-0.000 000 000 751 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 751(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
-0.000 000 000 751(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:

|-0.000 000 000 751| = 0.000 000 000 751


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: 0.000 000 000 751.

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) 0.000 000 000 751 × 2 = 0 + 0.000 000 001 502;
  • 2) 0.000 000 001 502 × 2 = 0 + 0.000 000 003 004;
  • 3) 0.000 000 003 004 × 2 = 0 + 0.000 000 006 008;
  • 4) 0.000 000 006 008 × 2 = 0 + 0.000 000 012 016;
  • 5) 0.000 000 012 016 × 2 = 0 + 0.000 000 024 032;
  • 6) 0.000 000 024 032 × 2 = 0 + 0.000 000 048 064;
  • 7) 0.000 000 048 064 × 2 = 0 + 0.000 000 096 128;
  • 8) 0.000 000 096 128 × 2 = 0 + 0.000 000 192 256;
  • 9) 0.000 000 192 256 × 2 = 0 + 0.000 000 384 512;
  • 10) 0.000 000 384 512 × 2 = 0 + 0.000 000 769 024;
  • 11) 0.000 000 769 024 × 2 = 0 + 0.000 001 538 048;
  • 12) 0.000 001 538 048 × 2 = 0 + 0.000 003 076 096;
  • 13) 0.000 003 076 096 × 2 = 0 + 0.000 006 152 192;
  • 14) 0.000 006 152 192 × 2 = 0 + 0.000 012 304 384;
  • 15) 0.000 012 304 384 × 2 = 0 + 0.000 024 608 768;
  • 16) 0.000 024 608 768 × 2 = 0 + 0.000 049 217 536;
  • 17) 0.000 049 217 536 × 2 = 0 + 0.000 098 435 072;
  • 18) 0.000 098 435 072 × 2 = 0 + 0.000 196 870 144;
  • 19) 0.000 196 870 144 × 2 = 0 + 0.000 393 740 288;
  • 20) 0.000 393 740 288 × 2 = 0 + 0.000 787 480 576;
  • 21) 0.000 787 480 576 × 2 = 0 + 0.001 574 961 152;
  • 22) 0.001 574 961 152 × 2 = 0 + 0.003 149 922 304;
  • 23) 0.003 149 922 304 × 2 = 0 + 0.006 299 844 608;
  • 24) 0.006 299 844 608 × 2 = 0 + 0.012 599 689 216;
  • 25) 0.012 599 689 216 × 2 = 0 + 0.025 199 378 432;
  • 26) 0.025 199 378 432 × 2 = 0 + 0.050 398 756 864;
  • 27) 0.050 398 756 864 × 2 = 0 + 0.100 797 513 728;
  • 28) 0.100 797 513 728 × 2 = 0 + 0.201 595 027 456;
  • 29) 0.201 595 027 456 × 2 = 0 + 0.403 190 054 912;
  • 30) 0.403 190 054 912 × 2 = 0 + 0.806 380 109 824;
  • 31) 0.806 380 109 824 × 2 = 1 + 0.612 760 219 648;
  • 32) 0.612 760 219 648 × 2 = 1 + 0.225 520 439 296;
  • 33) 0.225 520 439 296 × 2 = 0 + 0.451 040 878 592;
  • 34) 0.451 040 878 592 × 2 = 0 + 0.902 081 757 184;
  • 35) 0.902 081 757 184 × 2 = 1 + 0.804 163 514 368;
  • 36) 0.804 163 514 368 × 2 = 1 + 0.608 327 028 736;
  • 37) 0.608 327 028 736 × 2 = 1 + 0.216 654 057 472;
  • 38) 0.216 654 057 472 × 2 = 0 + 0.433 308 114 944;
  • 39) 0.433 308 114 944 × 2 = 0 + 0.866 616 229 888;
  • 40) 0.866 616 229 888 × 2 = 1 + 0.733 232 459 776;
  • 41) 0.733 232 459 776 × 2 = 1 + 0.466 464 919 552;
  • 42) 0.466 464 919 552 × 2 = 0 + 0.932 929 839 104;
  • 43) 0.932 929 839 104 × 2 = 1 + 0.865 859 678 208;
  • 44) 0.865 859 678 208 × 2 = 1 + 0.731 719 356 416;
  • 45) 0.731 719 356 416 × 2 = 1 + 0.463 438 712 832;
  • 46) 0.463 438 712 832 × 2 = 0 + 0.926 877 425 664;
  • 47) 0.926 877 425 664 × 2 = 1 + 0.853 754 851 328;
  • 48) 0.853 754 851 328 × 2 = 1 + 0.707 509 702 656;
  • 49) 0.707 509 702 656 × 2 = 1 + 0.415 019 405 312;
  • 50) 0.415 019 405 312 × 2 = 0 + 0.830 038 810 624;
  • 51) 0.830 038 810 624 × 2 = 1 + 0.660 077 621 248;
  • 52) 0.660 077 621 248 × 2 = 1 + 0.320 155 242 496;
  • 53) 0.320 155 242 496 × 2 = 0 + 0.640 310 484 992;
  • 54) 0.640 310 484 992 × 2 = 1 + 0.280 620 969 984;

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:


0.000 000 000 751(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1011 1011 1011 01(2)

6. Positive number before normalization:

0.000 000 000 751(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1011 1011 1011 01(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:


0.000 000 000 751(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1011 1011 1011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 1001 1011 1011 1011 01(2) × 20 =

1.1001 1100 1101 1101 1101 101(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 1101 1101 1101 101


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 1110 1110 1101 =

100 1110 0110 1110 1110 1101


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 1110 1110 1101


Decimal number -0.000 000 000 751 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1110 0110 1110 1110 1101

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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