-0.000 000 000 742 147 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 54(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
-0.000 000 000 742 147 54(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:

|-0.000 000 000 742 147 54| = 0.000 000 000 742 147 54


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: 0.000 000 000 742 147 54.

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) 0.000 000 000 742 147 54 × 2 = 0 + 0.000 000 001 484 295 08;
  • 2) 0.000 000 001 484 295 08 × 2 = 0 + 0.000 000 002 968 590 16;
  • 3) 0.000 000 002 968 590 16 × 2 = 0 + 0.000 000 005 937 180 32;
  • 4) 0.000 000 005 937 180 32 × 2 = 0 + 0.000 000 011 874 360 64;
  • 5) 0.000 000 011 874 360 64 × 2 = 0 + 0.000 000 023 748 721 28;
  • 6) 0.000 000 023 748 721 28 × 2 = 0 + 0.000 000 047 497 442 56;
  • 7) 0.000 000 047 497 442 56 × 2 = 0 + 0.000 000 094 994 885 12;
  • 8) 0.000 000 094 994 885 12 × 2 = 0 + 0.000 000 189 989 770 24;
  • 9) 0.000 000 189 989 770 24 × 2 = 0 + 0.000 000 379 979 540 48;
  • 10) 0.000 000 379 979 540 48 × 2 = 0 + 0.000 000 759 959 080 96;
  • 11) 0.000 000 759 959 080 96 × 2 = 0 + 0.000 001 519 918 161 92;
  • 12) 0.000 001 519 918 161 92 × 2 = 0 + 0.000 003 039 836 323 84;
  • 13) 0.000 003 039 836 323 84 × 2 = 0 + 0.000 006 079 672 647 68;
  • 14) 0.000 006 079 672 647 68 × 2 = 0 + 0.000 012 159 345 295 36;
  • 15) 0.000 012 159 345 295 36 × 2 = 0 + 0.000 024 318 690 590 72;
  • 16) 0.000 024 318 690 590 72 × 2 = 0 + 0.000 048 637 381 181 44;
  • 17) 0.000 048 637 381 181 44 × 2 = 0 + 0.000 097 274 762 362 88;
  • 18) 0.000 097 274 762 362 88 × 2 = 0 + 0.000 194 549 524 725 76;
  • 19) 0.000 194 549 524 725 76 × 2 = 0 + 0.000 389 099 049 451 52;
  • 20) 0.000 389 099 049 451 52 × 2 = 0 + 0.000 778 198 098 903 04;
  • 21) 0.000 778 198 098 903 04 × 2 = 0 + 0.001 556 396 197 806 08;
  • 22) 0.001 556 396 197 806 08 × 2 = 0 + 0.003 112 792 395 612 16;
  • 23) 0.003 112 792 395 612 16 × 2 = 0 + 0.006 225 584 791 224 32;
  • 24) 0.006 225 584 791 224 32 × 2 = 0 + 0.012 451 169 582 448 64;
  • 25) 0.012 451 169 582 448 64 × 2 = 0 + 0.024 902 339 164 897 28;
  • 26) 0.024 902 339 164 897 28 × 2 = 0 + 0.049 804 678 329 794 56;
  • 27) 0.049 804 678 329 794 56 × 2 = 0 + 0.099 609 356 659 589 12;
  • 28) 0.099 609 356 659 589 12 × 2 = 0 + 0.199 218 713 319 178 24;
  • 29) 0.199 218 713 319 178 24 × 2 = 0 + 0.398 437 426 638 356 48;
  • 30) 0.398 437 426 638 356 48 × 2 = 0 + 0.796 874 853 276 712 96;
  • 31) 0.796 874 853 276 712 96 × 2 = 1 + 0.593 749 706 553 425 92;
  • 32) 0.593 749 706 553 425 92 × 2 = 1 + 0.187 499 413 106 851 84;
  • 33) 0.187 499 413 106 851 84 × 2 = 0 + 0.374 998 826 213 703 68;
  • 34) 0.374 998 826 213 703 68 × 2 = 0 + 0.749 997 652 427 407 36;
  • 35) 0.749 997 652 427 407 36 × 2 = 1 + 0.499 995 304 854 814 72;
  • 36) 0.499 995 304 854 814 72 × 2 = 0 + 0.999 990 609 709 629 44;
  • 37) 0.999 990 609 709 629 44 × 2 = 1 + 0.999 981 219 419 258 88;
  • 38) 0.999 981 219 419 258 88 × 2 = 1 + 0.999 962 438 838 517 76;
  • 39) 0.999 962 438 838 517 76 × 2 = 1 + 0.999 924 877 677 035 52;
  • 40) 0.999 924 877 677 035 52 × 2 = 1 + 0.999 849 755 354 071 04;
  • 41) 0.999 849 755 354 071 04 × 2 = 1 + 0.999 699 510 708 142 08;
  • 42) 0.999 699 510 708 142 08 × 2 = 1 + 0.999 399 021 416 284 16;
  • 43) 0.999 399 021 416 284 16 × 2 = 1 + 0.998 798 042 832 568 32;
  • 44) 0.998 798 042 832 568 32 × 2 = 1 + 0.997 596 085 665 136 64;
  • 45) 0.997 596 085 665 136 64 × 2 = 1 + 0.995 192 171 330 273 28;
  • 46) 0.995 192 171 330 273 28 × 2 = 1 + 0.990 384 342 660 546 56;
  • 47) 0.990 384 342 660 546 56 × 2 = 1 + 0.980 768 685 321 093 12;
  • 48) 0.980 768 685 321 093 12 × 2 = 1 + 0.961 537 370 642 186 24;
  • 49) 0.961 537 370 642 186 24 × 2 = 1 + 0.923 074 741 284 372 48;
  • 50) 0.923 074 741 284 372 48 × 2 = 1 + 0.846 149 482 568 744 96;
  • 51) 0.846 149 482 568 744 96 × 2 = 1 + 0.692 298 965 137 489 92;
  • 52) 0.692 298 965 137 489 92 × 2 = 1 + 0.384 597 930 274 979 84;
  • 53) 0.384 597 930 274 979 84 × 2 = 0 + 0.769 195 860 549 959 68;
  • 54) 0.769 195 860 549 959 68 × 2 = 1 + 0.538 391 721 099 919 36;

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:


0.000 000 000 742 147 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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):
1.1001 0111 1111 1111 1111 101


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. 100 1011 1111 1111 1111 1101 =

100 1011 1111 1111 1111 1101


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) =
100 1011 1111 1111 1111 1101


Decimal number -0.000 000 000 742 147 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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