-0.000 000 000 742 146 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 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
-0.000 000 000 742 146 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:

|-0.000 000 000 742 146 7| = 0.000 000 000 742 146 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: 0.000 000 000 742 146 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) 0.000 000 000 742 146 7 × 2 = 0 + 0.000 000 001 484 293 4;
  • 2) 0.000 000 001 484 293 4 × 2 = 0 + 0.000 000 002 968 586 8;
  • 3) 0.000 000 002 968 586 8 × 2 = 0 + 0.000 000 005 937 173 6;
  • 4) 0.000 000 005 937 173 6 × 2 = 0 + 0.000 000 011 874 347 2;
  • 5) 0.000 000 011 874 347 2 × 2 = 0 + 0.000 000 023 748 694 4;
  • 6) 0.000 000 023 748 694 4 × 2 = 0 + 0.000 000 047 497 388 8;
  • 7) 0.000 000 047 497 388 8 × 2 = 0 + 0.000 000 094 994 777 6;
  • 8) 0.000 000 094 994 777 6 × 2 = 0 + 0.000 000 189 989 555 2;
  • 9) 0.000 000 189 989 555 2 × 2 = 0 + 0.000 000 379 979 110 4;
  • 10) 0.000 000 379 979 110 4 × 2 = 0 + 0.000 000 759 958 220 8;
  • 11) 0.000 000 759 958 220 8 × 2 = 0 + 0.000 001 519 916 441 6;
  • 12) 0.000 001 519 916 441 6 × 2 = 0 + 0.000 003 039 832 883 2;
  • 13) 0.000 003 039 832 883 2 × 2 = 0 + 0.000 006 079 665 766 4;
  • 14) 0.000 006 079 665 766 4 × 2 = 0 + 0.000 012 159 331 532 8;
  • 15) 0.000 012 159 331 532 8 × 2 = 0 + 0.000 024 318 663 065 6;
  • 16) 0.000 024 318 663 065 6 × 2 = 0 + 0.000 048 637 326 131 2;
  • 17) 0.000 048 637 326 131 2 × 2 = 0 + 0.000 097 274 652 262 4;
  • 18) 0.000 097 274 652 262 4 × 2 = 0 + 0.000 194 549 304 524 8;
  • 19) 0.000 194 549 304 524 8 × 2 = 0 + 0.000 389 098 609 049 6;
  • 20) 0.000 389 098 609 049 6 × 2 = 0 + 0.000 778 197 218 099 2;
  • 21) 0.000 778 197 218 099 2 × 2 = 0 + 0.001 556 394 436 198 4;
  • 22) 0.001 556 394 436 198 4 × 2 = 0 + 0.003 112 788 872 396 8;
  • 23) 0.003 112 788 872 396 8 × 2 = 0 + 0.006 225 577 744 793 6;
  • 24) 0.006 225 577 744 793 6 × 2 = 0 + 0.012 451 155 489 587 2;
  • 25) 0.012 451 155 489 587 2 × 2 = 0 + 0.024 902 310 979 174 4;
  • 26) 0.024 902 310 979 174 4 × 2 = 0 + 0.049 804 621 958 348 8;
  • 27) 0.049 804 621 958 348 8 × 2 = 0 + 0.099 609 243 916 697 6;
  • 28) 0.099 609 243 916 697 6 × 2 = 0 + 0.199 218 487 833 395 2;
  • 29) 0.199 218 487 833 395 2 × 2 = 0 + 0.398 436 975 666 790 4;
  • 30) 0.398 436 975 666 790 4 × 2 = 0 + 0.796 873 951 333 580 8;
  • 31) 0.796 873 951 333 580 8 × 2 = 1 + 0.593 747 902 667 161 6;
  • 32) 0.593 747 902 667 161 6 × 2 = 1 + 0.187 495 805 334 323 2;
  • 33) 0.187 495 805 334 323 2 × 2 = 0 + 0.374 991 610 668 646 4;
  • 34) 0.374 991 610 668 646 4 × 2 = 0 + 0.749 983 221 337 292 8;
  • 35) 0.749 983 221 337 292 8 × 2 = 1 + 0.499 966 442 674 585 6;
  • 36) 0.499 966 442 674 585 6 × 2 = 0 + 0.999 932 885 349 171 2;
  • 37) 0.999 932 885 349 171 2 × 2 = 1 + 0.999 865 770 698 342 4;
  • 38) 0.999 865 770 698 342 4 × 2 = 1 + 0.999 731 541 396 684 8;
  • 39) 0.999 731 541 396 684 8 × 2 = 1 + 0.999 463 082 793 369 6;
  • 40) 0.999 463 082 793 369 6 × 2 = 1 + 0.998 926 165 586 739 2;
  • 41) 0.998 926 165 586 739 2 × 2 = 1 + 0.997 852 331 173 478 4;
  • 42) 0.997 852 331 173 478 4 × 2 = 1 + 0.995 704 662 346 956 8;
  • 43) 0.995 704 662 346 956 8 × 2 = 1 + 0.991 409 324 693 913 6;
  • 44) 0.991 409 324 693 913 6 × 2 = 1 + 0.982 818 649 387 827 2;
  • 45) 0.982 818 649 387 827 2 × 2 = 1 + 0.965 637 298 775 654 4;
  • 46) 0.965 637 298 775 654 4 × 2 = 1 + 0.931 274 597 551 308 8;
  • 47) 0.931 274 597 551 308 8 × 2 = 1 + 0.862 549 195 102 617 6;
  • 48) 0.862 549 195 102 617 6 × 2 = 1 + 0.725 098 390 205 235 2;
  • 49) 0.725 098 390 205 235 2 × 2 = 1 + 0.450 196 780 410 470 4;
  • 50) 0.450 196 780 410 470 4 × 2 = 0 + 0.900 393 560 820 940 8;
  • 51) 0.900 393 560 820 940 8 × 2 = 1 + 0.800 787 121 641 881 6;
  • 52) 0.800 787 121 641 881 6 × 2 = 1 + 0.601 574 243 283 763 2;
  • 53) 0.601 574 243 283 763 2 × 2 = 1 + 0.203 148 486 567 526 4;
  • 54) 0.203 148 486 567 526 4 × 2 = 0 + 0.406 296 973 135 052 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:


0.000 000 000 742 146 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(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:


0.000 000 000 742 146 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) × 20 =

1.1001 0111 1111 1111 1101 110(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 1101 110


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 1110 1110 =

100 1011 1111 1111 1110 1110


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 1110 1110


Decimal number -0.000 000 000 742 146 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 1110

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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