-0.000 000 000 742 149 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 54(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
-0.000 000 000 742 149 54(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:

|-0.000 000 000 742 149 54| = 0.000 000 000 742 149 54


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: 0.000 000 000 742 149 54.

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) 0.000 000 000 742 149 54 × 2 = 0 + 0.000 000 001 484 299 08;
  • 2) 0.000 000 001 484 299 08 × 2 = 0 + 0.000 000 002 968 598 16;
  • 3) 0.000 000 002 968 598 16 × 2 = 0 + 0.000 000 005 937 196 32;
  • 4) 0.000 000 005 937 196 32 × 2 = 0 + 0.000 000 011 874 392 64;
  • 5) 0.000 000 011 874 392 64 × 2 = 0 + 0.000 000 023 748 785 28;
  • 6) 0.000 000 023 748 785 28 × 2 = 0 + 0.000 000 047 497 570 56;
  • 7) 0.000 000 047 497 570 56 × 2 = 0 + 0.000 000 094 995 141 12;
  • 8) 0.000 000 094 995 141 12 × 2 = 0 + 0.000 000 189 990 282 24;
  • 9) 0.000 000 189 990 282 24 × 2 = 0 + 0.000 000 379 980 564 48;
  • 10) 0.000 000 379 980 564 48 × 2 = 0 + 0.000 000 759 961 128 96;
  • 11) 0.000 000 759 961 128 96 × 2 = 0 + 0.000 001 519 922 257 92;
  • 12) 0.000 001 519 922 257 92 × 2 = 0 + 0.000 003 039 844 515 84;
  • 13) 0.000 003 039 844 515 84 × 2 = 0 + 0.000 006 079 689 031 68;
  • 14) 0.000 006 079 689 031 68 × 2 = 0 + 0.000 012 159 378 063 36;
  • 15) 0.000 012 159 378 063 36 × 2 = 0 + 0.000 024 318 756 126 72;
  • 16) 0.000 024 318 756 126 72 × 2 = 0 + 0.000 048 637 512 253 44;
  • 17) 0.000 048 637 512 253 44 × 2 = 0 + 0.000 097 275 024 506 88;
  • 18) 0.000 097 275 024 506 88 × 2 = 0 + 0.000 194 550 049 013 76;
  • 19) 0.000 194 550 049 013 76 × 2 = 0 + 0.000 389 100 098 027 52;
  • 20) 0.000 389 100 098 027 52 × 2 = 0 + 0.000 778 200 196 055 04;
  • 21) 0.000 778 200 196 055 04 × 2 = 0 + 0.001 556 400 392 110 08;
  • 22) 0.001 556 400 392 110 08 × 2 = 0 + 0.003 112 800 784 220 16;
  • 23) 0.003 112 800 784 220 16 × 2 = 0 + 0.006 225 601 568 440 32;
  • 24) 0.006 225 601 568 440 32 × 2 = 0 + 0.012 451 203 136 880 64;
  • 25) 0.012 451 203 136 880 64 × 2 = 0 + 0.024 902 406 273 761 28;
  • 26) 0.024 902 406 273 761 28 × 2 = 0 + 0.049 804 812 547 522 56;
  • 27) 0.049 804 812 547 522 56 × 2 = 0 + 0.099 609 625 095 045 12;
  • 28) 0.099 609 625 095 045 12 × 2 = 0 + 0.199 219 250 190 090 24;
  • 29) 0.199 219 250 190 090 24 × 2 = 0 + 0.398 438 500 380 180 48;
  • 30) 0.398 438 500 380 180 48 × 2 = 0 + 0.796 877 000 760 360 96;
  • 31) 0.796 877 000 760 360 96 × 2 = 1 + 0.593 754 001 520 721 92;
  • 32) 0.593 754 001 520 721 92 × 2 = 1 + 0.187 508 003 041 443 84;
  • 33) 0.187 508 003 041 443 84 × 2 = 0 + 0.375 016 006 082 887 68;
  • 34) 0.375 016 006 082 887 68 × 2 = 0 + 0.750 032 012 165 775 36;
  • 35) 0.750 032 012 165 775 36 × 2 = 1 + 0.500 064 024 331 550 72;
  • 36) 0.500 064 024 331 550 72 × 2 = 1 + 0.000 128 048 663 101 44;
  • 37) 0.000 128 048 663 101 44 × 2 = 0 + 0.000 256 097 326 202 88;
  • 38) 0.000 256 097 326 202 88 × 2 = 0 + 0.000 512 194 652 405 76;
  • 39) 0.000 512 194 652 405 76 × 2 = 0 + 0.001 024 389 304 811 52;
  • 40) 0.001 024 389 304 811 52 × 2 = 0 + 0.002 048 778 609 623 04;
  • 41) 0.002 048 778 609 623 04 × 2 = 0 + 0.004 097 557 219 246 08;
  • 42) 0.004 097 557 219 246 08 × 2 = 0 + 0.008 195 114 438 492 16;
  • 43) 0.008 195 114 438 492 16 × 2 = 0 + 0.016 390 228 876 984 32;
  • 44) 0.016 390 228 876 984 32 × 2 = 0 + 0.032 780 457 753 968 64;
  • 45) 0.032 780 457 753 968 64 × 2 = 0 + 0.065 560 915 507 937 28;
  • 46) 0.065 560 915 507 937 28 × 2 = 0 + 0.131 121 831 015 874 56;
  • 47) 0.131 121 831 015 874 56 × 2 = 0 + 0.262 243 662 031 749 12;
  • 48) 0.262 243 662 031 749 12 × 2 = 0 + 0.524 487 324 063 498 24;
  • 49) 0.524 487 324 063 498 24 × 2 = 1 + 0.048 974 648 126 996 48;
  • 50) 0.048 974 648 126 996 48 × 2 = 0 + 0.097 949 296 253 992 96;
  • 51) 0.097 949 296 253 992 96 × 2 = 0 + 0.195 898 592 507 985 92;
  • 52) 0.195 898 592 507 985 92 × 2 = 0 + 0.391 797 185 015 971 84;
  • 53) 0.391 797 185 015 971 84 × 2 = 0 + 0.783 594 370 031 943 68;
  • 54) 0.783 594 370 031 943 68 × 2 = 1 + 0.567 188 740 063 887 36;

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:


0.000 000 000 742 149 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 149 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 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:


0.000 000 000 742 149 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2) × 20 =

1.1001 1000 0000 0000 0100 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 1000 0000 0000 0100 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 1100 0000 0000 0010 0001 =

100 1100 0000 0000 0010 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 1100 0000 0000 0010 0001


Decimal number -0.000 000 000 742 149 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 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