-0.000 000 000 742 147 676 646 719 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 719 3(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
-0.000 000 000 742 147 676 646 719 3(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:

|-0.000 000 000 742 147 676 646 719 3| = 0.000 000 000 742 147 676 646 719 3


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: 0.000 000 000 742 147 676 646 719 3.

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) 0.000 000 000 742 147 676 646 719 3 × 2 = 0 + 0.000 000 001 484 295 353 293 438 6;
  • 2) 0.000 000 001 484 295 353 293 438 6 × 2 = 0 + 0.000 000 002 968 590 706 586 877 2;
  • 3) 0.000 000 002 968 590 706 586 877 2 × 2 = 0 + 0.000 000 005 937 181 413 173 754 4;
  • 4) 0.000 000 005 937 181 413 173 754 4 × 2 = 0 + 0.000 000 011 874 362 826 347 508 8;
  • 5) 0.000 000 011 874 362 826 347 508 8 × 2 = 0 + 0.000 000 023 748 725 652 695 017 6;
  • 6) 0.000 000 023 748 725 652 695 017 6 × 2 = 0 + 0.000 000 047 497 451 305 390 035 2;
  • 7) 0.000 000 047 497 451 305 390 035 2 × 2 = 0 + 0.000 000 094 994 902 610 780 070 4;
  • 8) 0.000 000 094 994 902 610 780 070 4 × 2 = 0 + 0.000 000 189 989 805 221 560 140 8;
  • 9) 0.000 000 189 989 805 221 560 140 8 × 2 = 0 + 0.000 000 379 979 610 443 120 281 6;
  • 10) 0.000 000 379 979 610 443 120 281 6 × 2 = 0 + 0.000 000 759 959 220 886 240 563 2;
  • 11) 0.000 000 759 959 220 886 240 563 2 × 2 = 0 + 0.000 001 519 918 441 772 481 126 4;
  • 12) 0.000 001 519 918 441 772 481 126 4 × 2 = 0 + 0.000 003 039 836 883 544 962 252 8;
  • 13) 0.000 003 039 836 883 544 962 252 8 × 2 = 0 + 0.000 006 079 673 767 089 924 505 6;
  • 14) 0.000 006 079 673 767 089 924 505 6 × 2 = 0 + 0.000 012 159 347 534 179 849 011 2;
  • 15) 0.000 012 159 347 534 179 849 011 2 × 2 = 0 + 0.000 024 318 695 068 359 698 022 4;
  • 16) 0.000 024 318 695 068 359 698 022 4 × 2 = 0 + 0.000 048 637 390 136 719 396 044 8;
  • 17) 0.000 048 637 390 136 719 396 044 8 × 2 = 0 + 0.000 097 274 780 273 438 792 089 6;
  • 18) 0.000 097 274 780 273 438 792 089 6 × 2 = 0 + 0.000 194 549 560 546 877 584 179 2;
  • 19) 0.000 194 549 560 546 877 584 179 2 × 2 = 0 + 0.000 389 099 121 093 755 168 358 4;
  • 20) 0.000 389 099 121 093 755 168 358 4 × 2 = 0 + 0.000 778 198 242 187 510 336 716 8;
  • 21) 0.000 778 198 242 187 510 336 716 8 × 2 = 0 + 0.001 556 396 484 375 020 673 433 6;
  • 22) 0.001 556 396 484 375 020 673 433 6 × 2 = 0 + 0.003 112 792 968 750 041 346 867 2;
  • 23) 0.003 112 792 968 750 041 346 867 2 × 2 = 0 + 0.006 225 585 937 500 082 693 734 4;
  • 24) 0.006 225 585 937 500 082 693 734 4 × 2 = 0 + 0.012 451 171 875 000 165 387 468 8;
  • 25) 0.012 451 171 875 000 165 387 468 8 × 2 = 0 + 0.024 902 343 750 000 330 774 937 6;
  • 26) 0.024 902 343 750 000 330 774 937 6 × 2 = 0 + 0.049 804 687 500 000 661 549 875 2;
  • 27) 0.049 804 687 500 000 661 549 875 2 × 2 = 0 + 0.099 609 375 000 001 323 099 750 4;
  • 28) 0.099 609 375 000 001 323 099 750 4 × 2 = 0 + 0.199 218 750 000 002 646 199 500 8;
  • 29) 0.199 218 750 000 002 646 199 500 8 × 2 = 0 + 0.398 437 500 000 005 292 399 001 6;
