-0.000 000 000 742 147 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 13(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 13(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 13| = 0.000 000 000 742 147 13


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 13.

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 13 × 2 = 0 + 0.000 000 001 484 294 26;
  • 2) 0.000 000 001 484 294 26 × 2 = 0 + 0.000 000 002 968 588 52;
  • 3) 0.000 000 002 968 588 52 × 2 = 0 + 0.000 000 005 937 177 04;
  • 4) 0.000 000 005 937 177 04 × 2 = 0 + 0.000 000 011 874 354 08;
  • 5) 0.000 000 011 874 354 08 × 2 = 0 + 0.000 000 023 748 708 16;
  • 6) 0.000 000 023 748 708 16 × 2 = 0 + 0.000 000 047 497 416 32;
  • 7) 0.000 000 047 497 416 32 × 2 = 0 + 0.000 000 094 994 832 64;
  • 8) 0.000 000 094 994 832 64 × 2 = 0 + 0.000 000 189 989 665 28;
  • 9) 0.000 000 189 989 665 28 × 2 = 0 + 0.000 000 379 979 330 56;
  • 10) 0.000 000 379 979 330 56 × 2 = 0 + 0.000 000 759 958 661 12;
  • 11) 0.000 000 759 958 661 12 × 2 = 0 + 0.000 001 519 917 322 24;
  • 12) 0.000 001 519 917 322 24 × 2 = 0 + 0.000 003 039 834 644 48;
  • 13) 0.000 003 039 834 644 48 × 2 = 0 + 0.000 006 079 669 288 96;
  • 14) 0.000 006 079 669 288 96 × 2 = 0 + 0.000 012 159 338 577 92;
  • 15) 0.000 012 159 338 577 92 × 2 = 0 + 0.000 024 318 677 155 84;
  • 16) 0.000 024 318 677 155 84 × 2 = 0 + 0.000 048 637 354 311 68;
  • 17) 0.000 048 637 354 311 68 × 2 = 0 + 0.000 097 274 708 623 36;
  • 18) 0.000 097 274 708 623 36 × 2 = 0 + 0.000 194 549 417 246 72;
  • 19) 0.000 194 549 417 246 72 × 2 = 0 + 0.000 389 098 834 493 44;
  • 20) 0.000 389 098 834 493 44 × 2 = 0 + 0.000 778 197 668 986 88;
  • 21) 0.000 778 197 668 986 88 × 2 = 0 + 0.001 556 395 337 973 76;
  • 22) 0.001 556 395 337 973 76 × 2 = 0 + 0.003 112 790 675 947 52;
  • 23) 0.003 112 790 675 947 52 × 2 = 0 + 0.006 225 581 351 895 04;
  • 24) 0.006 225 581 351 895 04 × 2 = 0 + 0.012 451 162 703 790 08;
  • 25) 0.012 451 162 703 790 08 × 2 = 0 + 0.024 902 325 407 580 16;
  • 26) 0.024 902 325 407 580 16 × 2 = 0 + 0.049 804 650 815 160 32;
  • 27) 0.049 804 650 815 160 32 × 2 = 0 + 0.099 609 301 630 320 64;
  • 28) 0.099 609 301 630 320 64 × 2 = 0 + 0.199 218 603 260 641 28;
  • 29) 0.199 218 603 260 641 28 × 2 = 0 + 0.398 437 206 521 282 56;
  • 30) 0.398 437 206 521 282 56 × 2 = 0 + 0.796 874 413 042 565 12;
  • 31) 0.796 874 413 042 565 12 × 2 = 1 + 0.593 748 826 085 130 24;
  • 32) 0.593 748 826 085 130 24 × 2 = 1 + 0.187 497 652 170 260 48;
  • 33) 0.187 497 652 170 260 48 × 2 = 0 + 0.374 995 304 340 520 96;
  • 34) 0.374 995 304 340 520 96 × 2 = 0 + 0.749 990 608 681 041 92;
  • 35) 0.749 990 608 681 041 92 × 2 = 1 + 0.499 981 217 362 083 84;
  • 36) 0.499 981 217 362 083 84 × 2 = 0 + 0.999 962 434 724 167 68;
  • 37) 0.999 962 434 724 167 68 × 2 = 1 + 0.999 924 869 448 335 36;
  • 38) 0.999 924 869 448 335 36 × 2 = 1 + 0.999 849 738 896 670 72;
  • 39) 0.999 849 738 896 670 72 × 2 = 1 + 0.999 699 477 793 341 44;
  • 40) 0.999 699 477 793 341 44 × 2 = 1 + 0.999 398 955 586 682 88;
  • 41) 0.999 398 955 586 682 88 × 2 = 1 + 0.998 797 911 173 365 76;
  • 42) 0.998 797 911 173 365 76 × 2 = 1 + 0.997 595 822 346 731 52;
  • 43) 0.997 595 822 346 731 52 × 2 = 1 + 0.995 191 644 693 463 04;
  • 44) 0.995 191 644 693 463 04 × 2 = 1 + 0.990 383 289 386 926 08;
  • 45) 0.990 383 289 386 926 08 × 2 = 1 + 0.980 766 578 773 852 16;
  • 46) 0.980 766 578 773 852 16 × 2 = 1 + 0.961 533 157 547 704 32;
  • 47) 0.961 533 157 547 704 32 × 2 = 1 + 0.923 066 315 095 408 64;
  • 48) 0.923 066 315 095 408 64 × 2 = 1 + 0.846 132 630 190 817 28;
  • 49) 0.846 132 630 190 817 28 × 2 = 1 + 0.692 265 260 381 634 56;
  • 50) 0.692 265 260 381 634 56 × 2 = 1 + 0.384 530 520 763 269 12;
  • 51) 0.384 530 520 763 269 12 × 2 = 0 + 0.769 061 041 526 538 24;
  • 52) 0.769 061 041 526 538 24 × 2 = 1 + 0.538 122 083 053 076 48;
  • 53) 0.538 122 083 053 076 48 × 2 = 1 + 0.076 244 166 106 152 96;
  • 54) 0.076 244 166 106 152 96 × 2 = 0 + 0.152 488 332 212 305 92;

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 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 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:


0.000 000 000 742 147 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 10(2) × 20 =

1.1001 0111 1111 1111 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 1111 1111 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 1111 1111 1111 0110 =

100 1011 1111 1111 1111 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 1111 1111 1111 0110


Decimal number -0.000 000 000 742 147 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 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