 201 984;
  • 9) 0.000 000 201 984 × 2 = 0 + 0.000 000 403 968;
  • 10) 0.000 000 403 968 × 2 = 0 + 0.000 000 807 936;
  • 11) 0.000 000 807 936 × 2 = 0 + 0.000 001 615 872;
  • 12) 0.000 001 615 872 × 2 = 0 + 0.000 003 231 744;
  • 13) 0.000 003 231 744 × 2 = 0 + 0.000 006 463 488;
  • 14) 0.000 006 463 488 × 2 = 0 + 0.000 012 926 976;
  • 15) 0.000 012 926 976 × 2 = 0 + 0.000 025 853 952;
  • 16) 0.000 025 853 952 × 2 = 0 + 0.000 051 707 904;
  • 17) 0.000 051 707 904 × 2 = 0 + 0.000 103 415 808;
  • 18) 0.000 103 415 808 × 2 = 0 + 0.000 206 831 616;
  • 19) 0.000 206 831 616 × 2 = 0 + 0.000 413 663 232;
  • 20) 0.000 413 663 232 × 2 = 0 + 0.000 827 326 464;
  • 21) 0.000 827 326 464 × 2 = 0 + 0.001 654 652 928;
  • 22) 0.001 654 652 928 × 2 = 0 + 0.003 309 305 856;
  • 23) 0.003 309 305 856 × 2 = 0 + 0.006 618 611 712;
  • 24) 0.006 618 611 712 × 2 = 0 + 0.013 237 223 424;
  • 25) 0.013 237 223 424 × 2 = 0 + 0.026 474 446 848;
  • 26) 0.026 474 446 848 × 2 = 0 + 0.052 948 893 696;
  • 27) 0.052 948 893 696 × 2 = 0 + 0.105 897 787 392;
  • 28) 0.105 897 787 392 × 2 = 0 + 0.211 795 574 784;
  • 29) 0.211 795 574 784 × 2 = 0 + 0.423 591 149 568;
  • 30) 0.423 591 149 568 × 2 = 0 + 0.847 182 299 136;
  • 31) 0.847 182 299 136 × 2 = 1 + 0.694 364 598 272;
  • 32) 0.694 364 598 272 × 2 = 1 + 0.388 729 196 544;
  • 33) 0.388 729 196 544 × 2 = 0 + 0.777 458 393 088;
  • 34) 0.777 458 393 088 × 2 = 1 + 0.554 916 786 176;
  • 35) 0.554 916 786 176 × 2 = 1 + 0.109 833 572 352;
  • 36) 0.109 833 572 352 × 2 = 0 + 0.219 667 144 704;
  • 37) 0.219 667 144 704 × 2 = 0 + 0.439 334 289 408;
  • 38) 0.439 334 289 408 × 2 = 0 + 0.878 668 578 816;
  • 39) 0.878 668 578 816 × 2 = 1 + 0.757 337 157 632;
  • 40) 0.757 337 157 632 × 2 = 1 + 0.514 674 315 264;
  • 41) 0.514 674 315 264 × 2 = 1 + 0.029 348 630 528;
  • 42) 0.029 348 630 528 × 2 = 0 + 0.058 697 261 056;
  • 43) 0.058 697 261 056 × 2 = 0 + 0.117 394 522 112;
  • 44) 0.117 394 522 112 × 2 = 0 + 0.234 789 044 224;
  • 45) 0.234 789 044 224 × 2 = 0 + 0.469 578 088 448;
  • 46) 0.469 578 088 448 × 2 = 0 + 0.939 156 176 896;
  • 47) 0.939 156 176 896 × 2 = 1 + 0.878 312 353 792;
  • 48) 0.878 312 353 792 × 2 = 1 + 0.756 624 707 584;
  • 49) 0.756 624 707 584 × 2 = 1 + 0.513 249 415 168;
  • 50) 0.513 249 415 168 × 2 = 1 + 0.026 498 830 336;
  • 51) 0.026 498 830 336 × 2 = 0 + 0.052 997 660 672;
  • 52) 0.052 997 660 672 × 2 = 0 + 0.105 995 321 344;
  • 53) 0.105 995 321 344 × 2 = 0 + 0.211 990 642 688;
  • 54) 0.211 990 642 688 × 2 = 0 + 0.423 981 285 376;

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:


0.000 000 000 789(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0011 1000 0011 1100 00(2)

6. Positive number before normalization:

0.000 000 000 789(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0011 1000 0011 1100 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:


0.000 000 000 789(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0011 1000 0011 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0011 1000 0011 1100 00(2) × 20 =

1.1011 0001 1100 0001 1110 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):
1.1011 0001 1100 0001 1110 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. 101 1000 1110 0000 1111 0000 =

101 1000 1110 0000 1111 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 0000

Mantissa (23 bits) =
101 1000 1110 0000 1111 0000


Decimal number -0.000 000 000 789 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1000 1110 0000 1111 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