-0.000 000 000 740 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 79(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
-0.000 000 000 740 79(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:

|-0.000 000 000 740 79| = 0.000 000 000 740 79


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: 0.000 000 000 740 79.

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) 0.000 000 000 740 79 × 2 = 0 + 0.000 000 001 481 58;
  • 2) 0.000 000 001 481 58 × 2 = 0 + 0.000 000 002 963 16;
  • 3) 0.000 000 002 963 16 × 2 = 0 + 0.000 000 005 926 32;
  • 4) 0.000 000 005 926 32 × 2 = 0 + 0.000 000 011 852 64;
  • 5) 0.000 000 011 852 64 × 2 = 0 + 0.000 000 023 705 28;
  • 6) 0.000 000 023 705 28 × 2 = 0 + 0.000 000 047 410 56;
  • 7) 0.000 000 047 410 56 × 2 = 0 + 0.000 000 094 821 12;
  • 8) 0.000 000 094 821 12 × 2 = 0 + 0.000 000 189 642 24;
  • 9) 0.000 000 189 642 24 × 2 = 0 + 0.000 000 379 284 48;
  • 10) 0.000 000 379 284 48 × 2 = 0 + 0.000 000 758 568 96;
  • 11) 0.000 000 758 568 96 × 2 = 0 + 0.000 001 517 137 92;
  • 12) 0.000 001 517 137 92 × 2 = 0 + 0.000 003 034 275 84;
  • 13) 0.000 003 034 275 84 × 2 = 0 + 0.000 006 068 551 68;
  • 14) 0.000 006 068 551 68 × 2 = 0 + 0.000 012 137 103 36;
  • 15) 0.000 012 137 103 36 × 2 = 0 + 0.000 024 274 206 72;
  • 16) 0.000 024 274 206 72 × 2 = 0 + 0.000 048 548 413 44;
  • 17) 0.000 048 548 413 44 × 2 = 0 + 0.000 097 096 826 88;
  • 18) 0.000 097 096 826 88 × 2 = 0 + 0.000 194 193 653 76;
  • 19) 0.000 194 193 653 76 × 2 = 0 + 0.000 388 387 307 52;
  • 20) 0.000 388 387 307 52 × 2 = 0 + 0.000 776 774 615 04;
  • 21) 0.000 776 774 615 04 × 2 = 0 + 0.001 553 549 230 08;
  • 22) 0.001 553 549 230 08 × 2 = 0 + 0.003 107 098 460 16;
  • 23) 0.003 107 098 460 16 × 2 = 0 + 0.006 214 196 920 32;
  • 24) 0.006 214 196 920 32 × 2 = 0 + 0.012 428 393 840 64;
  • 25) 0.012 428 393 840 64 × 2 = 0 + 0.024 856 787 681 28;
  • 26) 0.024 856 787 681 28 × 2 = 0 + 0.049 713 575 362 56;
  • 27) 0.049 713 575 362 56 × 2 = 0 + 0.099 427 150 725 12;
  • 28) 0.099 427 150 725 12 × 2 = 0 + 0.198 854 301 450 24;
  • 29) 0.198 854 301 450 24 × 2 = 0 + 0.397 708 602 900 48;
  • 30) 0.397 708 602 900 48 × 2 = 0 + 0.795 417 205 800 96;
  • 31) 0.795 417 205 800 96 × 2 = 1 + 0.590 834 411 601 92;
  • 32) 0.590 834 411 601 92 × 2 = 1 + 0.181 668 823 203 84;
  • 33) 0.181 668 823 203 84 × 2 = 0 + 0.363 337 646 407 68;
  • 34) 0.363 337 646 407 68 × 2 = 0 + 0.726 675 292 815 36;
  • 35) 0.726 675 292 815 36 × 2 = 1 + 0.453 350 585 630 72;
  • 36) 0.453 350 585 630 72 × 2 = 0 + 0.906 701 171 261 44;
  • 37) 0.906 701 171 261 44 × 2 = 1 + 0.813 402 342 522 88;
  • 38) 0.813 402 342 522 88 × 2 = 1 + 0.626 804 685 045 76;
  • 39) 0.626 804 685 045 76 × 2 = 1 + 0.253 609 370 091 52;
  • 40) 0.253 609 370 091 52 × 2 = 0 + 0.507 218 740 183 04;
  • 41) 0.507 218 740 183 04 × 2 = 1 + 0.014 437 480 366 08;
  • 42) 0.014 437 480 366 08 × 2 = 0 + 0.028 874 960 732 16;
  • 43) 0.028 874 960 732 16 × 2 = 0 + 0.057 749 921 464 32;
  • 44) 0.057 749 921 464 32 × 2 = 0 + 0.115 499 842 928 64;
  • 45) 0.115 499 842 928 64 × 2 = 0 + 0.230 999 685 857 28;
  • 46) 0.230 999 685 857 28 × 2 = 0 + 0.461 999 371 714 56;
  • 47) 0.461 999 371 714 56 × 2 = 0 + 0.923 998 743 429 12;
  • 48) 0.923 998 743 429 12 × 2 = 1 + 0.847 997 486 858 24;
  • 49) 0.847 997 486 858 24 × 2 = 1 + 0.695 994 973 716 48;
  • 50) 0.695 994 973 716 48 × 2 = 1 + 0.391 989 947 432 96;
  • 51) 0.391 989 947 432 96 × 2 = 0 + 0.783 979 894 865 92;
  • 52) 0.783 979 894 865 92 × 2 = 1 + 0.567 959 789 731 84;
  • 53) 0.567 959 789 731 84 × 2 = 1 + 0.135 919 579 463 68;
  • 54) 0.135 919 579 463 68 × 2 = 0 + 0.271 839 158 927 36;

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:


0.000 000 000 740 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1000 0001 1101 10(2)

6. Positive number before normalization:

0.000 000 000 740 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1000 0001 1101 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:


0.000 000 000 740 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1000 0001 1101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1000 0001 1101 10(2) × 20 =

1.1001 0111 0100 0000 1110 110(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 0111 0100 0000 1110 110


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 1011 1010 0000 0111 0110 =

100 1011 1010 0000 0111 0110


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 1011 1010 0000 0111 0110


Decimal number -0.000 000 000 740 79 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1010 0000 0111 0110

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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