-0.000 000 000 742 129 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 129 2(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
-0.000 000 000 742 129 2(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:

|-0.000 000 000 742 129 2| = 0.000 000 000 742 129 2


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: 0.000 000 000 742 129 2.

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) 0.000 000 000 742 129 2 × 2 = 0 + 0.000 000 001 484 258 4;
  • 2) 0.000 000 001 484 258 4 × 2 = 0 + 0.000 000 002 968 516 8;
  • 3) 0.000 000 002 968 516 8 × 2 = 0 + 0.000 000 005 937 033 6;
  • 4) 0.000 000 005 937 033 6 × 2 = 0 + 0.000 000 011 874 067 2;
  • 5) 0.000 000 011 874 067 2 × 2 = 0 + 0.000 000 023 748 134 4;
  • 6) 0.000 000 023 748 134 4 × 2 = 0 + 0.000 000 047 496 268 8;
  • 7) 0.000 000 047 496 268 8 × 2 = 0 + 0.000 000 094 992 537 6;
  • 8) 0.000 000 094 992 537 6 × 2 = 0 + 0.000 000 189 985 075 2;
  • 9) 0.000 000 189 985 075 2 × 2 = 0 + 0.000 000 379 970 150 4;
  • 10) 0.000 000 379 970 150 4 × 2 = 0 + 0.000 000 759 940 300 8;
  • 11) 0.000 000 759 940 300 8 × 2 = 0 + 0.000 001 519 880 601 6;
  • 12) 0.000 001 519 880 601 6 × 2 = 0 + 0.000 003 039 761 203 2;
  • 13) 0.000 003 039 761 203 2 × 2 = 0 + 0.000 006 079 522 406 4;
  • 14) 0.000 006 079 522 406 4 × 2 = 0 + 0.000 012 159 044 812 8;
  • 15) 0.000 012 159 044 812 8 × 2 = 0 + 0.000 024 318 089 625 6;
  • 16) 0.000 024 318 089 625 6 × 2 = 0 + 0.000 048 636 179 251 2;
  • 17) 0.000 048 636 179 251 2 × 2 = 0 + 0.000 097 272 358 502 4;
  • 18) 0.000 097 272 358 502 4 × 2 = 0 + 0.000 194 544 717 004 8;
  • 19) 0.000 194 544 717 004 8 × 2 = 0 + 0.000 389 089 434 009 6;
  • 20) 0.000 389 089 434 009 6 × 2 = 0 + 0.000 778 178 868 019 2;
  • 21) 0.000 778 178 868 019 2 × 2 = 0 + 0.001 556 357 736 038 4;
  • 22) 0.001 556 357 736 038 4 × 2 = 0 + 0.003 112 715 472 076 8;
  • 23) 0.003 112 715 472 076 8 × 2 = 0 + 0.006 225 430 944 153 6;
  • 24) 0.006 225 430 944 153 6 × 2 = 0 + 0.012 450 861 888 307 2;
  • 25) 0.012 450 861 888 307 2 × 2 = 0 + 0.024 901 723 776 614 4;
  • 26) 0.024 901 723 776 614 4 × 2 = 0 + 0.049 803 447 553 228 8;
  • 27) 0.049 803 447 553 228 8 × 2 = 0 + 0.099 606 895 106 457 6;
  • 28) 0.099 606 895 106 457 6 × 2 = 0 + 0.199 213 790 212 915 2;
  • 29) 0.199 213 790 212 915 2 × 2 = 0 + 0.398 427 580 425 830 4;
  • 30) 0.398 427 580 425 830 4 × 2 = 0 + 0.796 855 160 851 660 8;
  • 31) 0.796 855 160 851 660 8 × 2 = 1 + 0.593 710 321 703 321 6;
  • 32) 0.593 710 321 703 321 6 × 2 = 1 + 0.187 420 643 406 643 2;
  • 33) 0.187 420 643 406 643 2 × 2 = 0 + 0.374 841 286 813 286 4;
  • 34) 0.374 841 286 813 286 4 × 2 = 0 + 0.749 682 573 626 572 8;
  • 35) 0.749 682 573 626 572 8 × 2 = 1 + 0.499 365 147 253 145 6;
  • 36) 0.499 365 147 253 145 6 × 2 = 0 + 0.998 730 294 506 291 2;
  • 37) 0.998 730 294 506 291 2 × 2 = 1 + 0.997 460 589 012 582 4;
  • 38) 0.997 460 589 012 582 4 × 2 = 1 + 0.994 921 178 025 164 8;
  • 39) 0.994 921 178 025 164 8 × 2 = 1 + 0.989 842 356 050 329 6;
  • 40) 0.989 842 356 050 329 6 × 2 = 1 + 0.979 684 712 100 659 2;
  • 41) 0.979 684 712 100 659 2 × 2 = 1 + 0.959 369 424 201 318 4;
  • 42) 0.959 369 424 201 318 4 × 2 = 1 + 0.918 738 848 402 636 8;
  • 43) 0.918 738 848 402 636 8 × 2 = 1 + 0.837 477 696 805 273 6;
  • 44) 0.837 477 696 805 273 6 × 2 = 1 + 0.674 955 393 610 547 2;
  • 45) 0.674 955 393 610 547 2 × 2 = 1 + 0.349 910 787 221 094 4;
  • 46) 0.349 910 787 221 094 4 × 2 = 0 + 0.699 821 574 442 188 8;
  • 47) 0.699 821 574 442 188 8 × 2 = 1 + 0.399 643 148 884 377 6;
  • 48) 0.399 643 148 884 377 6 × 2 = 0 + 0.799 286 297 768 755 2;
  • 49) 0.799 286 297 768 755 2 × 2 = 1 + 0.598 572 595 537 510 4;
  • 50) 0.598 572 595 537 510 4 × 2 = 1 + 0.197 145 191 075 020 8;
  • 51) 0.197 145 191 075 020 8 × 2 = 0 + 0.394 290 382 150 041 6;
  • 52) 0.394 290 382 150 041 6 × 2 = 0 + 0.788 580 764 300 083 2;
  • 53) 0.788 580 764 300 083 2 × 2 = 1 + 0.577 161 528 600 166 4;
  • 54) 0.577 161 528 600 166 4 × 2 = 1 + 0.154 323 057 200 332 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:


0.000 000 000 742 129 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1100 11(2)

6. Positive number before normalization:

0.000 000 000 742 129 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1100 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:


0.000 000 000 742 129 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1100 11(2) × 20 =

1.1001 0111 1111 1101 0110 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 1101 0110 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 1110 1011 0011 =

100 1011 1111 1110 1011 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 1110 1011 0011


Decimal number -0.000 000 000 742 129 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1011 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