-0.000 000 100 100 147 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 100 100 147 4(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
-0.000 000 100 100 147 4(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:

|-0.000 000 100 100 147 4| = 0.000 000 100 100 147 4


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: 0.000 000 100 100 147 4.

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) 0.000 000 100 100 147 4 × 2 = 0 + 0.000 000 200 200 294 8;
  • 2) 0.000 000 200 200 294 8 × 2 = 0 + 0.000 000 400 400 589 6;
  • 3) 0.000 000 400 400 589 6 × 2 = 0 + 0.000 000 800 801 179 2;
  • 4) 0.000 000 800 801 179 2 × 2 = 0 + 0.000 001 601 602 358 4;
  • 5) 0.000 001 601 602 358 4 × 2 = 0 + 0.000 003 203 204 716 8;
  • 6) 0.000 003 203 204 716 8 × 2 = 0 + 0.000 006 406 409 433 6;
  • 7) 0.000 006 406 409 433 6 × 2 = 0 + 0.000 012 812 818 867 2;
  • 8) 0.000 012 812 818 867 2 × 2 = 0 + 0.000 025 625 637 734 4;
  • 9) 0.000 025 625 637 734 4 × 2 = 0 + 0.000 051 251 275 468 8;
  • 10) 0.000 051 251 275 468 8 × 2 = 0 + 0.000 102 502 550 937 6;
  • 11) 0.000 102 502 550 937 6 × 2 = 0 + 0.000 205 005 101 875 2;
  • 12) 0.000 205 005 101 875 2 × 2 = 0 + 0.000 410 010 203 750 4;
  • 13) 0.000 410 010 203 750 4 × 2 = 0 + 0.000 820 020 407 500 8;
  • 14) 0.000 820 020 407 500 8 × 2 = 0 + 0.001 640 040 815 001 6;
  • 15) 0.001 640 040 815 001 6 × 2 = 0 + 0.003 280 081 630 003 2;
  • 16) 0.003 280 081 630 003 2 × 2 = 0 + 0.006 560 163 260 006 4;
  • 17) 0.006 560 163 260 006 4 × 2 = 0 + 0.013 120 326 520 012 8;
  • 18) 0.013 120 326 520 012 8 × 2 = 0 + 0.026 240 653 040 025 6;
  • 19) 0.026 240 653 040 025 6 × 2 = 0 + 0.052 481 306 080 051 2;
  • 20) 0.052 481 306 080 051 2 × 2 = 0 + 0.104 962 612 160 102 4;
  • 21) 0.104 962 612 160 102 4 × 2 = 0 + 0.209 925 224 320 204 8;
  • 22) 0.209 925 224 320 204 8 × 2 = 0 + 0.419 850 448 640 409 6;
  • 23) 0.419 850 448 640 409 6 × 2 = 0 + 0.839 700 897 280 819 2;
  • 24) 0.839 700 897 280 819 2 × 2 = 1 + 0.679 401 794 561 638 4;
  • 25) 0.679 401 794 561 638 4 × 2 = 1 + 0.358 803 589 123 276 8;
  • 26) 0.358 803 589 123 276 8 × 2 = 0 + 0.717 607 178 246 553 6;
  • 27) 0.717 607 178 246 553 6 × 2 = 1 + 0.435 214 356 493 107 2;
  • 28) 0.435 214 356 493 107 2 × 2 = 0 + 0.870 428 712 986 214 4;
  • 29) 0.870 428 712 986 214 4 × 2 = 1 + 0.740 857 425 972 428 8;
  • 30) 0.740 857 425 972 428 8 × 2 = 1 + 0.481 714 851 944 857 6;
  • 31) 0.481 714 851 944 857 6 × 2 = 0 + 0.963 429 703 889 715 2;
  • 32) 0.963 429 703 889 715 2 × 2 = 1 + 0.926 859 407 779 430 4;
  • 33) 0.926 859 407 779 430 4 × 2 = 1 + 0.853 718 815 558 860 8;
  • 34) 0.853 718 815 558 860 8 × 2 = 1 + 0.707 437 631 117 721 6;
  • 35) 0.707 437 631 117 721 6 × 2 = 1 + 0.414 875 262 235 443 2;
  • 36) 0.414 875 262 235 443 2 × 2 = 0 + 0.829 750 524 470 886 4;
  • 37) 0.829 750 524 470 886 4 × 2 = 1 + 0.659 501 048 941 772 8;
  • 38) 0.659 501 048 941 772 8 × 2 = 1 + 0.319 002 097 883 545 6;
  • 39) 0.319 002 097 883 545 6 × 2 = 0 + 0.638 004 195 767 091 2;
  • 40) 0.638 004 195 767 091 2 × 2 = 1 + 0.276 008 391 534 182 4;
  • 41) 0.276 008 391 534 182 4 × 2 = 0 + 0.552 016 783 068 364 8;
  • 42) 0.552 016 783 068 364 8 × 2 = 1 + 0.104 033 566 136 729 6;
  • 43) 0.104 033 566 136 729 6 × 2 = 0 + 0.208 067 132 273 459 2;
  • 44) 0.208 067 132 273 459 2 × 2 = 0 + 0.416 134 264 546 918 4;
  • 45) 0.416 134 264 546 918 4 × 2 = 0 + 0.832 268 529 093 836 8;
  • 46) 0.832 268 529 093 836 8 × 2 = 1 + 0.664 537 058 187 673 6;
  • 47) 0.664 537 058 187 673 6 × 2 = 1 + 0.329 074 116 375 347 2;

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:


0.000 000 100 100 147 4(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 011(2)

6. Positive number before normalization:

0.000 000 100 100 147 4(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 100 100 147 4(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 011(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0100 011(2) × 20 =

1.1010 1101 1110 1101 0100 011(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 1110 1101 0100 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1111 0110 1010 0011 =

101 0110 1111 0110 1010 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 0111

Mantissa (23 bits) =
101 0110 1111 0110 1010 0011


Decimal number -0.000 000 100 100 147 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1111 0110 1010 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