-0.000 000 000 742 146 49 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 49(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 49(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 49| = 0.000 000 000 742 146 49


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 49.

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 49 × 2 = 0 + 0.000 000 001 484 292 98;
  • 2) 0.000 000 001 484 292 98 × 2 = 0 + 0.000 000 002 968 585 96;
  • 3) 0.000 000 002 968 585 96 × 2 = 0 + 0.000 000 005 937 171 92;
  • 4) 0.000 000 005 937 171 92 × 2 = 0 + 0.000 000 011 874 343 84;
  • 5) 0.000 000 011 874 343 84 × 2 = 0 + 0.000 000 023 748 687 68;
  • 6) 0.000 000 023 748 687 68 × 2 = 0 + 0.000 000 047 497 375 36;
  • 7) 0.000 000 047 497 375 36 × 2 = 0 + 0.000 000 094 994 750 72;
  • 8) 0.000 000 094 994 750 72 × 2 = 0 + 0.000 000 189 989 501 44;
  • 9) 0.000 000 189 989 501 44 × 2 = 0 + 0.000 000 379 979 002 88;
  • 10) 0.000 000 379 979 002 88 × 2 = 0 + 0.000 000 759 958 005 76;
  • 11) 0.000 000 759 958 005 76 × 2 = 0 + 0.000 001 519 916 011 52;
  • 12) 0.000 001 519 916 011 52 × 2 = 0 + 0.000 003 039 832 023 04;
  • 13) 0.000 003 039 832 023 04 × 2 = 0 + 0.000 006 079 664 046 08;
  • 14) 0.000 006 079 664 046 08 × 2 = 0 + 0.000 012 159 328 092 16;
  • 15) 0.000 012 159 328 092 16 × 2 = 0 + 0.000 024 318 656 184 32;
  • 16) 0.000 024 318 656 184 32 × 2 = 0 + 0.000 048 637 312 368 64;
  • 17) 0.000 048 637 312 368 64 × 2 = 0 + 0.000 097 274 624 737 28;
  • 18) 0.000 097 274 624 737 28 × 2 = 0 + 0.000 194 549 249 474 56;
  • 19) 0.000 194 549 249 474 56 × 2 = 0 + 0.000 389 098 498 949 12;
  • 20) 0.000 389 098 498 949 12 × 2 = 0 + 0.000 778 196 997 898 24;
  • 21) 0.000 778 196 997 898 24 × 2 = 0 + 0.001 556 393 995 796 48;
  • 22) 0.001 556 393 995 796 48 × 2 = 0 + 0.003 112 787 991 592 96;
  • 23) 0.003 112 787 991 592 96 × 2 = 0 + 0.006 225 575 983 185 92;
  • 24) 0.006 225 575 983 185 92 × 2 = 0 + 0.012 451 151 966 371 84;
  • 25) 0.012 451 151 966 371 84 × 2 = 0 + 0.024 902 303 932 743 68;
  • 26) 0.024 902 303 932 743 68 × 2 = 0 + 0.049 804 607 865 487 36;
  • 27) 0.049 804 607 865 487 36 × 2 = 0 + 0.099 609 215 730 974 72;
  • 28) 0.099 609 215 730 974 72 × 2 = 0 + 0.199 218 431 461 949 44;
  • 29) 0.199 218 431 461 949 44 × 2 = 0 + 0.398 436 862 923 898 88;
  • 30) 0.398 436 862 923 898 88 × 2 = 0 + 0.796 873 725 847 797 76;
  • 31) 0.796 873 725 847 797 76 × 2 = 1 + 0.593 747 451 695 595 52;
  • 32) 0.593 747 451 695 595 52 × 2 = 1 + 0.187 494 903 391 191 04;
  • 33) 0.187 494 903 391 191 04 × 2 = 0 + 0.374 989 806 782 382 08;
  • 34) 0.374 989 806 782 382 08 × 2 = 0 + 0.749 979 613 564 764 16;
  • 35) 0.749 979 613 564 764 16 × 2 = 1 + 0.499 959 227 129 528 32;
  • 36) 0.499 959 227 129 528 32 × 2 = 0 + 0.999 918 454 259 056 64;
  • 37) 0.999 918 454 259 056 64 × 2 = 1 + 0.999 836 908 518 113 28;
  • 38) 0.999 836 908 518 113 28 × 2 = 1 + 0.999 673 817 036 226 56;
  • 39) 0.999 673 817 036 226 56 × 2 = 1 + 0.999 347 634 072 453 12;
  • 40) 0.999 347 634 072 453 12 × 2 = 1 + 0.998 695 268 144 906 24;
  • 41) 0.998 695 268 144 906 24 × 2 = 1 + 0.997 390 536 289 812 48;
  • 42) 0.997 390 536 289 812 48 × 2 = 1 + 0.994 781 072 579 624 96;
  • 43) 0.994 781 072 579 624 96 × 2 = 1 + 0.989 562 145 159 249 92;
  • 44) 0.989 562 145 159 249 92 × 2 = 1 + 0.979 124 290 318 499 84;
  • 45) 0.979 124 290 318 499 84 × 2 = 1 + 0.958 248 580 636 999 68;
  • 46) 0.958 248 580 636 999 68 × 2 = 1 + 0.916 497 161 273 999 36;
  • 47) 0.916 497 161 273 999 36 × 2 = 1 + 0.832 994 322 547 998 72;
  • 48) 0.832 994 322 547 998 72 × 2 = 1 + 0.665 988 645 095 997 44;
  • 49) 0.665 988 645 095 997 44 × 2 = 1 + 0.331 977 290 191 994 88;
  • 50) 0.331 977 290 191 994 88 × 2 = 0 + 0.663 954 580 383 989 76;
  • 51) 0.663 954 580 383 989 76 × 2 = 1 + 0.327 909 160 767 979 52;
  • 52) 0.327 909 160 767 979 52 × 2 = 0 + 0.655 818 321 535 959 04;
  • 53) 0.655 818 321 535 959 04 × 2 = 1 + 0.311 636 643 071 918 08;
  • 54) 0.311 636 643 071 918 08 × 2 = 0 + 0.623 273 286 143 836 16;

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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2) × 20 =

1.1001 0111 1111 1111 1101 010(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 010


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 1010 =

100 1011 1111 1111 1110 1010


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 1010


Decimal number -0.000 000 000 742 146 49 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 1010

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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