-0.000 000 000 742 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 63(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 63(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 63| = 0.000 000 000 742 63


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 63.

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 63 × 2 = 0 + 0.000 000 001 485 26;
  • 2) 0.000 000 001 485 26 × 2 = 0 + 0.000 000 002 970 52;
  • 3) 0.000 000 002 970 52 × 2 = 0 + 0.000 000 005 941 04;
  • 4) 0.000 000 005 941 04 × 2 = 0 + 0.000 000 011 882 08;
  • 5) 0.000 000 011 882 08 × 2 = 0 + 0.000 000 023 764 16;
  • 6) 0.000 000 023 764 16 × 2 = 0 + 0.000 000 047 528 32;
  • 7) 0.000 000 047 528 32 × 2 = 0 + 0.000 000 095 056 64;
  • 8) 0.000 000 095 056 64 × 2 = 0 + 0.000 000 190 113 28;
  • 9) 0.000 000 190 113 28 × 2 = 0 + 0.000 000 380 226 56;
  • 10) 0.000 000 380 226 56 × 2 = 0 + 0.000 000 760 453 12;
  • 11) 0.000 000 760 453 12 × 2 = 0 + 0.000 001 520 906 24;
  • 12) 0.000 001 520 906 24 × 2 = 0 + 0.000 003 041 812 48;
  • 13) 0.000 003 041 812 48 × 2 = 0 + 0.000 006 083 624 96;
  • 14) 0.000 006 083 624 96 × 2 = 0 + 0.000 012 167 249 92;
  • 15) 0.000 012 167 249 92 × 2 = 0 + 0.000 024 334 499 84;
  • 16) 0.000 024 334 499 84 × 2 = 0 + 0.000 048 668 999 68;
  • 17) 0.000 048 668 999 68 × 2 = 0 + 0.000 097 337 999 36;
  • 18) 0.000 097 337 999 36 × 2 = 0 + 0.000 194 675 998 72;
  • 19) 0.000 194 675 998 72 × 2 = 0 + 0.000 389 351 997 44;
  • 20) 0.000 389 351 997 44 × 2 = 0 + 0.000 778 703 994 88;
  • 21) 0.000 778 703 994 88 × 2 = 0 + 0.001 557 407 989 76;
  • 22) 0.001 557 407 989 76 × 2 = 0 + 0.003 114 815 979 52;
  • 23) 0.003 114 815 979 52 × 2 = 0 + 0.006 229 631 959 04;
  • 24) 0.006 229 631 959 04 × 2 = 0 + 0.012 459 263 918 08;
  • 25) 0.012 459 263 918 08 × 2 = 0 + 0.024 918 527 836 16;
  • 26) 0.024 918 527 836 16 × 2 = 0 + 0.049 837 055 672 32;
  • 27) 0.049 837 055 672 32 × 2 = 0 + 0.099 674 111 344 64;
  • 28) 0.099 674 111 344 64 × 2 = 0 + 0.199 348 222 689 28;
  • 29) 0.199 348 222 689 28 × 2 = 0 + 0.398 696 445 378 56;
  • 30) 0.398 696 445 378 56 × 2 = 0 + 0.797 392 890 757 12;
  • 31) 0.797 392 890 757 12 × 2 = 1 + 0.594 785 781 514 24;
  • 32) 0.594 785 781 514 24 × 2 = 1 + 0.189 571 563 028 48;
  • 33) 0.189 571 563 028 48 × 2 = 0 + 0.379 143 126 056 96;
  • 34) 0.379 143 126 056 96 × 2 = 0 + 0.758 286 252 113 92;
  • 35) 0.758 286 252 113 92 × 2 = 1 + 0.516 572 504 227 84;
  • 36) 0.516 572 504 227 84 × 2 = 1 + 0.033 145 008 455 68;
  • 37) 0.033 145 008 455 68 × 2 = 0 + 0.066 290 016 911 36;
  • 38) 0.066 290 016 911 36 × 2 = 0 + 0.132 580 033 822 72;
  • 39) 0.132 580 033 822 72 × 2 = 0 + 0.265 160 067 645 44;
  • 40) 0.265 160 067 645 44 × 2 = 0 + 0.530 320 135 290 88;
  • 41) 0.530 320 135 290 88 × 2 = 1 + 0.060 640 270 581 76;
  • 42) 0.060 640 270 581 76 × 2 = 0 + 0.121 280 541 163 52;
  • 43) 0.121 280 541 163 52 × 2 = 0 + 0.242 561 082 327 04;
  • 44) 0.242 561 082 327 04 × 2 = 0 + 0.485 122 164 654 08;
  • 45) 0.485 122 164 654 08 × 2 = 0 + 0.970 244 329 308 16;
  • 46) 0.970 244 329 308 16 × 2 = 1 + 0.940 488 658 616 32;
  • 47) 0.940 488 658 616 32 × 2 = 1 + 0.880 977 317 232 64;
  • 48) 0.880 977 317 232 64 × 2 = 1 + 0.761 954 634 465 28;
  • 49) 0.761 954 634 465 28 × 2 = 1 + 0.523 909 268 930 56;
  • 50) 0.523 909 268 930 56 × 2 = 1 + 0.047 818 537 861 12;
  • 51) 0.047 818 537 861 12 × 2 = 0 + 0.095 637 075 722 24;
  • 52) 0.095 637 075 722 24 × 2 = 0 + 0.191 274 151 444 48;
  • 53) 0.191 274 151 444 48 × 2 = 0 + 0.382 548 302 888 96;
  • 54) 0.382 548 302 888 96 × 2 = 0 + 0.765 096 605 777 92;

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 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0111 1100 00(2)

6. Positive number before normalization:

0.000 000 000 742 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0111 1100 00(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 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0111 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0111 1100 00(2) × 20 =

1.1001 1000 0100 0011 1110 000(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 0100 0011 1110 000


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 0010 0001 1111 0000 =

100 1100 0010 0001 1111 0000


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 0010 0001 1111 0000


Decimal number -0.000 000 000 742 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0010 0001 1111 0000

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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