-0.000 000 000 736 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 736 9(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
-0.000 000 000 736 9(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:

|-0.000 000 000 736 9| = 0.000 000 000 736 9


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: 0.000 000 000 736 9.

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) 0.000 000 000 736 9 × 2 = 0 + 0.000 000 001 473 8;
  • 2) 0.000 000 001 473 8 × 2 = 0 + 0.000 000 002 947 6;
  • 3) 0.000 000 002 947 6 × 2 = 0 + 0.000 000 005 895 2;
  • 4) 0.000 000 005 895 2 × 2 = 0 + 0.000 000 011 790 4;
  • 5) 0.000 000 011 790 4 × 2 = 0 + 0.000 000 023 580 8;
  • 6) 0.000 000 023 580 8 × 2 = 0 + 0.000 000 047 161 6;
  • 7) 0.000 000 047 161 6 × 2 = 0 + 0.000 000 094 323 2;
  • 8) 0.000 000 094 323 2 × 2 = 0 + 0.000 000 188 646 4;
  • 9) 0.000 000 188 646 4 × 2 = 0 + 0.000 000 377 292 8;
  • 10) 0.000 000 377 292 8 × 2 = 0 + 0.000 000 754 585 6;
  • 11) 0.000 000 754 585 6 × 2 = 0 + 0.000 001 509 171 2;
  • 12) 0.000 001 509 171 2 × 2 = 0 + 0.000 003 018 342 4;
  • 13) 0.000 003 018 342 4 × 2 = 0 + 0.000 006 036 684 8;
  • 14) 0.000 006 036 684 8 × 2 = 0 + 0.000 012 073 369 6;
  • 15) 0.000 012 073 369 6 × 2 = 0 + 0.000 024 146 739 2;
  • 16) 0.000 024 146 739 2 × 2 = 0 + 0.000 048 293 478 4;
  • 17) 0.000 048 293 478 4 × 2 = 0 + 0.000 096 586 956 8;
  • 18) 0.000 096 586 956 8 × 2 = 0 + 0.000 193 173 913 6;
  • 19) 0.000 193 173 913 6 × 2 = 0 + 0.000 386 347 827 2;
  • 20) 0.000 386 347 827 2 × 2 = 0 + 0.000 772 695 654 4;
  • 21) 0.000 772 695 654 4 × 2 = 0 + 0.001 545 391 308 8;
  • 22) 0.001 545 391 308 8 × 2 = 0 + 0.003 090 782 617 6;
  • 23) 0.003 090 782 617 6 × 2 = 0 + 0.006 181 565 235 2;
  • 24) 0.006 181 565 235 2 × 2 = 0 + 0.012 363 130 470 4;
  • 25) 0.012 363 130 470 4 × 2 = 0 + 0.024 726 260 940 8;
  • 26) 0.024 726 260 940 8 × 2 = 0 + 0.049 452 521 881 6;
  • 27) 0.049 452 521 881 6 × 2 = 0 + 0.098 905 043 763 2;
  • 28) 0.098 905 043 763 2 × 2 = 0 + 0.197 810 087 526 4;
  • 29) 0.197 810 087 526 4 × 2 = 0 + 0.395 620 175 052 8;
  • 30) 0.395 620 175 052 8 × 2 = 0 + 0.791 240 350 105 6;
  • 31) 0.791 240 350 105 6 × 2 = 1 + 0.582 480 700 211 2;
  • 32) 0.582 480 700 211 2 × 2 = 1 + 0.164 961 400 422 4;
  • 33) 0.164 961 400 422 4 × 2 = 0 + 0.329 922 800 844 8;
  • 34) 0.329 922 800 844 8 × 2 = 0 + 0.659 845 601 689 6;
  • 35) 0.659 845 601 689 6 × 2 = 1 + 0.319 691 203 379 2;
  • 36) 0.319 691 203 379 2 × 2 = 0 + 0.639 382 406 758 4;
  • 37) 0.639 382 406 758 4 × 2 = 1 + 0.278 764 813 516 8;
  • 38) 0.278 764 813 516 8 × 2 = 0 + 0.557 529 627 033 6;
  • 39) 0.557 529 627 033 6 × 2 = 1 + 0.115 059 254 067 2;
  • 40) 0.115 059 254 067 2 × 2 = 0 + 0.230 118 508 134 4;
  • 41) 0.230 118 508 134 4 × 2 = 0 + 0.460 237 016 268 8;
  • 42) 0.460 237 016 268 8 × 2 = 0 + 0.920 474 032 537 6;
  • 43) 0.920 474 032 537 6 × 2 = 1 + 0.840 948 065 075 2;
  • 44) 0.840 948 065 075 2 × 2 = 1 + 0.681 896 130 150 4;
  • 45) 0.681 896 130 150 4 × 2 = 1 + 0.363 792 260 300 8;
  • 46) 0.363 792 260 300 8 × 2 = 0 + 0.727 584 520 601 6;
  • 47) 0.727 584 520 601 6 × 2 = 1 + 0.455 169 041 203 2;
  • 48) 0.455 169 041 203 2 × 2 = 0 + 0.910 338 082 406 4;
  • 49) 0.910 338 082 406 4 × 2 = 1 + 0.820 676 164 812 8;
  • 50) 0.820 676 164 812 8 × 2 = 1 + 0.641 352 329 625 6;
  • 51) 0.641 352 329 625 6 × 2 = 1 + 0.282 704 659 251 2;
  • 52) 0.282 704 659 251 2 × 2 = 0 + 0.565 409 318 502 4;
  • 53) 0.565 409 318 502 4 × 2 = 1 + 0.130 818 637 004 8;
  • 54) 0.130 818 637 004 8 × 2 = 0 + 0.261 637 274 009 6;

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:


0.000 000 000 736 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 0011 1010 1110 10(2)

6. Positive number before normalization:

0.000 000 000 736 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 0011 1010 1110 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:


0.000 000 000 736 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 0011 1010 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 0011 1010 1110 10(2) × 20 =

1.1001 0101 0001 1101 0111 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 0101 0001 1101 0111 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 1010 1000 1110 1011 1010 =

100 1010 1000 1110 1011 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 1010 1000 1110 1011 1010


Decimal number -0.000 000 000 736 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 1000 1110 1011 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