-0.000 000 000 742 117 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 117 7(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
-0.000 000 000 742 117 7(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:

|-0.000 000 000 742 117 7| = 0.000 000 000 742 117 7


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: 0.000 000 000 742 117 7.

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) 0.000 000 000 742 117 7 × 2 = 0 + 0.000 000 001 484 235 4;
  • 2) 0.000 000 001 484 235 4 × 2 = 0 + 0.000 000 002 968 470 8;
  • 3) 0.000 000 002 968 470 8 × 2 = 0 + 0.000 000 005 936 941 6;
  • 4) 0.000 000 005 936 941 6 × 2 = 0 + 0.000 000 011 873 883 2;
  • 5) 0.000 000 011 873 883 2 × 2 = 0 + 0.000 000 023 747 766 4;
  • 6) 0.000 000 023 747 766 4 × 2 = 0 + 0.000 000 047 495 532 8;
  • 7) 0.000 000 047 495 532 8 × 2 = 0 + 0.000 000 094 991 065 6;
  • 8) 0.000 000 094 991 065 6 × 2 = 0 + 0.000 000 189 982 131 2;
  • 9) 0.000 000 189 982 131 2 × 2 = 0 + 0.000 000 379 964 262 4;
  • 10) 0.000 000 379 964 262 4 × 2 = 0 + 0.000 000 759 928 524 8;
  • 11) 0.000 000 759 928 524 8 × 2 = 0 + 0.000 001 519 857 049 6;
  • 12) 0.000 001 519 857 049 6 × 2 = 0 + 0.000 003 039 714 099 2;
  • 13) 0.000 003 039 714 099 2 × 2 = 0 + 0.000 006 079 428 198 4;
  • 14) 0.000 006 079 428 198 4 × 2 = 0 + 0.000 012 158 856 396 8;
  • 15) 0.000 012 158 856 396 8 × 2 = 0 + 0.000 024 317 712 793 6;
  • 16) 0.000 024 317 712 793 6 × 2 = 0 + 0.000 048 635 425 587 2;
  • 17) 0.000 048 635 425 587 2 × 2 = 0 + 0.000 097 270 851 174 4;
  • 18) 0.000 097 270 851 174 4 × 2 = 0 + 0.000 194 541 702 348 8;
  • 19) 0.000 194 541 702 348 8 × 2 = 0 + 0.000 389 083 404 697 6;
  • 20) 0.000 389 083 404 697 6 × 2 = 0 + 0.000 778 166 809 395 2;
  • 21) 0.000 778 166 809 395 2 × 2 = 0 + 0.001 556 333 618 790 4;
  • 22) 0.001 556 333 618 790 4 × 2 = 0 + 0.003 112 667 237 580 8;
  • 23) 0.003 112 667 237 580 8 × 2 = 0 + 0.006 225 334 475 161 6;
  • 24) 0.006 225 334 475 161 6 × 2 = 0 + 0.012 450 668 950 323 2;
  • 25) 0.012 450 668 950 323 2 × 2 = 0 + 0.024 901 337 900 646 4;
  • 26) 0.024 901 337 900 646 4 × 2 = 0 + 0.049 802 675 801 292 8;
  • 27) 0.049 802 675 801 292 8 × 2 = 0 + 0.099 605 351 602 585 6;
  • 28) 0.099 605 351 602 585 6 × 2 = 0 + 0.199 210 703 205 171 2;
  • 29) 0.199 210 703 205 171 2 × 2 = 0 + 0.398 421 406 410 342 4;
  • 30) 0.398 421 406 410 342 4 × 2 = 0 + 0.796 842 812 820 684 8;
  • 31) 0.796 842 812 820 684 8 × 2 = 1 + 0.593 685 625 641 369 6;
  • 32) 0.593 685 625 641 369 6 × 2 = 1 + 0.187 371 251 282 739 2;
  • 33) 0.187 371 251 282 739 2 × 2 = 0 + 0.374 742 502 565 478 4;
  • 34) 0.374 742 502 565 478 4 × 2 = 0 + 0.749 485 005 130 956 8;
  • 35) 0.749 485 005 130 956 8 × 2 = 1 + 0.498 970 010 261 913 6;
  • 36) 0.498 970 010 261 913 6 × 2 = 0 + 0.997 940 020 523 827 2;
  • 37) 0.997 940 020 523 827 2 × 2 = 1 + 0.995 880 041 047 654 4;
  • 38) 0.995 880 041 047 654 4 × 2 = 1 + 0.991 760 082 095 308 8;
  • 39) 0.991 760 082 095 308 8 × 2 = 1 + 0.983 520 164 190 617 6;
  • 40) 0.983 520 164 190 617 6 × 2 = 1 + 0.967 040 328 381 235 2;
  • 41) 0.967 040 328 381 235 2 × 2 = 1 + 0.934 080 656 762 470 4;
  • 42) 0.934 080 656 762 470 4 × 2 = 1 + 0.868 161 313 524 940 8;
  • 43) 0.868 161 313 524 940 8 × 2 = 1 + 0.736 322 627 049 881 6;
  • 44) 0.736 322 627 049 881 6 × 2 = 1 + 0.472 645 254 099 763 2;
  • 45) 0.472 645 254 099 763 2 × 2 = 0 + 0.945 290 508 199 526 4;
  • 46) 0.945 290 508 199 526 4 × 2 = 1 + 0.890 581 016 399 052 8;
  • 47) 0.890 581 016 399 052 8 × 2 = 1 + 0.781 162 032 798 105 6;
  • 48) 0.781 162 032 798 105 6 × 2 = 1 + 0.562 324 065 596 211 2;
  • 49) 0.562 324 065 596 211 2 × 2 = 1 + 0.124 648 131 192 422 4;
  • 50) 0.124 648 131 192 422 4 × 2 = 0 + 0.249 296 262 384 844 8;
  • 51) 0.249 296 262 384 844 8 × 2 = 0 + 0.498 592 524 769 689 6;
  • 52) 0.498 592 524 769 689 6 × 2 = 0 + 0.997 185 049 539 379 2;
  • 53) 0.997 185 049 539 379 2 × 2 = 1 + 0.994 370 099 078 758 4;
  • 54) 0.994 370 099 078 758 4 × 2 = 1 + 0.988 740 198 157 516 8;

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:


0.000 000 000 742 117 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 1000 11(2)

6. Positive number before normalization:

0.000 000 000 742 117 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 1000 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:


0.000 000 000 742 117 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 1000 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 1000 11(2) × 20 =

1.1001 0111 1111 1011 1100 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):
1.1001 0111 1111 1011 1100 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. 100 1011 1111 1101 1110 0011 =

100 1011 1111 1101 1110 0011


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 1101 1110 0011


Decimal number -0.000 000 000 742 117 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1101 1110 0011

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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