-0.000 000 000 742 149 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 7(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
-0.000 000 000 742 149 7(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:

|-0.000 000 000 742 149 7| = 0.000 000 000 742 149 7


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: 0.000 000 000 742 149 7.

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) 0.000 000 000 742 149 7 × 2 = 0 + 0.000 000 001 484 299 4;
  • 2) 0.000 000 001 484 299 4 × 2 = 0 + 0.000 000 002 968 598 8;
  • 3) 0.000 000 002 968 598 8 × 2 = 0 + 0.000 000 005 937 197 6;
  • 4) 0.000 000 005 937 197 6 × 2 = 0 + 0.000 000 011 874 395 2;
  • 5) 0.000 000 011 874 395 2 × 2 = 0 + 0.000 000 023 748 790 4;
  • 6) 0.000 000 023 748 790 4 × 2 = 0 + 0.000 000 047 497 580 8;
  • 7) 0.000 000 047 497 580 8 × 2 = 0 + 0.000 000 094 995 161 6;
  • 8) 0.000 000 094 995 161 6 × 2 = 0 + 0.000 000 189 990 323 2;
  • 9) 0.000 000 189 990 323 2 × 2 = 0 + 0.000 000 379 980 646 4;
  • 10) 0.000 000 379 980 646 4 × 2 = 0 + 0.000 000 759 961 292 8;
  • 11) 0.000 000 759 961 292 8 × 2 = 0 + 0.000 001 519 922 585 6;
  • 12) 0.000 001 519 922 585 6 × 2 = 0 + 0.000 003 039 845 171 2;
  • 13) 0.000 003 039 845 171 2 × 2 = 0 + 0.000 006 079 690 342 4;
  • 14) 0.000 006 079 690 342 4 × 2 = 0 + 0.000 012 159 380 684 8;
  • 15) 0.000 012 159 380 684 8 × 2 = 0 + 0.000 024 318 761 369 6;
  • 16) 0.000 024 318 761 369 6 × 2 = 0 + 0.000 048 637 522 739 2;
  • 17) 0.000 048 637 522 739 2 × 2 = 0 + 0.000 097 275 045 478 4;
  • 18) 0.000 097 275 045 478 4 × 2 = 0 + 0.000 194 550 090 956 8;
  • 19) 0.000 194 550 090 956 8 × 2 = 0 + 0.000 389 100 181 913 6;
  • 20) 0.000 389 100 181 913 6 × 2 = 0 + 0.000 778 200 363 827 2;
  • 21) 0.000 778 200 363 827 2 × 2 = 0 + 0.001 556 400 727 654 4;
  • 22) 0.001 556 400 727 654 4 × 2 = 0 + 0.003 112 801 455 308 8;
  • 23) 0.003 112 801 455 308 8 × 2 = 0 + 0.006 225 602 910 617 6;
  • 24) 0.006 225 602 910 617 6 × 2 = 0 + 0.012 451 205 821 235 2;
  • 25) 0.012 451 205 821 235 2 × 2 = 0 + 0.024 902 411 642 470 4;
  • 26) 0.024 902 411 642 470 4 × 2 = 0 + 0.049 804 823 284 940 8;
  • 27) 0.049 804 823 284 940 8 × 2 = 0 + 0.099 609 646 569 881 6;
  • 28) 0.099 609 646 569 881 6 × 2 = 0 + 0.199 219 293 139 763 2;
  • 29) 0.199 219 293 139 763 2 × 2 = 0 + 0.398 438 586 279 526 4;
  • 30) 0.398 438 586 279 526 4 × 2 = 0 + 0.796 877 172 559 052 8;
  • 31) 0.796 877 172 559 052 8 × 2 = 1 + 0.593 754 345 118 105 6;
  • 32) 0.593 754 345 118 105 6 × 2 = 1 + 0.187 508 690 236 211 2;
  • 33) 0.187 508 690 236 211 2 × 2 = 0 + 0.375 017 380 472 422 4;
  • 34) 0.375 017 380 472 422 4 × 2 = 0 + 0.750 034 760 944 844 8;
  • 35) 0.750 034 760 944 844 8 × 2 = 1 + 0.500 069 521 889 689 6;
  • 36) 0.500 069 521 889 689 6 × 2 = 1 + 0.000 139 043 779 379 2;
  • 37) 0.000 139 043 779 379 2 × 2 = 0 + 0.000 278 087 558 758 4;
  • 38) 0.000 278 087 558 758 4 × 2 = 0 + 0.000 556 175 117 516 8;
  • 39) 0.000 556 175 117 516 8 × 2 = 0 + 0.001 112 350 235 033 6;
  • 40) 0.001 112 350 235 033 6 × 2 = 0 + 0.002 224 700 470 067 2;
  • 41) 0.002 224 700 470 067 2 × 2 = 0 + 0.004 449 400 940 134 4;
  • 42) 0.004 449 400 940 134 4 × 2 = 0 + 0.008 898 801 880 268 8;
  • 43) 0.008 898 801 880 268 8 × 2 = 0 + 0.017 797 603 760 537 6;
  • 44) 0.017 797 603 760 537 6 × 2 = 0 + 0.035 595 207 521 075 2;
  • 45) 0.035 595 207 521 075 2 × 2 = 0 + 0.071 190 415 042 150 4;
  • 46) 0.071 190 415 042 150 4 × 2 = 0 + 0.142 380 830 084 300 8;
  • 47) 0.142 380 830 084 300 8 × 2 = 0 + 0.284 761 660 168 601 6;
  • 48) 0.284 761 660 168 601 6 × 2 = 0 + 0.569 523 320 337 203 2;
  • 49) 0.569 523 320 337 203 2 × 2 = 1 + 0.139 046 640 674 406 4;
  • 50) 0.139 046 640 674 406 4 × 2 = 0 + 0.278 093 281 348 812 8;
  • 51) 0.278 093 281 348 812 8 × 2 = 0 + 0.556 186 562 697 625 6;
  • 52) 0.556 186 562 697 625 6 × 2 = 1 + 0.112 373 125 395 251 2;
  • 53) 0.112 373 125 395 251 2 × 2 = 0 + 0.224 746 250 790 502 4;
  • 54) 0.224 746 250 790 502 4 × 2 = 0 + 0.449 492 501 581 004 8;

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:


0.000 000 000 742 149 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 00(2)

6. Positive number before normalization:

0.000 000 000 742 149 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 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:


0.000 000 000 742 149 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 00(2) × 20 =

1.1001 1000 0000 0000 0100 100(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 0000 0000 0100 100


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 0000 0000 0010 0100 =

100 1100 0000 0000 0010 0100


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 0000 0000 0010 0100


Decimal number -0.000 000 000 742 149 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 0100

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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