-0.000 000 000 787 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 787(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
-0.000 000 000 787(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:

|-0.000 000 000 787| = 0.000 000 000 787


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: 0.000 000 000 787.

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) 0.000 000 000 787 × 2 = 0 + 0.000 000 001 574;
  • 2) 0.000 000 001 574 × 2 = 0 + 0.000 000 003 148;
  • 3) 0.000 000 003 148 × 2 = 0 + 0.000 000 006 296;
  • 4) 0.000 000 006 296 × 2 = 0 + 0.000 000 012 592;
  • 5) 0.000 000 012 592 × 2 = 0 + 0.000 000 025 184;
  • 6) 0.000 000 025 184 × 2 = 0 + 0.000 000 050 368;
  • 7) 0.000 000 050 368 × 2 = 0 + 0.000 000 100 736;
  • 8) 0.000 000 100 736 × 2 = 0 + 0.000 000 201 472;
  • 9) 0.000 000 201 472 × 2 = 0 + 0.000 000 402 944;
  • 10) 0.000 000 402 944 × 2 = 0 + 0.000 000 805 888;
  • 11) 0.000 000 805 888 × 2 = 0 + 0.000 001 611 776;
  • 12) 0.000 001 611 776 × 2 = 0 + 0.000 003 223 552;
  • 13) 0.000 003 223 552 × 2 = 0 + 0.000 006 447 104;
  • 14) 0.000 006 447 104 × 2 = 0 + 0.000 012 894 208;
  • 15) 0.000 012 894 208 × 2 = 0 + 0.000 025 788 416;
  • 16) 0.000 025 788 416 × 2 = 0 + 0.000 051 576 832;
  • 17) 0.000 051 576 832 × 2 = 0 + 0.000 103 153 664;
  • 18) 0.000 103 153 664 × 2 = 0 + 0.000 206 307 328;
  • 19) 0.000 206 307 328 × 2 = 0 + 0.000 412 614 656;
  • 20) 0.000 412 614 656 × 2 = 0 + 0.000 825 229 312;
  • 21) 0.000 825 229 312 × 2 = 0 + 0.001 650 458 624;
  • 22) 0.001 650 458 624 × 2 = 0 + 0.003 300 917 248;
  • 23) 0.003 300 917 248 × 2 = 0 + 0.006 601 834 496;
  • 24) 0.006 601 834 496 × 2 = 0 + 0.013 203 668 992;
  • 25) 0.013 203 668 992 × 2 = 0 + 0.026 407 337 984;
  • 26) 0.026 407 337 984 × 2 = 0 + 0.052 814 675 968;
  • 27) 0.052 814 675 968 × 2 = 0 + 0.105 629 351 936;
  • 28) 0.105 629 351 936 × 2 = 0 + 0.211 258 703 872;
  • 29) 0.211 258 703 872 × 2 = 0 + 0.422 517 407 744;
  • 30) 0.422 517 407 744 × 2 = 0 + 0.845 034 815 488;
  • 31) 0.845 034 815 488 × 2 = 1 + 0.690 069 630 976;
  • 32) 0.690 069 630 976 × 2 = 1 + 0.380 139 261 952;
  • 33) 0.380 139 261 952 × 2 = 0 + 0.760 278 523 904;
  • 34) 0.760 278 523 904 × 2 = 1 + 0.520 557 047 808;
  • 35) 0.520 557 047 808 × 2 = 1 + 0.041 114 095 616;
  • 36) 0.041 114 095 616 × 2 = 0 + 0.082 228 191 232;
  • 37) 0.082 228 191 232 × 2 = 0 + 0.164 456 382 464;
  • 38) 0.164 456 382 464 × 2 = 0 + 0.328 912 764 928;
  • 39) 0.328 912 764 928 × 2 = 0 + 0.657 825 529 856;
  • 40) 0.657 825 529 856 × 2 = 1 + 0.315 651 059 712;
  • 41) 0.315 651 059 712 × 2 = 0 + 0.631 302 119 424;
  • 42) 0.631 302 119 424 × 2 = 1 + 0.262 604 238 848;
  • 43) 0.262 604 238 848 × 2 = 0 + 0.525 208 477 696;
  • 44) 0.525 208 477 696 × 2 = 1 + 0.050 416 955 392;
  • 45) 0.050 416 955 392 × 2 = 0 + 0.100 833 910 784;
  • 46) 0.100 833 910 784 × 2 = 0 + 0.201 667 821 568;
  • 47) 0.201 667 821 568 × 2 = 0 + 0.403 335 643 136;
  • 48) 0.403 335 643 136 × 2 = 0 + 0.806 671 286 272;
  • 49) 0.806 671 286 272 × 2 = 1 + 0.613 342 572 544;
  • 50) 0.613 342 572 544 × 2 = 1 + 0.226 685 145 088;
  • 51) 0.226 685 145 088 × 2 = 0 + 0.453 370 290 176;
  • 52) 0.453 370 290 176 × 2 = 0 + 0.906 740 580 352;
  • 53) 0.906 740 580 352 × 2 = 1 + 0.813 481 160 704;
  • 54) 0.813 481 160 704 × 2 = 1 + 0.626 962 321 408;

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:


0.000 000 000 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0001 0101 0000 1100 11(2)

6. Positive number before normalization:

0.000 000 000 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0001 0101 0000 1100 11(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:


0.000 000 000 787(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0001 0101 0000 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0001 0101 0000 1100 11(2) × 20 =

1.1011 0000 1010 1000 0110 011(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):
1.1011 0000 1010 1000 0110 011


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. 101 1000 0101 0100 0011 0011 =

101 1000 0101 0100 0011 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 0000

Mantissa (23 bits) =
101 1000 0101 0100 0011 0011


Decimal number -0.000 000 000 787 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1000 0101 0100 0011 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