-0.000 000 000 742 145 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 145 1(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
-0.000 000 000 742 145 1(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:

|-0.000 000 000 742 145 1| = 0.000 000 000 742 145 1


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: 0.000 000 000 742 145 1.

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) 0.000 000 000 742 145 1 × 2 = 0 + 0.000 000 001 484 290 2;
  • 2) 0.000 000 001 484 290 2 × 2 = 0 + 0.000 000 002 968 580 4;
  • 3) 0.000 000 002 968 580 4 × 2 = 0 + 0.000 000 005 937 160 8;
  • 4) 0.000 000 005 937 160 8 × 2 = 0 + 0.000 000 011 874 321 6;
  • 5) 0.000 000 011 874 321 6 × 2 = 0 + 0.000 000 023 748 643 2;
  • 6) 0.000 000 023 748 643 2 × 2 = 0 + 0.000 000 047 497 286 4;
  • 7) 0.000 000 047 497 286 4 × 2 = 0 + 0.000 000 094 994 572 8;
  • 8) 0.000 000 094 994 572 8 × 2 = 0 + 0.000 000 189 989 145 6;
  • 9) 0.000 000 189 989 145 6 × 2 = 0 + 0.000 000 379 978 291 2;
  • 10) 0.000 000 379 978 291 2 × 2 = 0 + 0.000 000 759 956 582 4;
  • 11) 0.000 000 759 956 582 4 × 2 = 0 + 0.000 001 519 913 164 8;
  • 12) 0.000 001 519 913 164 8 × 2 = 0 + 0.000 003 039 826 329 6;
  • 13) 0.000 003 039 826 329 6 × 2 = 0 + 0.000 006 079 652 659 2;
  • 14) 0.000 006 079 652 659 2 × 2 = 0 + 0.000 012 159 305 318 4;
  • 15) 0.000 012 159 305 318 4 × 2 = 0 + 0.000 024 318 610 636 8;
  • 16) 0.000 024 318 610 636 8 × 2 = 0 + 0.000 048 637 221 273 6;
  • 17) 0.000 048 637 221 273 6 × 2 = 0 + 0.000 097 274 442 547 2;
  • 18) 0.000 097 274 442 547 2 × 2 = 0 + 0.000 194 548 885 094 4;
  • 19) 0.000 194 548 885 094 4 × 2 = 0 + 0.000 389 097 770 188 8;
  • 20) 0.000 389 097 770 188 8 × 2 = 0 + 0.000 778 195 540 377 6;
  • 21) 0.000 778 195 540 377 6 × 2 = 0 + 0.001 556 391 080 755 2;
  • 22) 0.001 556 391 080 755 2 × 2 = 0 + 0.003 112 782 161 510 4;
  • 23) 0.003 112 782 161 510 4 × 2 = 0 + 0.006 225 564 323 020 8;
  • 24) 0.006 225 564 323 020 8 × 2 = 0 + 0.012 451 128 646 041 6;
  • 25) 0.012 451 128 646 041 6 × 2 = 0 + 0.024 902 257 292 083 2;
  • 26) 0.024 902 257 292 083 2 × 2 = 0 + 0.049 804 514 584 166 4;
  • 27) 0.049 804 514 584 166 4 × 2 = 0 + 0.099 609 029 168 332 8;
  • 28) 0.099 609 029 168 332 8 × 2 = 0 + 0.199 218 058 336 665 6;
  • 29) 0.199 218 058 336 665 6 × 2 = 0 + 0.398 436 116 673 331 2;
  • 30) 0.398 436 116 673 331 2 × 2 = 0 + 0.796 872 233 346 662 4;
  • 31) 0.796 872 233 346 662 4 × 2 = 1 + 0.593 744 466 693 324 8;
  • 32) 0.593 744 466 693 324 8 × 2 = 1 + 0.187 488 933 386 649 6;
  • 33) 0.187 488 933 386 649 6 × 2 = 0 + 0.374 977 866 773 299 2;
  • 34) 0.374 977 866 773 299 2 × 2 = 0 + 0.749 955 733 546 598 4;
  • 35) 0.749 955 733 546 598 4 × 2 = 1 + 0.499 911 467 093 196 8;
  • 36) 0.499 911 467 093 196 8 × 2 = 0 + 0.999 822 934 186 393 6;
  • 37) 0.999 822 934 186 393 6 × 2 = 1 + 0.999 645 868 372 787 2;
  • 38) 0.999 645 868 372 787 2 × 2 = 1 + 0.999 291 736 745 574 4;
  • 39) 0.999 291 736 745 574 4 × 2 = 1 + 0.998 583 473 491 148 8;
  • 40) 0.998 583 473 491 148 8 × 2 = 1 + 0.997 166 946 982 297 6;
  • 41) 0.997 166 946 982 297 6 × 2 = 1 + 0.994 333 893 964 595 2;
  • 42) 0.994 333 893 964 595 2 × 2 = 1 + 0.988 667 787 929 190 4;
  • 43) 0.988 667 787 929 190 4 × 2 = 1 + 0.977 335 575 858 380 8;
  • 44) 0.977 335 575 858 380 8 × 2 = 1 + 0.954 671 151 716 761 6;
  • 45) 0.954 671 151 716 761 6 × 2 = 1 + 0.909 342 303 433 523 2;
  • 46) 0.909 342 303 433 523 2 × 2 = 1 + 0.818 684 606 867 046 4;
  • 47) 0.818 684 606 867 046 4 × 2 = 1 + 0.637 369 213 734 092 8;
  • 48) 0.637 369 213 734 092 8 × 2 = 1 + 0.274 738 427 468 185 6;
  • 49) 0.274 738 427 468 185 6 × 2 = 0 + 0.549 476 854 936 371 2;
  • 50) 0.549 476 854 936 371 2 × 2 = 1 + 0.098 953 709 872 742 4;
  • 51) 0.098 953 709 872 742 4 × 2 = 0 + 0.197 907 419 745 484 8;
  • 52) 0.197 907 419 745 484 8 × 2 = 0 + 0.395 814 839 490 969 6;
  • 53) 0.395 814 839 490 969 6 × 2 = 0 + 0.791 629 678 981 939 2;
  • 54) 0.791 629 678 981 939 2 × 2 = 1 + 0.583 259 357 963 878 4;

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:


0.000 000 000 742 145 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0100 01(2)

6. Positive number before normalization:

0.000 000 000 742 145 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0100 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:


0.000 000 000 742 145 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0100 01(2) × 20 =

1.1001 0111 1111 1111 1010 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):
1.1001 0111 1111 1111 1010 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. 100 1011 1111 1111 1101 0001 =

100 1011 1111 1111 1101 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) =
100 1011 1111 1111 1101 0001


Decimal number -0.000 000 000 742 145 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1101 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