  • 30) 0.398 437 500 000 005 292 399 001 6 × 2 = 0 + 0.796 875 000 000 010 584 798 003 2;
  • 31) 0.796 875 000 000 010 584 798 003 2 × 2 = 1 + 0.593 750 000 000 021 169 596 006 4;
  • 32) 0.593 750 000 000 021 169 596 006 4 × 2 = 1 + 0.187 500 000 000 042 339 192 012 8;
  • 33) 0.187 500 000 000 042 339 192 012 8 × 2 = 0 + 0.375 000 000 000 084 678 384 025 6;
  • 34) 0.375 000 000 000 084 678 384 025 6 × 2 = 0 + 0.750 000 000 000 169 356 768 051 2;
  • 35) 0.750 000 000 000 169 356 768 051 2 × 2 = 1 + 0.500 000 000 000 338 713 536 102 4;
  • 36) 0.500 000 000 000 338 713 536 102 4 × 2 = 1 + 0.000 000 000 000 677 427 072 204 8;
  • 37) 0.000 000 000 000 677 427 072 204 8 × 2 = 0 + 0.000 000 000 001 354 854 144 409 6;
  • 38) 0.000 000 000 001 354 854 144 409 6 × 2 = 0 + 0.000 000 000 002 709 708 288 819 2;
  • 39) 0.000 000 000 002 709 708 288 819 2 × 2 = 0 + 0.000 000 000 005 419 416 577 638 4;
  • 40) 0.000 000 000 005 419 416 577 638 4 × 2 = 0 + 0.000 000 000 010 838 833 155 276 8;
  • 41) 0.000 000 000 010 838 833 155 276 8 × 2 = 0 + 0.000 000 000 021 677 666 310 553 6;
  • 42) 0.000 000 000 021 677 666 310 553 6 × 2 = 0 + 0.000 000 000 043 355 332 621 107 2;
  • 43) 0.000 000 000 043 355 332 621 107 2 × 2 = 0 + 0.000 000 000 086 710 665 242 214 4;
  • 44) 0.000 000 000 086 710 665 242 214 4 × 2 = 0 + 0.000 000 000 173 421 330 484 428 8;
  • 45) 0.000 000 000 173 421 330 484 428 8 × 2 = 0 + 0.000 000 000 346 842 660 968 857 6;
  • 46) 0.000 000 000 346 842 660 968 857 6 × 2 = 0 + 0.000 000 000 693 685 321 937 715 2;
  • 47) 0.000 000 000 693 685 321 937 715 2 × 2 = 0 + 0.000 000 001 387 370 643 875 430 4;
  • 48) 0.000 000 001 387 370 643 875 430 4 × 2 = 0 + 0.000 000 002 774 741 287 750 860 8;
  • 49) 0.000 000 002 774 741 287 750 860 8 × 2 = 0 + 0.000 000 005 549 482 575 501 721 6;
  • 50) 0.000 000 005 549 482 575 501 721 6 × 2 = 0 + 0.000 000 011 098 965 151 003 443 2;
  • 51) 0.000 000 011 098 965 151 003 443 2 × 2 = 0 + 0.000 000 022 197 930 302 006 886 4;
  • 52) 0.000 000 022 197 930 302 006 886 4 × 2 = 0 + 0.000 000 044 395 860 604 013 772 8;
  • 53) 0.000 000 044 395 860 604 013 772 8 × 2 = 0 + 0.000 000 088 791 721 208 027 545 6;
  • 54) 0.000 000 088 791 721 208 027 545 6 × 2 = 0 + 0.000 000 177 583 442 416 055 091 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:


0.000 000 000 742 147 676 646 719 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 719 3(10) =

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


0.000 000 000 742 147 676 646 719 3(10) =

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

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

1.1001 1000 0000 0000 0000 000(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 0000 000


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 0000 0000 =

100 1100 0000 0000 0000 0000


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 0000 0000


Decimal number -0.000 000 000 742 147 676 646 719 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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