-0.000 000 000 750 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 750 8(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
-0.000 000 000 750 8(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:

|-0.000 000 000 750 8| = 0.000 000 000 750 8


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: 0.000 000 000 750 8.

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) 0.000 000 000 750 8 × 2 = 0 + 0.000 000 001 501 6;
  • 2) 0.000 000 001 501 6 × 2 = 0 + 0.000 000 003 003 2;
  • 3) 0.000 000 003 003 2 × 2 = 0 + 0.000 000 006 006 4;
  • 4) 0.000 000 006 006 4 × 2 = 0 + 0.000 000 012 012 8;
  • 5) 0.000 000 012 012 8 × 2 = 0 + 0.000 000 024 025 6;
  • 6) 0.000 000 024 025 6 × 2 = 0 + 0.000 000 048 051 2;
  • 7) 0.000 000 048 051 2 × 2 = 0 + 0.000 000 096 102 4;
  • 8) 0.000 000 096 102 4 × 2 = 0 + 0.000 000 192 204 8;
  • 9) 0.000 000 192 204 8 × 2 = 0 + 0.000 000 384 409 6;
  • 10) 0.000 000 384 409 6 × 2 = 0 + 0.000 000 768 819 2;
  • 11) 0.000 000 768 819 2 × 2 = 0 + 0.000 001 537 638 4;
  • 12) 0.000 001 537 638 4 × 2 = 0 + 0.000 003 075 276 8;
  • 13) 0.000 003 075 276 8 × 2 = 0 + 0.000 006 150 553 6;
  • 14) 0.000 006 150 553 6 × 2 = 0 + 0.000 012 301 107 2;
  • 15) 0.000 012 301 107 2 × 2 = 0 + 0.000 024 602 214 4;
  • 16) 0.000 024 602 214 4 × 2 = 0 + 0.000 049 204 428 8;
  • 17) 0.000 049 204 428 8 × 2 = 0 + 0.000 098 408 857 6;
  • 18) 0.000 098 408 857 6 × 2 = 0 + 0.000 196 817 715 2;
  • 19) 0.000 196 817 715 2 × 2 = 0 + 0.000 393 635 430 4;
  • 20) 0.000 393 635 430 4 × 2 = 0 + 0.000 787 270 860 8;
  • 21) 0.000 787 270 860 8 × 2 = 0 + 0.001 574 541 721 6;
  • 22) 0.001 574 541 721 6 × 2 = 0 + 0.003 149 083 443 2;
  • 23) 0.003 149 083 443 2 × 2 = 0 + 0.006 298 166 886 4;
  • 24) 0.006 298 166 886 4 × 2 = 0 + 0.012 596 333 772 8;
  • 25) 0.012 596 333 772 8 × 2 = 0 + 0.025 192 667 545 6;
  • 26) 0.025 192 667 545 6 × 2 = 0 + 0.050 385 335 091 2;
  • 27) 0.050 385 335 091 2 × 2 = 0 + 0.100 770 670 182 4;
  • 28) 0.100 770 670 182 4 × 2 = 0 + 0.201 541 340 364 8;
  • 29) 0.201 541 340 364 8 × 2 = 0 + 0.403 082 680 729 6;
  • 30) 0.403 082 680 729 6 × 2 = 0 + 0.806 165 361 459 2;
  • 31) 0.806 165 361 459 2 × 2 = 1 + 0.612 330 722 918 4;
  • 32) 0.612 330 722 918 4 × 2 = 1 + 0.224 661 445 836 8;
  • 33) 0.224 661 445 836 8 × 2 = 0 + 0.449 322 891 673 6;
  • 34) 0.449 322 891 673 6 × 2 = 0 + 0.898 645 783 347 2;
  • 35) 0.898 645 783 347 2 × 2 = 1 + 0.797 291 566 694 4;
  • 36) 0.797 291 566 694 4 × 2 = 1 + 0.594 583 133 388 8;
  • 37) 0.594 583 133 388 8 × 2 = 1 + 0.189 166 266 777 6;
  • 38) 0.189 166 266 777 6 × 2 = 0 + 0.378 332 533 555 2;
  • 39) 0.378 332 533 555 2 × 2 = 0 + 0.756 665 067 110 4;
  • 40) 0.756 665 067 110 4 × 2 = 1 + 0.513 330 134 220 8;
  • 41) 0.513 330 134 220 8 × 2 = 1 + 0.026 660 268 441 6;
  • 42) 0.026 660 268 441 6 × 2 = 0 + 0.053 320 536 883 2;
  • 43) 0.053 320 536 883 2 × 2 = 0 + 0.106 641 073 766 4;
  • 44) 0.106 641 073 766 4 × 2 = 0 + 0.213 282 147 532 8;
  • 45) 0.213 282 147 532 8 × 2 = 0 + 0.426 564 295 065 6;
  • 46) 0.426 564 295 065 6 × 2 = 0 + 0.853 128 590 131 2;
  • 47) 0.853 128 590 131 2 × 2 = 1 + 0.706 257 180 262 4;
  • 48) 0.706 257 180 262 4 × 2 = 1 + 0.412 514 360 524 8;
  • 49) 0.412 514 360 524 8 × 2 = 0 + 0.825 028 721 049 6;
  • 50) 0.825 028 721 049 6 × 2 = 1 + 0.650 057 442 099 2;
  • 51) 0.650 057 442 099 2 × 2 = 1 + 0.300 114 884 198 4;
  • 52) 0.300 114 884 198 4 × 2 = 0 + 0.600 229 768 396 8;
  • 53) 0.600 229 768 396 8 × 2 = 1 + 0.200 459 536 793 6;
  • 54) 0.200 459 536 793 6 × 2 = 0 + 0.400 919 073 587 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:


0.000 000 000 750 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1000 0011 0110 10(2)

6. Positive number before normalization:

0.000 000 000 750 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1000 0011 0110 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:


0.000 000 000 750 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1000 0011 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1000 0011 0110 10(2) × 20 =

1.1001 1100 1100 0001 1011 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 1100 1100 0001 1011 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 1110 0110 0000 1101 1010 =

100 1110 0110 0000 1101 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 1110 0110 0000 1101 1010


Decimal number -0.000 000 000 750 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1110 0110 0000 1101 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