-0.000 000 000 743 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 743(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
-0.000 000 000 743(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:

|-0.000 000 000 743| = 0.000 000 000 743


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: 0.000 000 000 743.

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) 0.000 000 000 743 × 2 = 0 + 0.000 000 001 486;
  • 2) 0.000 000 001 486 × 2 = 0 + 0.000 000 002 972;
  • 3) 0.000 000 002 972 × 2 = 0 + 0.000 000 005 944;
  • 4) 0.000 000 005 944 × 2 = 0 + 0.000 000 011 888;
  • 5) 0.000 000 011 888 × 2 = 0 + 0.000 000 023 776;
  • 6) 0.000 000 023 776 × 2 = 0 + 0.000 000 047 552;
  • 7) 0.000 000 047 552 × 2 = 0 + 0.000 000 095 104;
  • 8) 0.000 000 095 104 × 2 = 0 + 0.000 000 190 208;
  • 9) 0.000 000 190 208 × 2 = 0 + 0.000 000 380 416;
  • 10) 0.000 000 380 416 × 2 = 0 + 0.000 000 760 832;
  • 11) 0.000 000 760 832 × 2 = 0 + 0.000 001 521 664;
  • 12) 0.000 001 521 664 × 2 = 0 + 0.000 003 043 328;
  • 13) 0.000 003 043 328 × 2 = 0 + 0.000 006 086 656;
  • 14) 0.000 006 086 656 × 2 = 0 + 0.000 012 173 312;
  • 15) 0.000 012 173 312 × 2 = 0 + 0.000 024 346 624;
  • 16) 0.000 024 346 624 × 2 = 0 + 0.000 048 693 248;
  • 17) 0.000 048 693 248 × 2 = 0 + 0.000 097 386 496;
  • 18) 0.000 097 386 496 × 2 = 0 + 0.000 194 772 992;
  • 19) 0.000 194 772 992 × 2 = 0 + 0.000 389 545 984;
  • 20) 0.000 389 545 984 × 2 = 0 + 0.000 779 091 968;
  • 21) 0.000 779 091 968 × 2 = 0 + 0.001 558 183 936;
  • 22) 0.001 558 183 936 × 2 = 0 + 0.003 116 367 872;
  • 23) 0.003 116 367 872 × 2 = 0 + 0.006 232 735 744;
  • 24) 0.006 232 735 744 × 2 = 0 + 0.012 465 471 488;
  • 25) 0.012 465 471 488 × 2 = 0 + 0.024 930 942 976;
  • 26) 0.024 930 942 976 × 2 = 0 + 0.049 861 885 952;
  • 27) 0.049 861 885 952 × 2 = 0 + 0.099 723 771 904;
  • 28) 0.099 723 771 904 × 2 = 0 + 0.199 447 543 808;
  • 29) 0.199 447 543 808 × 2 = 0 + 0.398 895 087 616;
  • 30) 0.398 895 087 616 × 2 = 0 + 0.797 790 175 232;
  • 31) 0.797 790 175 232 × 2 = 1 + 0.595 580 350 464;
  • 32) 0.595 580 350 464 × 2 = 1 + 0.191 160 700 928;
  • 33) 0.191 160 700 928 × 2 = 0 + 0.382 321 401 856;
  • 34) 0.382 321 401 856 × 2 = 0 + 0.764 642 803 712;
  • 35) 0.764 642 803 712 × 2 = 1 + 0.529 285 607 424;
  • 36) 0.529 285 607 424 × 2 = 1 + 0.058 571 214 848;
  • 37) 0.058 571 214 848 × 2 = 0 + 0.117 142 429 696;
  • 38) 0.117 142 429 696 × 2 = 0 + 0.234 284 859 392;
  • 39) 0.234 284 859 392 × 2 = 0 + 0.468 569 718 784;
  • 40) 0.468 569 718 784 × 2 = 0 + 0.937 139 437 568;
  • 41) 0.937 139 437 568 × 2 = 1 + 0.874 278 875 136;
  • 42) 0.874 278 875 136 × 2 = 1 + 0.748 557 750 272;
  • 43) 0.748 557 750 272 × 2 = 1 + 0.497 115 500 544;
  • 44) 0.497 115 500 544 × 2 = 0 + 0.994 231 001 088;
  • 45) 0.994 231 001 088 × 2 = 1 + 0.988 462 002 176;
  • 46) 0.988 462 002 176 × 2 = 1 + 0.976 924 004 352;
  • 47) 0.976 924 004 352 × 2 = 1 + 0.953 848 008 704;
  • 48) 0.953 848 008 704 × 2 = 1 + 0.907 696 017 408;
  • 49) 0.907 696 017 408 × 2 = 1 + 0.815 392 034 816;
  • 50) 0.815 392 034 816 × 2 = 1 + 0.630 784 069 632;
  • 51) 0.630 784 069 632 × 2 = 1 + 0.261 568 139 264;
  • 52) 0.261 568 139 264 × 2 = 0 + 0.523 136 278 528;
  • 53) 0.523 136 278 528 × 2 = 1 + 0.046 272 557 056;
  • 54) 0.046 272 557 056 × 2 = 0 + 0.092 545 114 112;

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:


0.000 000 000 743(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1110 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 743(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1110 1111 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:


0.000 000 000 743(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1110 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1110 1111 1110 10(2) × 20 =

1.1001 1000 0111 0111 1111 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 1000 0111 0111 1111 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 1100 0011 1011 1111 1010 =

100 1100 0011 1011 1111 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 1100 0011 1011 1111 1010


Decimal number -0.000 000 000 743 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0011 1011 1111 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